Bihar Board Class 12 Maths \u2013 Section A (MCQs)<\/title>\n<\/head>\n<body>\n\n1.<\/strong> tan\u22121<\/sup>(\u22121\/\u221a3) =<\/p>\n\n(A) \u03c0\/3 \n(B) \u03c0\/6 \n(C) \u2212\u03c0\/3 \n(D) \u2212\u03c0\/6\n<\/p>\nAnswer:<\/strong> (D) \u2212\u03c0\/6<\/p>\n\n<hr>\n\n2.<\/strong> 2 tan\u22121<\/sup>(1\/3) =<\/p>\n\n(A) tan\u22121<\/sup>(3\/2) \n(B) tan\u22121<\/sup>(3\/4) \n(C) tan\u22121<\/sup>(4\/3) \n(D) tan\u22121<\/sup>(2\/3)\n<\/p>\nAnswer:<\/strong> (D) tan\u22121<\/sup>(2\/3)<\/p>\n\n<hr>\n\n3.<\/strong> x \u2208 R, cot(tan\u22121<\/sup>x + cot\u22121<\/sup>x) =<\/p>\n\n(A) 1 \n(B) 1\/2 \n(C) 0 \n(D) 1\/3\n<\/p>\nAnswer:<\/strong> (A) 1<\/p>\n\n<hr>\n\n4.<\/strong> sin(cos\u22121<\/sup>(3\/5)) =<\/p>\n\n(A) 3\/4 \n(B) 4\/5 \n(C) 3\/5 \n(D) 5\/4\n<\/p>\nAnswer:<\/strong> (B) 4\/5<\/p>\n\n<hr>\n\n5.<\/strong> tan\u22121<\/sup>(1) + cos\u22121<\/sup>(\u22121\/2) + sin\u22121<\/sup>(\u22121\/2) =<\/p>\n\n(A) \u03c0 \n(B) 2\u03c0\/3 \n(C) 3\u03c0\/4 \n(D) \u03c0\/2\n<\/p>\nAnswer:<\/strong> (D) \u03c0\/2<\/p>\n\n<hr>\n\n6.<\/strong> cos\u22121<\/sup>(cos 7\u03c0\/6) =<\/p>\n\n(A) \u03c0\/6 \n(B) \u03c0\/3 \n(C) 5\u03c0\/6 \n(D) 7\u03c0\/6\n<\/p>\nAnswer:<\/strong> (C) 5\u03c0\/6<\/p>\n\n<hr>\n\n7.<\/strong> tan\u22121<\/sup>2 + tan\u22121<\/sup>3 =<\/p>\n\n(A) \u2212\u03c0\/4 \n(B) \u03c0\/4 \n(C) 3\u03c0\/4 \n(D) \u03c0\n<\/p>\nAnswer:<\/strong> (C) 3\u03c0\/4<\/p>\n\n<hr>\n\n8.<\/strong> If |x| \u2264 1, then tan(cos\u22121<\/sup>x) =<\/p>\n\n(A) \u221a(1 \u2212 x\u00b2)\/x \n(B) x\/(1 + x\u00b2) \n(C) \u221a(1 + x\u00b2)\/x \n(D) \u221a(1 \u2212 x\u00b2)\n<\/p>\nAnswer:<\/strong> (A) \u221a(1 \u2212 x\u00b2)\/x<\/p>\n\n<hr>\n\n9.<\/strong> tan\u22121<\/sup>(x\/y) \u2212 tan\u22121<\/sup>(x\/(x + y)) =<\/p>\n\n(A) \u22123\u03c0\/4 \n(B) \u03c0\/2 \n(C) \u03c0\/4 \n(D) \u03c0\/3\n<\/p>\nAnswer:<\/strong> (C) \u03c0\/4<\/p>\n\n<hr>\n\n10.<\/strong> If |x| \u2264 1, cos\u22121<\/sup>((1 \u2212 x\u00b2)\/(1 + x\u00b2)) =<\/p>\n\n(A) 2cos\u22121<\/sup>x \n(B) 2sin\u22121<\/sup>x \n(C) 2tan\u22121<\/sup>x \n(D) tan\u22121<\/sup>2x\n<\/p>\nAnswer:<\/strong> (A) 2cos\u22121<\/sup>x<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<figure class=\"wp-block-table\">\n<table class=\"has-fixed-layout\" style=\"border: 1px solid #000000; width: 100%; border-collapse: collapse;\">\n<tbody>\n<tr>\n<td style=\"border: 1px solid #000000; width: 62.6551%; padding: 8px;\">Join Our Telegram Channel<\/strong><\/span><\/td>\n<td style=\"border: 1px solid #000000; background-color: #24a1de; text-align: center; width: 37.3449%; padding: 8px;\">\n\n<a style=\"color: #ffffff;\" href=\"https:\/\/t.me\/examdhara_school\">Join <\/strong><\/a>\n<\/span>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/figure>\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Bihar Board Class 12 Maths \u2013 Section A (MCQs)<\/title>\n<\/head>\n<body>\n\n11.<\/strong> | 23 12 11 | \n | 36 10 26 | = \n | 63 26 37 |<\/p>\n\n(A) 1 \n(B) \u22121 \n(C) 0 \n(D) 2\n<\/p>\nAnswer:<\/strong> (C) 0<\/p>\n\n<hr>\n\n12.<\/strong> | a\u2212b b\u2212c c\u2212a | \n | b\u2212c c\u2212a a\u2212b | = \n | c\u2212a a\u2212b b\u2212c |<\/p>\n\n(A) 1 \n(B) 0 \n(C) \u22121 \n(D) a + b + c\n<\/p>\nAnswer:<\/strong> (B) 0<\/p>\n\n<hr>\n\n13.<\/strong> | x 4 | = 0 \u21d2 x = \n | 2 2x |<\/p>\n\n(A) \u00b12 \n(B) \u00b11 \n(C) \u00b13 \n(D) 0\n<\/p>\nAnswer:<\/strong> (A) \u00b12<\/p>\n\n<hr>\n\n14.<\/strong> | cos15\u00b0 sin15\u00b0 | \n | sin75\u00b0 cos75\u00b0 | =<\/p>\n\n(A) 1 \n(B) 0 \n(C) \u22121 \n(D) 1\/2\n<\/p>\nAnswer:<\/strong> (B) 0<\/p>\n\n<hr>\n\n15.<\/strong> 1 \u2212 P(A\u2032 \u2229 B\u2032) =<\/p>\n\n(A) P(A \u2229 B) \n(B) P(A \u222a B) \n(C) P(A) \n(D) P(B)\n<\/p>\nAnswer:<\/strong> (B) P(A \u222a B)<\/p>\n\n<hr>\n\n16.<\/strong> If A, B and C are three independent events, then P(A \u2229 B \u2229 C) =<\/p>\n\n(A) P(A) + P(B) + P(C) \n(B) P(A) \u2212 P(B) \u2212 P(C) \n(C) P(A)\u00b7P(B)\u00b7P(C) \n(D) None of these\n<\/p>\nAnswer:<\/strong> (C) P(A)\u00b7P(B)\u00b7P(C)<\/p>\n\n<hr>\n\n17.<\/strong> If P(A)=7\/13, P(B)=9\/13 and P(A \u2229 B)=4\/13, then P(A|B) =<\/p>\n\n(A) 4\/9 \n(B) 4\/7 \n(C) 12\/13 \n(D) 6\n<\/p>\nAnswer:<\/strong> (A) 4\/9<\/p>\n\n<hr>\n\n18.<\/strong> If the odds against the event A is 3 : 7, then P(A) =<\/p>\n\n(A) 3\/10 \n(B) 7\/10 \n(C) 3\/7 \n(D) 7\/3\n<\/p>\nAnswer:<\/strong> (B) 7\/10<\/p>\n\n<hr>\n\n19.<\/strong> d\/dx (log xn<\/sup>) =<\/p>\n\n(A) 1\/xn<\/sup> \n(B) n \n(C) 1\/x \n(D) n\/x\n<\/p>\nAnswer:<\/strong> (D) n\/x<\/p>\n\n<hr>\n\n20.<\/strong> d2<\/sup>\/dx2<\/sup> (sin 2x) =<\/p>\n\n(A) 4 sin 2x \n(B) 4 cos\u00b2 2x \n(C) \u22124 sin 2x \n(D) 2 sin 4x\n<\/p>\nAnswer:<\/strong> (C) \u22124 sin 2x<\/p>\n\n<hr>\n\n21.<\/strong> d\/dx (ex\u2212a<\/sup>) =<\/p>\n\n(A) ex\u2212a<\/sup> \n(B) (x\u2212a)ex\u2212a<\/sup> \n(C) ex<\/sup> \n(D) \u2212ex\u2212a<\/sup>\n<\/p>\nAnswer:<\/strong> (A) ex\u2212a<\/sup><\/p>\n\n<hr>\n\n22.<\/strong> d\/dx \u221a(x\u00b2 + ax + 1) =<\/p>\n\n(A) (x + a) \/ (2\u221a(x\u00b2 + ax + 1)) \n(B) (2x + a) \/ (2\u221a(x\u00b2 + ax + 1)) \n(C) (2x + a) \/ \u221a(x\u00b2 + ax + 1) \n(D) 1 \/ (2\u221a(x\u00b2 + ax + 1))\n<\/p>\nAnswer:<\/strong> (B) (2x + a) \/ (2\u221a(x\u00b2 + ax + 1))<\/p>\n\n<hr>\n\n23.<\/strong> d\/dx (sin x\u00b2) =<\/p>\n\n(A) 2x cos x\u00b2 \n(B) cos x\u00b2 \n(C) x\u00b2 cos x\u00b2 \n(D) x cos x\u00b2\n<\/p>\nAnswer:<\/strong> (A) 2x cos x\u00b2<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Bihar Board Class 12 Maths \u2013 Section A (Q.24\u201337)<\/title>\n<\/head>\n<body>\n\n24.<\/strong> d\/dx (\u221acot x) =<\/p>\n\n(A) 1 \/ (2\u221acot x) \n(B) \u221acosec\u00b2x \n(C) \u2212cosec\u00b2x \/ (2\u221acot x) \n(D) cosec\u00b2x \/ (2\u221acot x)\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n25.<\/strong> d\/dx (tan\u22121<\/sup>\u221ax + cot\u22121<\/sup>\u221ax) =<\/p>\n\n(A) \u03c0\/2 \n(B) 0 \n(C) 1 \n(D) \u03c0\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n26.<\/strong> d\/dx (2 tan\u22121<\/sup> x) =<\/p>\n\n(A) 1 \/ (1 + x\u00b2) \n(B) 2 \/ (1 + x\u00b2) \n(C) 1 \/ (2(1 + x\u00b2)) \n(D) 1 \/ (1 \u2212 x\u00b2)\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n27.<\/strong> d\/dx (cos \u221ax) =<\/p>\n\n(A) sin \u221ax \n(B) \u2212 sin \u221ax \/ (2\u221ax) \n(C) sin \u221ax \/ (2\u221ax) \n(D) 1 \/ (2\u221ax)\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n28.<\/strong> d\/dx { limx\u21920<\/sub> (x5<\/sup> \u2212 a5<\/sup>) \/ (x \u2212 a) } =<\/p>\n\n(A) a \n(B) 0 \n(C) 5a4<\/sup> \n(D) 5\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n29.<\/strong> d\/dx (cot\u22121<\/sup> x) =<\/p>\n\n(A) 1 \/ (1 + x\u00b2) \n(B) \u22121 \/ (1 + x\u00b2) \n(C) 1 \/ x \n(D) \u22121 \/ x\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n30.<\/strong> The equation of the tangent to the curve y = x\u00b2 + 4x + 1 at x = 3 is<\/p>\n\n(A) x + 10y = 8 \n(B) 10x + y = 8 \n(C) 10x \u2212 y = 8 \n(D) x \u2212 10y = 8\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n31.<\/strong> If y = \u221a(sin x + \u221a(sin x + \u221a(sin x + \u2026))), then dy\/dx =<\/p>\n\n(A) 1 \/ (2y \u2212 1) \n(B) cos x \/ (2y \u2212 1) \n(C) sin x \/ (2y \u2212 1) \n(D) (2y \u2212 1) \/ cos x\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n32.<\/strong> If xn<\/sup> + yn<\/sup> = an<\/sup>, then dy\/dx =<\/p>\n\n(A) \u2212 xn\u22121<\/sup> \/ yn\u22121<\/sup> \n(B) xn\u22121<\/sup> \/ yn\u22121<\/sup> \n(C) \u2212 yn\u22121<\/sup> \/ xn\u22121<\/sup> \n(D) nxn\u22121<\/sup>\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n33.<\/strong> If x = a(1 \u2212 cos\u03b8), y = a(\u03b8 + sin\u03b8), then dy\/dx =<\/p>\n\n(A) tan(\u03b8\/2) \n(B) \u2212tan(\u03b8\/2) \n(C) cot(\u03b8\/2) \n(D) \u2212cot(\u03b8\/2)\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n34.<\/strong> If y = xx<\/sup>, then dy\/dx =<\/p>\n\n(A) xx<\/sup>(log x + 1) \n(B) log x \n(C) log x + 1 \n(D) nxn\u22121<\/sup>\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n35.<\/strong> d\/dx [sin\u22121<\/sup>(3x \u2212 4x\u00b3)] =<\/p>\n\n(A) 3 \/ \u221a(1 \u2212 x\u00b2) \n(B) 1 \/ \u221a(1 \u2212 x\u00b2) \n(C) \u22121 \/ \u221a(1 \u2212 x\u00b2) \n(D) 1 \/ (1 \u2212 x\u00b2)\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n36.<\/strong> d\/dx (sin 3x \u00b7 cos 5x) =<\/p>\n\n(A) 4 cos 8x \n(B) 4 cos 8x \u2212 cos 2x \n(C) cos 2x \n(D) cos 2x \u2212 4 cos 8x\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n37.<\/strong> d\/dx (cos x\u00b3) =<\/p>\n\n(A) \u22123x\u00b2 sin x\u00b3 \n(B) sin x\u00b3 \n(C) 3x\u00b2 sin x\u00b3 \n(D) 3x\u00b2\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Bihar Board Class 12 Maths \u2013 Section A (Q.38\u201350)<\/title>\n<\/head>\n<body>\n\n38.<\/strong> If f : R \u2192 R, where f(x) = 3x \u2212 5, then f\u22121<\/sup>(x) =<\/p>\n\n(A) (1\/3)(x + 5) \n(B) (1\/3)(x \u2212 5) \n(C) 3x \u2212 5 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n39.<\/strong> \u222b (x + 2) dx =<\/p>\n\n(A) (x + 2)3<\/sup> + k \n(B) x2<\/sup>\/2 + k \n(C) x2<\/sup>\/2 + 2x + k \n(D) log(x + 2) + k\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n40.<\/strong> \u222b cos 2x \/ (sin x + cos x)2<\/sup> dx =<\/p>\n\n(A) 2 log(sin x + cos x) + k \n(B) log(sin x + cos x) + k \n(C) log(sin x \u2212 cos x) + k \n(D) \u22121\/(sin x + cos x) + k\n<\/p>\nAnswer:<\/strong> (D)<\/p>\n\n<hr>\n\n41.<\/strong> \u222b (1 \u2212 cos 2x) \/ (1 + cos 2x) dx =<\/p>\n\n(A) tan x + x + k \n(B) tan x \u2212 x + k \n(C) x \u2212 tan\u00b2x + k \n(D) tan(x\/2) + k\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n42.<\/strong> \u222b dx \/ (x \u2212 1) =<\/p>\n\n(A) log|x + 1| + k \n(B) \u2212log|x + 1| + k \n(C) log|x \u2212 1| + k \n(D) log x + k\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n43.<\/strong> \u222b log x dx =<\/p>\n\n(A) 1\/x + k \n(B) x log x + k \n(C) x log x \u2212 x + k \n(D) x log x + x + k\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n44.<\/strong> \u222b cos \u221ax dx =<\/p>\n\n(A) sin\u221ax + cos\u221ax + k \n(B) 1\/2 (\u221ax sin\u221ax \u2212 cos\u221ax) + k \n(C) 2(\u221ax sin\u221ax + cos\u221ax) + k \n(D) sin\u221ax + k\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n45.<\/strong> \u222b ex<\/sup>(tan\u22121<\/sup>x + 1\/(1 + x2<\/sup>)) dx =<\/p>\n\n(A) ex<\/sup> tan\u22121<\/sup>x + k \n(B) ex<\/sup> \/ (1 + x2<\/sup>) + k \n(C) ex<\/sup> + k \n(D) tan\u22121<\/sup>x + k\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n46.<\/strong> \u222b dx \/ (x2<\/sup> \u2212 a2<\/sup>) =<\/p>\n\n(A) (1\/a) tan\u22121<\/sup>(x\/a) + k \n(B) (1\/2a) log|(x \u2212 a)\/(x + a)| + k \n(C) (1\/2a) log|(a + x)\/(a \u2212 x)| + k \n(D) (1\/a) log|(x \u2212 a)\/(x + a)| + k\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n47.<\/strong> \u222b sin2<\/sup>(x\/2) dx =<\/p>\n\n(A) 1\/2 x \u2212 1\/2 sin x + k \n(B) 1\/2 x \u2212 1\/2 cos x + k \n(C) 1\/2 sin x + k \n(D) \u22121\/2 sin x + k\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n48.<\/strong> \u222b tan(sin\u22121<\/sup>x) \/ \u221a(1 \u2212 x2<\/sup>) dx =<\/p>\n\n(A) log|sec(sin\u22121<\/sup>x)| + k \n(B) log|cos(sin\u22121<\/sup>x)| + k \n(C) tan(sin\u22121<\/sup>x) + k \n(D) log|sin\u22121<\/sup>x| + k\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n49.<\/strong> \u222b dx \/ (ex<\/sup> + e\u2212x<\/sup>) =<\/p>\n\n(A) cot\u22121<\/sup>(ex<\/sup>) + k \n(B) tan\u22121<\/sup>(ex<\/sup>) + k \n(C) log|ex<\/sup> + 1| + k \n(D) sin\u22121<\/sup>(ex<\/sup>) + k\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n50.<\/strong> \u222b0<\/sub>1<\/sup> dx \/ (1 + x2<\/sup>) =<\/p>\n\n(A) \u03c0\/2 \n(B) \u03c0\/3 \n(C) \u03c0\/4 \n(D) \u03c0\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Bihar Board Class 12 Maths \u2013 Section A (Q.51\u201363)<\/title>\n<\/head>\n<body>\n\n51.<\/strong> \u222b0<\/sub>1<\/sup> x\u00b3 \/ (1 + x\u2078) dx =<\/p>\n\n(A) \u03c0\/2 \n(B) \u03c0\/4 \n(C) \u03c0\/8 \n(D) \u03c0\/16\n<\/p>\nAnswer:<\/strong> (C) \u03c0\/8<\/p>\n\n<hr>\n\n52.<\/strong> \u222b1<\/sub>e<\/sup> (log x)\u00b2 \/ x dx =<\/p>\n\n(A) 1\/3 \n(B) (1\/3)e\u00b3 \n(C) (1\/3)(e\u00b3 \u2212 1) \n(D) e\u00b3\n<\/p>\nAnswer:<\/strong> (B) (1\/3)e\u00b3<\/p>\n\n<hr>\n\n53.<\/strong> If A = [4 2 3], then A\u2032 =<\/p>\n\n(A) [4 2 3] \n(B) [3 2 4] \n(C) [3 2 4] \n(D) [4 2 3]T<\/sup>\n<\/p>\nAnswer:<\/strong> (D)<\/p>\n\n<hr>\n\n54.<\/strong> \u222b0<\/sub>\u03c0\/4<\/sup> etan x<\/sup> \/ cos\u00b2x dx =<\/p>\n\n(A) e \u2212 1 \n(B) e + 1 \n(C) 1\/e + 1 \n(D) 1\/e \u2212 1\n<\/p>\nAnswer:<\/strong> (A) e \u2212 1<\/p>\n\n<hr>\n\n55.<\/strong> \u222b0<\/sub>\u03c0\/2<\/sup> cos\u00b2x dx =<\/p>\n\n(A) \u03c0\/2 \n(B) \u03c0 \n(C) \u03c0\/4 \n(D) 1\n<\/p>\nAnswer:<\/strong> (C) \u03c0\/4<\/p>\n\n<hr>\n\n56.<\/strong> \u222b0<\/sub>\u03c0\/3<\/sup> cos\u00b3x dx =<\/p>\n\n(A) 3\u221a3\/8 \n(B) \u221a3\/8 \n(C) 3\/8 \n(D) 1\/8\n<\/p>\nAnswer:<\/strong> (B) \u221a3\/8<\/p>\n\n<hr>\n\n57.<\/strong> \u222b0<\/sub>2\u03c0<\/sup> |sin x| dx =<\/p>\n\n(A) 2 \n(B) 4 \n(C) 1 \n(D) 3\n<\/p>\nAnswer:<\/strong> (B) 4<\/p>\n\n<hr>\n\n58.<\/strong> \u222b0<\/sub>\u03c0\/2<\/sup> sin x \/ (sin x + cos x) dx =<\/p>\n\n(A) \u03c0 \n(B) \u03c0\/2 \n(C) 0 \n(D) \u03c0\/4\n<\/p>\nAnswer:<\/strong> (D) \u03c0\/4<\/p>\n\n<hr>\n\n59.<\/strong> \u222b\u2212\u03c0<\/sub>\u03c0<\/sup> sin\u2075x dx =<\/p>\n\n(A) 3\u03c0\/4 \n(B) 2\u03c0 \n(C) 5\u03c0\/6 \n(D) 0\n<\/p>\nAnswer:<\/strong> (D) 0<\/p>\n\n<hr>\n\n60.<\/strong> \u222b0<\/sub>a<\/sup> \u221ax \/ \u221a(a \u2212 x + \u221ax) dx =<\/p>\n\n(A) a \n(B) a\/2 \n(C) 2a \n(D) 3a\n<\/p>\nAnswer:<\/strong> (A) a<\/p>\n\n<hr>\n\n61.<\/strong> \u222b0<\/sub>1<\/sup> e\u221ax<\/sup> \/ \u221ax dx =<\/p>\n\n(A) 2(e \u2212 1) \n(B) e \u2212 1 \n(C) 2(e + 1) \n(D) e + 1\n<\/p>\nAnswer:<\/strong> (A) 2(e \u2212 1)<\/p>\n\n<hr>\n\n62.<\/strong> \u222b0<\/sub>1<\/sup> x ex<\/sup> dx =<\/p>\n\n(A) 1 \n(B) 0 \n(C) 2 \n(D) \u22121\n<\/p>\nAnswer:<\/strong> (B) 0<\/p>\n\n<hr>\n\n63.<\/strong> \u222b1<\/sub>4<\/sup> dx \/ \u221ax =<\/p>\n\n(A) 1 \n(B) \u22122 \n(C) 2 \n(D) \u22121\n<\/p>\nAnswer:<\/strong> (C) 2<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Bihar Board Class 12 Maths \u2013 Section A (Q.64\u201372)<\/title>\n<\/head>\n<body>\n\n64.<\/strong> \u222b0<\/sub>a<\/sup> \u221a(a\u00b2 \u2212 x\u00b2) dx =<\/p>\n\n(A) \u03c0\/4 \n(B) a\u00b2\/4 \n(C) \u03c0a\u00b2\/4 \n(D) \u03c0\n<\/p>\nAnswer:<\/strong> (C) \u03c0a\u00b2\/4<\/p>\n\n<hr>\n\n65.<\/strong> \u222b\u22122<\/sub>2<\/sup> |x| dx =<\/p>\n\n(A) 4 \n(B) 3 \n(C) 2 \n(D) 0\n<\/p>\nAnswer:<\/strong> (A) 4<\/p>\n\n<hr>\n\n66.<\/strong> The order and degree of the differential equation \n(d\u00b2s\/dt\u00b2)\u00b2 + (ds\/dt)\u00b3 + 4 = 0 is<\/p>\n\n(A) order = 2, degree = 1 \n(B) order = 2, degree = 2 \n(C) order = 1, degree = 2 \n(D) order = 1, degree = 1\n<\/p>\nAnswer:<\/strong> (B) order = 2, degree = 2<\/p>\n\n<hr>\n\n67.<\/strong> The integrating factor of the differential equation \n(1 + x\u00b2) dy\/dx + y = etan\u22121<\/sup>x<\/sup> is<\/p>\n\n(A) etan\u22121<\/sup>x<\/sup> \n(B) esin\u22121<\/sup>x<\/sup> \n(C) tan\u22121<\/sup>x \n(D) sin\u22121<\/sup>x\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n68.<\/strong> The solution of the differential equation dy\/dx = ex+y<\/sup> is<\/p>\n\n(A) ex<\/sup> + e\u2212y<\/sup> = k \n(B) ex<\/sup> + ey<\/sup> = k \n(C) e\u2212x<\/sup> + ey<\/sup> = k \n(D) e\u2212x<\/sup> + e\u2212y<\/sup> = k\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n69.<\/strong> The solution of the differential equation \nx (dy\/dx) = cot y is<\/p>\n\n(A) x cos y = k \n(B) x tan y = k \n(C) x sec y = k \n(D) x sin y = k\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n70.<\/strong> If the operation * is defined as a * b = a + 2b, then (2 * 3) * 4 is<\/p>\n\n(A) 30 \n(B) 20 \n(C) 16 \n(D) 15\n<\/p>\nAnswer:<\/strong> (B) 20<\/p>\n\n<hr>\n\n71.<\/strong> If A = \n[\n1 \u22121 \n\u22121 1\n], then A\u00b3 =<\/p>\n\n(A) 3A \n(B) 4A \n(C) 2A \n(D) None of these\n<\/p>\nAnswer:<\/strong> (D)<\/p>\n\n<hr>\n\n72.<\/strong> If A = \n[\ncos\u03b1 sin\u03b1 \n\u2212sin\u03b1 cos\u03b1\n] and A + A\u2032 = I, then \u03b1 =<\/p>\n\n(A) \u03c0 \n(B) \u03c0\/3 \n(C) 3\u03c0\/2 \n(D) \u03c0\/6\n<\/p>\nAnswer:<\/strong> (D) \u03c0\/6<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Bihar Board Class 12 Maths \u2013 Section A (Q.73\u201380)<\/title>\n<\/head>\n<body>\n\n73.<\/strong> If A =\n[\n3 \u22125 \n\u22121 2\n], then adj A =<\/p>\n\n(A)\n[\n2 5 \n1 3\n] \n(B)\n[\n2 3 \n1 5\n] \n(C)\n[\n1 3 \n2 5\n] \n(D) None of these\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n74.<\/strong> If A = [1 2 3 4] and B =\n[\n1 \n2 \n3 \n4\n], then AB =<\/p>\n\n(A) [30] \n(B) [10] \n(C) [20] \n(D) [40]\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n75.<\/strong> A = [aij<\/sub>]m\u00d7n<\/sub> is a square matrix if<\/p>\n\n(A) m = n \n(B) m < n \n(C) m > n \n(D) None of these\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n76.<\/strong> If A =\n[\n3 \u22124 \n1 \u22121\n], then A + A\u2032 =<\/p>\n\n(A)\n[\n6 \u22123 \n\u22123 \u22122\n] \n(B)\n[\n6 3 \n3 2\n] \n(C)\n[\n6 3 \n\u22123 \u22122\n] \n(D)\n[\n6 \u22123 \n\u22123 2\n]\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n77.<\/strong> f : A \u2192 B will be an onto function if<\/p>\n\n(A) f(A) \u2282 B \n(B) f(A) = B \n(C) f(A) \u2283 B \n(D) None of these\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n78.<\/strong> What type of relation is \u201cless than\u201d in the set of real numbers?<\/p>\n\n(A) Only symmetric \n(B) Only transitive \n(C) Only reflexive \n(D) Equivalence\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n79.<\/strong> | 3i + 4j + 7k | =<\/p>\n\n(A) \u221a14 \n(B) \u221a74 \n(C) \u221a61 \n(D) \u221a94\n<\/p>\nAnswer:<\/strong> (D)<\/p>\n\n<hr>\n\n80.<\/strong> i \u00b7 ( j \u00d7 k ) =<\/p>\n\n(A) 1 \n(B) 0 \n(C) \u22121 \n(D) i\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Bihar Board Class 12 Maths \u2013 Section A (Q.81\u201389)<\/title>\n<\/head>\n<body>\n\n81.<\/strong> If | a\u20d7 + b\u20d7 | = | a\u20d7 \u2212 b\u20d7 |, then<\/p>\n\n(A) |a\u20d7| = |b\u20d7| \n(B) a\u20d7 \u2225 b\u20d7 \n(C) a\u20d7 \u27c2 b\u20d7 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n82.<\/strong> a\u20d7 \u00b7 (a\u20d7 \u00d7 a\u20d7) =<\/p>\n\n(A) 1 \n(B) 0 \n(C) a\u20d7 \n(D) \u22121\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n83.<\/strong> If 3i\u20d7 + j\u20d7 \u2212 2k\u20d7 and i\u20d7 + \u03bbj\u20d7 \u2212 3k\u20d7 are perpendicular to each other, then the value of \u03bb is<\/p>\n\n(A) \u22123 \n(B) \u22126 \n(C) \u22129 \n(D) \u22121\n<\/p>\nAnswer:<\/strong> (D)<\/p>\n\n<hr>\n\n84.<\/strong> The projection of the vector i\u20d7 + 3j\u20d7 + 7k\u20d7 on the vector 2i\u20d7 \u2212 3j\u20d7 + 6k\u20d7 is<\/p>\n\n(A) 5 \n(B) 25 \n(C) 6 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n85.<\/strong> If |a\u20d7| = 2, |b\u20d7| = 3 and a\u20d7 \u00b7 b\u20d7 = 4, then |a\u20d7 \u2212 b\u20d7| =<\/p>\n\n(A) 5 \n(B) \u221a5 \n(C) 4 \n(D) 2\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n86.<\/strong> If a\u20d7 = i\u20d7 \u2212 j\u20d7 + 2k\u20d7 and b\u20d7 = 2i\u20d7 + 3j\u20d7 \u2212 4k\u20d7, then |a\u20d7 \u00d7 b\u20d7| =<\/p>\n\n(A) \u221a174 \n(B) \u221a87 \n(C) \u221a93 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n87.<\/strong> If |a\u20d7 + b\u20d7| = |a\u20d7 \u2212 b\u20d7|, then<\/p>\n\n(A) |a\u20d7| = |b\u20d7| \n(B) a\u20d7 \u2225 b\u20d7 \n(C) a\u20d7 \u27c2 b\u20d7 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n88.<\/strong> 2j\u20d7 \u00b7 (\u22123k\u20d7) =<\/p>\n\n(A) 6 \n(B) \u22126 \n(C) 0 \n(D) \u22126i\u20d7\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n89.<\/strong> The direction ratios of a straight line are 2, 6, \u22123. Then its direction cosines are<\/p>\n\n(A) 1\/7, 2\/7, 3\/7 \n(B) 2\/7, \u22126\/7, 3\/7 \n(C) 2\/7, 6\/7, \u22123\/7 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Bihar Board Class 12 Maths \u2013 Section A (Q.90\u201395)<\/title>\n<\/head>\n<body>\n\n90.<\/strong> If a line makes angles \u03b1, \u03b2 and \u03b3 with the positive directions of x, y and z axes respectively, then<\/p>\n\n(A) cos\u00b2\u03b1 + cos\u00b2\u03b2 + cos\u00b2\u03b3 = 1 \n(B) sin\u00b2\u03b1 + sin\u00b2\u03b2 + sin\u00b2\u03b3 = 4 \n(C) cos\u00b2\u03b1 + cos\u00b2\u03b2 + cos\u00b2\u03b3 = 2 \n(D) sin\u00b2\u03b1 + sin\u00b2\u03b2 + sin\u00b2\u03b3 = 1\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n91.<\/strong> The angle between the straight lines \n(x \u2212 2)\/2 = (y \u2212 1)\/7 = (z + 3)\/(\u22123) \nand \n(x + 2)\/(\u22121) = (y \u2212 4)\/2 = (z \u2212 5)\/4 \nis<\/p>\n\n(A) \u03c0\/2 \n(B) 0 \n(C) \u03c0\/6 \n(D) \u03c0\/4\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n92.<\/strong> The distance of the plane x + 2y \u2212 2z = 9 from the point (2, 3, \u22125) is<\/p>\n\n(A) 1 \n(B) 2 \n(C) 3 \n(D) 4\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n93.<\/strong> If two planes x \u2212 4y + \u03bbz + 3 = 0 and 2x + 2y + 3z = 5 are perpendicular to each other, then \u03bb =<\/p>\n\n(A) 1 \n(B) 2 \n(C) 3 \n(D) 4\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n94.<\/strong> If the line \n(x \u2212 x\u2081)\/a\u2081 = (y \u2212 y\u2081)\/b\u2081 = (z \u2212 z\u2081)\/c\u2081 \nis parallel to the plane a\u2082x + b\u2082y + c\u2082z + d = 0, then<\/p>\n\n(A) a\u2081\/a\u2082 = b\u2081\/b\u2082 = c\u2081\/c\u2082 \n(B) a\u2081x + b\u2081y + c\u2081z + d = 0 \n(C) a\u2081a\u2082 + b\u2081b\u2082 + c\u2081c\u2082 = 0 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n95.<\/strong> The equation of the plane whose intercepts on the x, y and z axes are \u22122, 3 and 4 respectively is<\/p>\n\n(A) 6x \u2212 4y \u2212 3z + 12 = 0 \n(B) 6x + 4y + 3z + 12 = 0 \n(C) 6x \u2212 4y \u2212 3z = 0 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Bihar Board Class 12 Maths \u2013 Section A (Q.96\u2013100)<\/title>\n<\/head>\n<body>\n\n96.<\/strong> If n(A) = 4 and n(B) = 2, then n(A \u00d7 B) =<\/p>\n\n(A) 6 \n(B) 8 \n(C) 16 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (B) 8<\/p>\n\n<hr>\n\n97.<\/strong> The solution of the differential equation \nx (dy\/dx) = cot y is<\/p>\n\n(A) x cos y = k \n(B) x tan y = k \n(C) x sec y = k \n(D) x sin y = k\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n98.<\/strong> | a + ib\u2003\u2003c + id | \n | \u2212c + id\u2003a \u2212 ib | =<\/p>\n\n(A) a\u00b2 + b\u00b2 + c\u00b2 + d\u00b2 \n(B) a\u00b2 \u2212 b\u00b2 \u2212 c\u00b2 \u2212 d\u00b2 \n(C) a\u00b2 \u2212 b\u00b2 + c\u00b2 + d\u00b2 \n(D) a\u00b2 + b\u00b2 + c\u00b2 \u2212 d\u00b2\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n99.<\/strong> The maximum value of Z = 4x + y subject to the constraints \nx + y \u2264 50, x \u2265 0, y \u2265 0 is<\/p>\n\n(A) 50 \n(B) 250 \n(C) 0 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (B) 250<\/p>\n\n<hr>\n\n100.<\/strong> If n(A) = 4 and n(B) = 2, then n(A \u00d7 B) =<\/p>\n\n(A) 6 \n(B) 8 \n(C) 16 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (B) 8<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<h2 class=\"wp-block-heading\"><\/span>Section B Bihar Board 12th Math 2026 Question<\/span><\/h2>\n\n\n\n<h3 class=\"wp-block-heading\"><\/span>Short Answer Type Question<\/span><\/h3>\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Section B \u2013 Short Answer Solutions<\/title>\n<\/head>\n<body>\n\n1. Prove that<\/strong> \ntan\u22121<\/sup>1 + tan\u22121<\/sup>2 + tan\u22121<\/sup>3 = \u03c0<\/p>\n\nSolution:<\/strong> \nUsing the identity \ntan\u22121<\/sup>a + tan\u22121<\/sup>b = tan\u22121<\/sup>((a+b)\/(1\u2212ab))<\/p>\n\n\ntan\u22121<\/sup>1 + tan\u22121<\/sup>2 = tan\u22121<\/sup>(3 \/ \u22121) = tan\u22121<\/sup>(\u22123)\n<\/p>\n\n\ntan\u22121<\/sup>(\u22123) + tan\u22121<\/sup>3 = \u03c0\n<\/p>\n\nHence proved.<\/p>\n\n<hr>\n\n2. Find the value of<\/strong> \n2tan\u22121<\/sup>(1\/3) + tan\u22121<\/sup>(1\/7)<\/p>\n\nSolution:<\/strong> \n2tan\u22121<\/sup>x = tan\u22121<\/sup>(2x \/ (1 \u2212 x\u00b2))<\/p>\n\n\n2tan\u22121<\/sup>(1\/3) = tan\u22121<\/sup>(3\/4)\n<\/p>\n\n\ntan\u22121<\/sup>(3\/4) + tan\u22121<\/sup>(1\/7) = tan\u22121<\/sup>(1)\n<\/p>\n\n\n= \u03c0\/4\n<\/p>\n\n<hr>\n\n3. Prove that<\/strong> \nsec2<\/sup>(tan\u22121<\/sup>2) + cosec2<\/sup>(cot\u22121<\/sup>3) = 15<\/p>\n\nSolution:<\/strong> \nLet tan\u03b8 = 2 \u21d2 sec2<\/sup>\u03b8 = 1 + tan2<\/sup>\u03b8 = 5<\/p>\n\n\nLet cot\u03c6 = 3 \u21d2 cosec2<\/sup>\u03c6 = 1 + cot2<\/sup>\u03c6 = 10\n<\/p>\n\n\n\u2234 Required sum = 5 + 10 = 15\n<\/p>\n\nHence proved.<\/p>\n\n<hr>\n\n4. Prove that the value of the determinant is zero:<\/strong><\/p>\n\n\n| 102 18 36 | \n| 1 3 4 | \n| 17 3 6 |\n<\/p>\n\nSolution:<\/strong> \nFirst row = 6 \u00d7 Third row<\/p>\n\n\nSince two rows are proportional, the determinant is zero.\n<\/p>\n\nHence proved.<\/p>\n\n<hr>\n\n5. Prove that<\/strong><\/p>\n\n\n| 0 a b | \n| c 0 b | = 2abc \n| c a 0 |\n<\/p>\n\nSolution:<\/strong> \nExpanding along the first row:<\/p>\n\n\n= \u2212a(\u2212bc) + b(ca) = abc + abc = 2abc\n<\/p>\n\nHence proved.<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Section B \u2013 Short Answer Solutions (Q.6\u201311)<\/title>\n<\/head>\n<body>\n\n6. Find the value of x, when<\/strong><\/p>\n\n| 2x 8 | \n| 4 x | = 0\n<\/p>\n\nSolution:<\/strong> \nDeterminant = (2x)(x) \u2212 (8)(4)<\/p>\n\n\n2x\u00b2 \u2212 32 = 0 \nx\u00b2 = 16 \nx = \u00b14\n<\/p>\n\n<hr>\n\n7. Find the slope at the point (1, \u221a2) of the curve<\/strong> \nx\u00b2 + y\u00b2 = 3<\/p>\n\nSolution:<\/strong> \nDifferentiate implicitly:<\/p>\n\n\n2x + 2y (dy\/dx) = 0 \ndy\/dx = \u2212x \/ y\n<\/p>\n\n\nAt (1, \u221a2): \ndy\/dx = \u22121 \/ \u221a2\n<\/p>\n\n<hr>\n\n8. Find dy\/dx when<\/strong> \nx = at\u00b2, y = 2at<\/p>\n\nSolution:<\/strong> \ndx\/dt = 2at \ndy\/dt = 2a<\/p>\n\n\ndy\/dx = (dy\/dt)\/(dx\/dt) = 2a \/ (2at) = 1\/t\n<\/p>\n\n<hr>\n\n9. If<\/strong> y = sec(tan\u221ax), find dy\/dx<\/strong><\/p>\n\nSolution:<\/strong> \nLet u = tan\u221ax<\/p>\n\n\ndy\/dx = sec(u)tan(u) \u00b7 du\/dx\n<\/p>\n\n\ndu\/dx = sec\u00b2\u221ax \u00b7 (1 \/ 2\u221ax)\n<\/p>\n\n\n\u2234 dy\/dx = sec(tan\u221ax) \u00b7 tan(tan\u221ax) \u00b7 sec\u00b2\u221ax \/ (2\u221ax)\n<\/p>\n\n<hr>\n\n10. Find dy\/dx when<\/strong> \nxsin y<\/sup> = ysin x<\/sup><\/p>\n\nSolution:<\/strong> \nTaking log on both sides:<\/p>\n\n\nsin y \u00b7 log x = sin x \u00b7 log y\n<\/p>\n\n\nDifferentiate implicitly:<\/p>\n\n\ncos y \u00b7 dy\/dx \u00b7 log x + sin y \u00b7 (1\/x) \n= cos x \u00b7 log y + sin x \u00b7 (1\/y) \u00b7 dy\/dx\n<\/p>\n\n\nCollecting dy\/dx terms and simplifying gives the required dy\/dx.\n<\/p>\n\n<hr>\n\n11. Find the inverse matrix of<\/strong><\/p>\n\n[ 4 1 ] \n[ 2 3 ]\n<\/p>\n\nSolution:<\/strong> \nDeterminant = (4)(3) \u2212 (1)(2) = 10<\/p>\n\n\nInverse matrix = (1\/10)\n<\/p>\n\n\n[ 3 \u22121 ] \n[ \u22122 4 ]\n<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Section B \u2013 Short Answer Solutions (Q.12\u201316)<\/title>\n<\/head>\n<body>\n\n12. If<\/strong> \nA = \n[\n1 2 \n\u22123 4\n], \nB =\n[\n0 7 \n2 \u22124\n], \nfind the value of (4A \u2212 B).<\/p>\n\nSolution:<\/strong> \n4A =\n[\n4 8 \n\u221212 16\n]<\/p>\n\n\n4A \u2212 B =\n[\n4\u22120 8\u22127 \n\u221212\u22122 16\u2212(\u22124)\n]\n<\/p>\n\n\n=\n[\n4 1 \n\u221214 20\n]\n<\/p>\n\n<hr>\n\n13. Integrate:<\/strong> \n\u222b dx \/ (1 + cos x)<\/p>\n\nSolution:<\/strong> \nMultiply numerator and denominator by (1 \u2212 cos x):<\/p>\n\n\n\u222b (1 \u2212 cos x) \/ (1 \u2212 cos\u00b2x) dx \n= \u222b (1 \u2212 cos x) \/ sin\u00b2x dx\n<\/p>\n\n\n= \u222b (cosec\u00b2x \u2212 cot x cosec x) dx\n<\/p>\n\n\n= \u2212cot x + cosec x + C\n<\/p>\n\n<hr>\n\n14. Integrate:<\/strong> \n\u222b sec\u2074x dx<\/p>\n\nSolution:<\/strong> \nsec\u2074x = sec\u00b2x \u00b7 sec\u00b2x<\/p>\n\n\n\u222b sec\u00b2x (1 + tan\u00b2x) dx\n<\/p>\n\n\n= \u222b (1 + tan\u00b2x) d(tan x)\n<\/p>\n\n\n= tan x + (tan\u00b3x)\/3 + C\n<\/p>\n\n<hr>\n\n15. Integrate:<\/strong> \n\u222b sin\u22121<\/sup>x \/ \u221a(1 \u2212 x\u00b2) dx<\/p>\n\nSolution:<\/strong> \nLet t = sin\u22121<\/sup>x<\/p>\n\n\ndt = dx \/ \u221a(1 \u2212 x\u00b2)\n<\/p>\n\n\n\u2234 Integral = \u222b t dt = t\u00b2\/2 + C\n<\/p>\n\n\n= (sin\u22121<\/sup>x)\u00b2 \/ 2 + C\n<\/p>\n\n<hr>\n\n16. Integrate:<\/strong> \n\u222b dx \/ (x\u00b2 + 6x + 13)<\/p>\n\nSolution:<\/strong> \nx\u00b2 + 6x + 13 = (x + 3)\u00b2 + 4<\/p>\n\n\n\u222b dx \/ [(x + 3)\u00b2 + 2\u00b2]\n<\/p>\n\n\n= (1\/2) tan\u22121<\/sup>((x + 3)\/2) + C\n<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Section B \u2013 Short Answer Solutions (Q.17\u201321)<\/title>\n<\/head>\n<body>\n\n17. Find the value of<\/strong> \n\u222b0<\/sub>a<\/sup> x \/ \u221a(a\u00b2 + x\u00b2) dx<\/p>\n\nSolution:<\/strong> \nLet u = a\u00b2 + x\u00b2 \u21d2 du = 2x dx<\/p>\n\n\n\u222b x \/ \u221a(a\u00b2 + x\u00b2) dx = \u00bd \u222b du \/ \u221au = \u221au\n<\/p>\n\n\n= \u221a(a\u00b2 + x\u00b2)\n<\/p>\n\n\nApplying limits 0 to a:\n<\/p>\n\n\n= \u221a(a\u00b2 + a\u00b2) \u2212 \u221a(a\u00b2) = a(\u221a2 \u2212 1)\n<\/p>\n\n<hr>\n\n18. Solve:<\/strong> \nloge<\/sub>(dy\/dx) = ax + by<\/p>\n\nSolution:<\/strong> \ndy\/dx = eax+by<\/sup><\/p>\n\n\ne\u2212by<\/sup> dy = eax<\/sup> dx\n<\/p>\n\n\n\u222b e\u2212by<\/sup> dy = \u222b eax<\/sup> dx\n<\/p>\n\n\n(\u22121\/b)e\u2212by<\/sup> = (1\/a)eax<\/sup> + C\n<\/p>\n\n\nThis is the required solution.\n<\/p>\n\n<hr>\n\n19. Solve:<\/strong> \n\u222b1<\/sub>2<\/sup> (log x)\/x dx<\/p>\n\nSolution:<\/strong> \nLet t = log x \u21d2 dt = dx\/x\n<\/p>\n\n\nIntegral = \u222b t dt = t\u00b2\/2\n<\/p>\n\n\nApplying limits:\n<\/p>\n\n\n= (log 2)\u00b2\/2 \u2212 (log 1)\u00b2\/2 = (log 2)\u00b2 \/ 2\n<\/p>\n\n<hr>\n\n20. Find the value of<\/strong> \n\u222b0<\/sub>\u03c0\/2<\/sup> \u221acos x \/ (\u221acos x + \u221asin x) dx<\/p>\n\nSolution:<\/strong> \nLet I = \u222b \u221acos x \/ (\u221acos x + \u221asin x) dx<\/p>\n\n\nReplace x by (\u03c0\/2 \u2212 x):\n<\/p>\n\n\nI = \u222b \u221asin x \/ (\u221asin x + \u221acos x) dx\n<\/p>\n\n\nAdding both:\n<\/p>\n\n\n2I = \u222b dx = \u03c0\/2\n<\/p>\n\n\n\u2234 I = \u03c0\/4\n<\/p>\n\n<hr>\n\n21. Solve:<\/strong> \ndy\/dx + 2y tan x = sin x<\/p>\n\nSolution:<\/strong> \nThis is a linear differential equation<\/p>\n\n\nIF = e\u222b2 tan x dx<\/sup> = sec\u00b2x\n<\/p>\n\n\nMultiplying throughout by IF:\n<\/p>\n\n\nd\/dx (y sec\u00b2x) = sin x sec\u00b2x\n<\/p>\n\n\nIntegrating:\n<\/p>\n\n\ny sec\u00b2x = sec x + C\n<\/p>\n\n\ny = cos x + C cos\u00b2x\n<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Section B \u2013 Short Answer Solutions (Q.22\u201326)<\/title>\n<\/head>\n<body>\n\n22. If<\/strong> a<\/b> + b<\/b> + c<\/b> = 0, prove that \na \u00d7 b = b \u00d7 c = c \u00d7 a<\/b>.<\/p>\n\nSolution:<\/strong> \nGiven: a + b + c = 0<\/b> \u21d2 a = \u2212(b + c)<\/b><\/p>\n\n\na \u00d7 b = \u2212(b + c) \u00d7 b \n= \u2212(b \u00d7 b + c \u00d7 b) \n= \u2212(0 + c \u00d7 b) \n= b \u00d7 c\n<\/p>\n\n\nSimilarly, b \u00d7 c = c \u00d7 a\n<\/p>\n\n\n\u2234 a \u00d7 b = b \u00d7 c = c \u00d7 a. \u2714\n<\/p>\n\n<hr>\n\n23. Find the area of the parallelogram whose adjacent sides are<\/strong> \n(i + 2j + 3k)<\/b> and (\u22123i \u2212 2j + k)<\/b>.<\/p>\n\nSolution:<\/strong> \nArea = |a \u00d7 b|<\/p>\n\n\na \u00d7 b =\n| i j k | \n| 1 2 3 | \n| \u22123 \u22122 1 |\n<\/p>\n\n\n= i(2\u00d71 \u2212 3\u00d7(\u22122)) \u2212 j(1\u00d71 \u2212 3\u00d7(\u22123)) + k(1\u00d7(\u22122) \u2212 2\u00d7(\u22123))\n<\/p>\n\n\n= 8i \u2212 10j + 4k\n<\/p>\n\n\n|a \u00d7 b| = \u221a(8\u00b2 + (\u221210)\u00b2 + 4\u00b2) = \u221a180 = 6\u221a5\n<\/p>\n\n\n\u2234 Area = 6\u221a5 square units<\/b>.\n<\/p>\n\n<hr>\n\n24. Find the sine of the angle between the vectors<\/strong> \n3i + j + 2k<\/b> and 2i \u2212 2j + 4k<\/b>.<\/p>\n\nSolution:<\/strong> \nsin\u03b8 = |a \u00d7 b| \/ (|a||b|)<\/p>\n\n\na \u00d7 b =\n| i j k | \n| 3 1 2 | \n| 2 \u22122 4 |\n<\/p>\n\n\n= 8i \u2212 8j \u2212 8k\n<\/p>\n\n\n|a \u00d7 b| = 8\u221a3 \n|a| = \u221a14, |b| = \u221a24 = 2\u221a6\n<\/p>\n\n\nsin\u03b8 = (8\u221a3) \/ (\u221a14 \u00d7 2\u221a6) = 2 \/ \u221a7<\/b>\n<\/p>\n\n<hr>\n\n25. If<\/strong> P(A)=1\/4, P(B)=1\/3 and P(A\u222aB)=1\/2, prove that A and B are independent.<\/p>\n\nSolution:<\/strong> \nP(A\u222aB) = P(A) + P(B) \u2212 P(A\u2229B)<\/p>\n\n\n1\/2 = 1\/4 + 1\/3 \u2212 P(A\u2229B)\n<\/p>\n\n\n\u21d2 P(A\u2229B) = 1\/12\n<\/p>\n\n\nP(A)\u00b7P(B) = (1\/4)(1\/3) = 1\/12\n<\/p>\n\n\n\u2234 P(A\u2229B) = P(A)\u00b7P(B) \nHence, A and B are independent events. \u2714\n<\/p>\n\n<hr>\n\n26. Two dice are thrown. Find the probability that the sum is 8 if the second die always shows 4.<\/strong><\/p>\n\nSolution:<\/strong> \nGiven: second die = 4<\/p>\n\n\nSum = 8 \u21d2 first die must be 4\n<\/p>\n\n\nTotal possible outcomes = 6 \nFavourable outcomes = 1\n<\/p>\n\n\nProbability = 1\/6\n<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Section B \u2013 Short Answer Solutions (Q.27\u201330)<\/title>\n<\/head>\n<body>\n\n27. Maximize<\/strong> Z = 3x + 2y \nsubject to: \n3x + y \u2264 15, \nx \u2265 0, y \u2265 0<\/p>\n\nSolution:<\/strong> \nThe feasible region is bounded by the axes and the line 3x + y = 15.<\/p>\n\n\nCorner points are: \n(0,0), (5,0), (0,15)\n<\/p>\n\n\nEvaluate Z at these points:\n<\/p>\n\n\nZ(0,0) = 0 \nZ(5,0) = 15 \nZ(0,15) = 30\n<\/p>\n\n\n\u2234 Maximum value of Z is 30<\/strong> at point (0,15)<\/strong>.\n<\/p>\n\n<hr>\n\n28. Find the distance between the planes<\/strong> \n2x \u2212 y + 3z + 4 = 0 and 6x \u2212 3y + 9z \u2212 3 = 0<\/p>\n\nSolution:<\/strong> \nDivide the second equation by 3:<\/p>\n\n\n2x \u2212 y + 3z \u2212 1 = 0\n<\/p>\n\n\nDistance between parallel planes is:\n<\/p>\n\n\n|d2<\/sub> \u2212 d1<\/sub>| \/ \u221a(a\u00b2 + b\u00b2 + c\u00b2)\n<\/p>\n\n\n= |\u22121 \u2212 4| \/ \u221a(2\u00b2 + (\u22121)\u00b2 + 3\u00b2)\n<\/p>\n\n\n= 5 \/ \u221a14\n<\/p>\n\n\n\u2234 Required distance = 5\/\u221a14<\/strong>.\n<\/p>\n\n<hr>\n\n29. Find the value of p so that the straight lines<\/strong> \n(x \u2212 1)\/(-1) = (y + 3)\/p = (z \u2212 7)\/2 \nand \n(x + 1)\/p = y\/2 = (z + 6)\/(-3) \nare perpendicular.<\/p>\n\nSolution:<\/strong> \nDirection ratios of first line: (\u22121, p, 2)<\/p>\n\n\nDirection ratios of second line: (p, 2, \u22123)\n<\/p>\n\n\nFor perpendicular lines, dot product = 0:\n<\/p>\n\n\n(\u22121)(p) + (p)(2) + (2)(\u22123) = 0\n<\/p>\n\n\n\u2212p + 2p \u2212 6 = 0 \u21d2 p = 6\n<\/p>\n\n\n\u2234 Value of p = 6<\/strong>.\n<\/p>\n\n<hr>\n\n30. If<\/strong> f : R \u2192 R, f(x) = (3 \u2212 x\u00b3)1\/3<\/sup>, prove that (f \u2218 f)(x) = x.<\/p>\n\nSolution:<\/strong> \nGiven f(x) = (3 \u2212 x\u00b3)1\/3<\/sup><\/p>\n\n\n(f \u2218 f)(x) = f(f(x))\n<\/p>\n\n\n= [3 \u2212 {(3 \u2212 x\u00b3)1\/3<\/sup>}\u00b3]1\/3<\/sup>\n<\/p>\n\n\n= (3 \u2212 (3 \u2212 x\u00b3))1\/3<\/sup>\n<\/p>\n\n\n= (x\u00b3)1\/3<\/sup> = x\n<\/p>\n\n\n\u2234 (f \u2218 f)(x) = x. \u2714\n<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<h3 class=\"wp-block-heading\"><\/span>Long Answer Type Question<\/span><\/h3>\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Section C \u2013 Long Answer Type Questions<\/title>\n<\/head>\n<body>\n\n31. Prove that<\/strong><\/p>\n\n\n| x y z | \n| x\u00b2 y\u00b2 z\u00b2 | = (x \u2212 y)(y \u2212 z)(z \u2212 x)(xy + yz + zx) \n| yz zx xy |\n<\/p>\n\nSolution:<\/strong><\/p>\n\n\nLet \u0394 =\n<\/p>\n\n\n| x y z | \n| x\u00b2 y\u00b2 z\u00b2 | \n| yz zx xy |\n<\/p>\n\n\nApply the operation: \nR2<\/sub> \u2192 R2<\/sub> \u2212 xR1<\/sub> \nR3<\/sub> \u2192 R3<\/sub> \u2212 yzR1<\/sub>\n<\/p>\n\n\nThen the determinant simplifies and common factors (x\u2212y), (y\u2212z), (z\u2212x) appear.\n<\/p>\n\n\nAfter simplification, we get:\n<\/p>\n\n\n\u0394 = (x \u2212 y)(y \u2212 z)(z \u2212 x)(xy + yz + zx)\n<\/p>\n\n\nHence proved.\n<\/p>\n\n<hr>\n\n32. If<\/strong> sin\u22121<\/sup>x + sin\u22121<\/sup>y + sin\u22121<\/sup>z = \u03c0, prove that<\/strong><\/p>\n\n\nx\u221a(1 \u2212 x\u00b2) + y\u221a(1 \u2212 y\u00b2) + z\u221a(1 \u2212 z\u00b2) = 2xyz\n<\/p>\n\nSolution:<\/strong><\/p>\n\n\nLet sin\u22121<\/sup>x = A, sin\u22121<\/sup>y = B, sin\u22121<\/sup>z = C\n<\/p>\n\n\nThen A + B + C = \u03c0\n<\/p>\n\n\nSo, sin(A + B) = sin(\u03c0 \u2212 C) = sin C\n<\/p>\n\n\nUsing identity:\n<\/p>\n\n\nsin(A + B) = sinA cosB + cosA sinB\n<\/p>\n\n\n\u21d2 x\u221a(1 \u2212 y\u00b2) + y\u221a(1 \u2212 x\u00b2) = z\n<\/p>\n\n\nMultiplying both sides appropriately and adding cyclic terms, we get:\n<\/p>\n\n\nx\u221a(1 \u2212 x\u00b2) + y\u221a(1 \u2212 y\u00b2) + z\u221a(1 \u2212 z\u00b2) = 2xyz\n<\/p>\n\n\nHence proved.\n<\/p>\n\n<hr>\n\n33. Solve:<\/strong><\/p>\n\n\n(x\u00b2 \u2212 y\u00b2) dy\/dx = 2xy\n<\/p>\n\nSolution:<\/strong><\/p>\n\n\nRewriting the equation:\n<\/p>\n\n\n(x\u00b2 \u2212 y\u00b2) dy = 2xy dx\n<\/p>\n\n\nSeparate variables:\n<\/p>\n\n\n(x\u00b2 \u2212 y\u00b2)\/y dy = 2x dx\n<\/p>\n\n\n\u21d2 (x\u00b2\/y \u2212 y) dy = 2x dx\n<\/p>\n\n\nIntegrate both sides:\n<\/p>\n\n\n\u222b(x\u00b2\/y \u2212 y) dy = \u222b2x dx\n<\/p>\n\n\n\u21d2 x\u00b2 ln|y| \u2212 y\u00b2\/2 = x\u00b2 + C\n<\/p>\n\n\nThis is the required solution.\n<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Section C \u2013 Long Answer Solutions (Q.34\u201338)<\/title>\n<\/head>\n<body>\n\n34. Prove that<\/strong><\/p>\n\n\n\u222b0<\/sub>\u03c0\/2<\/sup> ( \u221atan x + \u221acot x ) dx = \u03c0\u221a2\n<\/p>\n\nSolution:<\/strong><\/p>\n\n\nLet I = \u222b0<\/sub>\u03c0\/2<\/sup> ( \u221atan x + \u221acot x ) dx\n<\/p>\n\n\nUsing the property: \n\u222b0<\/sub>\u03c0\/2<\/sup> f(x) dx = \u222b0<\/sub>\u03c0\/2<\/sup> f(\u03c0\/2 \u2212 x) dx\n<\/p>\n\n\n\u21d2 I = \u222b0<\/sub>\u03c0\/2<\/sup> ( \u221acot x + \u221atan x ) dx\n<\/p>\n\n\nAdding both expressions:\n<\/p>\n\n\n2I = 2 \u222b0<\/sub>\u03c0\/2<\/sup> \u221atan x dx\n<\/p>\n\n\n\u21d2 I = \u222b0<\/sub>\u03c0\/2<\/sup> \u221atan x dx\n<\/p>\n\n\nUsing standard result:\n<\/p>\n\n\n\u222b0<\/sub>\u03c0\/2<\/sup> \u221atan x dx = \u03c0\/\u221a2\n<\/p>\n\n\n\u2234 I = \u03c0\u221a2\n<\/p>\n\n\nHence proved.\n<\/p>\n\n<hr>\n\n35. Find<\/strong> dy\/dx, when<\/strong> xy<\/sup> + yx<\/sup> = 1.<\/p>\n\nSolution:<\/strong><\/p>\n\n\nDifferentiate both sides w.r.t. x:\n<\/p>\n\n\nd\/dx (xy<\/sup>) + d\/dx (yx<\/sup>) = 0\n<\/p>\n\n\nxy<\/sup>(y’\/ln x + y\/x) + yx<\/sup>(ln y + x\u00b7y’\/y) = 0\n<\/p>\n\n\nCollect y’:\n<\/p>\n\n\ny’ [ xy<\/sup> ln x + yx<\/sup> x\/y ] = \u2212 [ xy<\/sup> y\/x + yx<\/sup> ln y ]\n<\/p>\n\n\n\u2234 dy\/dx = \u2212 ( xy<\/sup> y\/x + yx<\/sup> ln y ) \/ ( xy<\/sup> ln x + yx<\/sup> x\/y )\n<\/p>\n\n<hr>\n\n36. Prove that<\/strong><\/p>\n\n\n| a \u00d7 b |2<\/sup> + (a \u00b7 b)2<\/sup> = a2<\/sup>b2<\/sup>\n<\/p>\n\nSolution:<\/strong><\/p>\n\n\nWe know:\n<\/p>\n\n\n|a \u00d7 b| = ab sin\u03b8 \na \u00b7 b = ab cos\u03b8\n<\/p>\n\n\nSquaring and adding:\n<\/p>\n\n\n|a \u00d7 b|2<\/sup> + (a \u00b7 b)2<\/sup> \n= a2<\/sup>b2<\/sup>(sin\u00b2\u03b8 + cos\u00b2\u03b8)\n<\/p>\n\n\n= a2<\/sup>b2<\/sup>\n<\/p>\n\n\nHence proved.\n<\/p>\n\n<hr>\n\n37. Maximize<\/strong> Z = x + y<\/p>\n\n\nsubject to: \nx \u2212 y \u2264 \u22121 \n\u2212x + y \u2264 0 \nx \u2265 0, y \u2265 0\n<\/p>\n\nSolution:<\/strong><\/p>\n\n\nThe feasible region is bounded by the lines: \nx \u2212 y = \u22121, y = x, x = 0\n<\/p>\n\n\nCorner points are:\n<\/p>\n\n\n(0,1), (0,0), (1,1)\n<\/p>\n\n\nEvaluate Z:\n<\/p>\n\n\nZ(0,1) = 1 \nZ(0,0) = 0 \nZ(1,1) = 2\n<\/p>\n\n\n\u2234 Maximum value of Z is 2<\/strong> at point (1,1)<\/strong>.\n<\/p>\n\n<hr>\n\n38. 10 coins are tossed. Find the probability that exactly 5 heads appear.<\/strong><\/p>\n\nSolution:<\/strong><\/p>\n\n\nTotal number of outcomes = 210<\/sup>\n<\/p>\n\n\nNumber of favourable outcomes = 10<\/sup>C5<\/sub>\n<\/p>\n\n\nProbability = 10<\/sup>C5<\/sub> \/ 210<\/sup>\n<\/p>\n\n\n= 252 \/ 1024 = 63 \/ 256<\/strong>\n<\/p>\n\n<\/body>\n<\/html>\n\n","protected":false},"excerpt":{"rendered":"","protected":false},"author":1,"featured_media":2426,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[196,311,193],"tags":[],"class_list":["post-2413","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-12th","category-answer-key","category-previous-year-papers"],"_links":{"self":[{"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/posts\/2413","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/comments?post=2413"}],"version-history":[{"count":12,"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/posts\/2413\/revisions"}],"predecessor-version":[{"id":2428,"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/posts\/2413\/revisions\/2428"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/media\/2426"}],"wp:attachment":[{"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/media?parent=2413"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/categories?post=2413"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/tags?post=2413"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}

Table of Contents<\/p>\nToggle<\/span>

Section A (MCQ) Bihar Board 12th Math 2026 Question<\/a><\/li>
Section B Bihar Board 12th Math 2026 Question<\/a>
- Short Answer Type Question<\/a><\/li>
- Long Answer Type Question<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n
 <\/span>Section A (MCQ) Bihar Board 12th Math 2026 Question<\/span><\/h2>\n\n\n\n\n\n\n\nBihar Board Class 12 Maths \u2013 Section A (MCQs)<\/title>\n<\/head>\n<body>\n\n1.<\/strong> tan\u22121<\/sup>(\u22121\/\u221a3) =<\/p>\n\n(A) \u03c0\/3 \n(B) \u03c0\/6 \n(C) \u2212\u03c0\/3 \n(D) \u2212\u03c0\/6\n<\/p>\nAnswer:<\/strong> (D) \u2212\u03c0\/6<\/p>\n\n<hr>\n\n2.<\/strong> 2 tan\u22121<\/sup>(1\/3) =<\/p>\n\n(A) tan\u22121<\/sup>(3\/2) \n(B) tan\u22121<\/sup>(3\/4) \n(C) tan\u22121<\/sup>(4\/3) \n(D) tan\u22121<\/sup>(2\/3)\n<\/p>\nAnswer:<\/strong> (D) tan\u22121<\/sup>(2\/3)<\/p>\n\n<hr>\n\n3.<\/strong> x \u2208 R, cot(tan\u22121<\/sup>x + cot\u22121<\/sup>x) =<\/p>\n\n(A) 1 \n(B) 1\/2 \n(C) 0 \n(D) 1\/3\n<\/p>\nAnswer:<\/strong> (A) 1<\/p>\n\n<hr>\n\n4.<\/strong> sin(cos\u22121<\/sup>(3\/5)) =<\/p>\n\n(A) 3\/4 \n(B) 4\/5 \n(C) 3\/5 \n(D) 5\/4\n<\/p>\nAnswer:<\/strong> (B) 4\/5<\/p>\n\n<hr>\n\n5.<\/strong> tan\u22121<\/sup>(1) + cos\u22121<\/sup>(\u22121\/2) + sin\u22121<\/sup>(\u22121\/2) =<\/p>\n\n(A) \u03c0 \n(B) 2\u03c0\/3 \n(C) 3\u03c0\/4 \n(D) \u03c0\/2\n<\/p>\nAnswer:<\/strong> (D) \u03c0\/2<\/p>\n\n<hr>\n\n6.<\/strong> cos\u22121<\/sup>(cos 7\u03c0\/6) =<\/p>\n\n(A) \u03c0\/6 \n(B) \u03c0\/3 \n(C) 5\u03c0\/6 \n(D) 7\u03c0\/6\n<\/p>\nAnswer:<\/strong> (C) 5\u03c0\/6<\/p>\n\n<hr>\n\n7.<\/strong> tan\u22121<\/sup>2 + tan\u22121<\/sup>3 =<\/p>\n\n(A) \u2212\u03c0\/4 \n(B) \u03c0\/4 \n(C) 3\u03c0\/4 \n(D) \u03c0\n<\/p>\nAnswer:<\/strong> (C) 3\u03c0\/4<\/p>\n\n<hr>\n\n8.<\/strong> If |x| \u2264 1, then tan(cos\u22121<\/sup>x) =<\/p>\n\n(A) \u221a(1 \u2212 x\u00b2)\/x \n(B) x\/(1 + x\u00b2) \n(C) \u221a(1 + x\u00b2)\/x \n(D) \u221a(1 \u2212 x\u00b2)\n<\/p>\nAnswer:<\/strong> (A) \u221a(1 \u2212 x\u00b2)\/x<\/p>\n\n<hr>\n\n9.<\/strong> tan\u22121<\/sup>(x\/y) \u2212 tan\u22121<\/sup>(x\/(x + y)) =<\/p>\n\n(A) \u22123\u03c0\/4 \n(B) \u03c0\/2 \n(C) \u03c0\/4 \n(D) \u03c0\/3\n<\/p>\nAnswer:<\/strong> (C) \u03c0\/4<\/p>\n\n<hr>\n\n10.<\/strong> If |x| \u2264 1, cos\u22121<\/sup>((1 \u2212 x\u00b2)\/(1 + x\u00b2)) =<\/p>\n\n(A) 2cos\u22121<\/sup>x \n(B) 2sin\u22121<\/sup>x \n(C) 2tan\u22121<\/sup>x \n(D) tan\u22121<\/sup>2x\n<\/p>\nAnswer:<\/strong> (A) 2cos\u22121<\/sup>x<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<figure class=\"wp-block-table\">\n<table class=\"has-fixed-layout\" style=\"border: 1px solid #000000; width: 100%; border-collapse: collapse;\">\n<tbody>\n<tr>\n<td style=\"border: 1px solid #000000; width: 62.6551%; padding: 8px;\">Join Our Telegram Channel<\/strong><\/span><\/td>\n<td style=\"border: 1px solid #000000; background-color: #24a1de; text-align: center; width: 37.3449%; padding: 8px;\">\n\n<a style=\"color: #ffffff;\" href=\"https:\/\/t.me\/examdhara_school\">Join <\/strong><\/a>\n<\/span>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/figure>\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Bihar Board Class 12 Maths \u2013 Section A (MCQs)<\/title>\n<\/head>\n<body>\n\n11.<\/strong> | 23 12 11 | \n | 36 10 26 | = \n | 63 26 37 |<\/p>\n\n(A) 1 \n(B) \u22121 \n(C) 0 \n(D) 2\n<\/p>\nAnswer:<\/strong> (C) 0<\/p>\n\n<hr>\n\n12.<\/strong> | a\u2212b b\u2212c c\u2212a | \n | b\u2212c c\u2212a a\u2212b | = \n | c\u2212a a\u2212b b\u2212c |<\/p>\n\n(A) 1 \n(B) 0 \n(C) \u22121 \n(D) a + b + c\n<\/p>\nAnswer:<\/strong> (B) 0<\/p>\n\n<hr>\n\n13.<\/strong> | x 4 | = 0 \u21d2 x = \n | 2 2x |<\/p>\n\n(A) \u00b12 \n(B) \u00b11 \n(C) \u00b13 \n(D) 0\n<\/p>\nAnswer:<\/strong> (A) \u00b12<\/p>\n\n<hr>\n\n14.<\/strong> | cos15\u00b0 sin15\u00b0 | \n | sin75\u00b0 cos75\u00b0 | =<\/p>\n\n(A) 1 \n(B) 0 \n(C) \u22121 \n(D) 1\/2\n<\/p>\nAnswer:<\/strong> (B) 0<\/p>\n\n<hr>\n\n15.<\/strong> 1 \u2212 P(A\u2032 \u2229 B\u2032) =<\/p>\n\n(A) P(A \u2229 B) \n(B) P(A \u222a B) \n(C) P(A) \n(D) P(B)\n<\/p>\nAnswer:<\/strong> (B) P(A \u222a B)<\/p>\n\n<hr>\n\n16.<\/strong> If A, B and C are three independent events, then P(A \u2229 B \u2229 C) =<\/p>\n\n(A) P(A) + P(B) + P(C) \n(B) P(A) \u2212 P(B) \u2212 P(C) \n(C) P(A)\u00b7P(B)\u00b7P(C) \n(D) None of these\n<\/p>\nAnswer:<\/strong> (C) P(A)\u00b7P(B)\u00b7P(C)<\/p>\n\n<hr>\n\n17.<\/strong> If P(A)=7\/13, P(B)=9\/13 and P(A \u2229 B)=4\/13, then P(A|B) =<\/p>\n\n(A) 4\/9 \n(B) 4\/7 \n(C) 12\/13 \n(D) 6\n<\/p>\nAnswer:<\/strong> (A) 4\/9<\/p>\n\n<hr>\n\n18.<\/strong> If the odds against the event A is 3 : 7, then P(A) =<\/p>\n\n(A) 3\/10 \n(B) 7\/10 \n(C) 3\/7 \n(D) 7\/3\n<\/p>\nAnswer:<\/strong> (B) 7\/10<\/p>\n\n<hr>\n\n19.<\/strong> d\/dx (log xn<\/sup>) =<\/p>\n\n(A) 1\/xn<\/sup> \n(B) n \n(C) 1\/x \n(D) n\/x\n<\/p>\nAnswer:<\/strong> (D) n\/x<\/p>\n\n<hr>\n\n20.<\/strong> d2<\/sup>\/dx2<\/sup> (sin 2x) =<\/p>\n\n(A) 4 sin 2x \n(B) 4 cos\u00b2 2x \n(C) \u22124 sin 2x \n(D) 2 sin 4x\n<\/p>\nAnswer:<\/strong> (C) \u22124 sin 2x<\/p>\n\n<hr>\n\n21.<\/strong> d\/dx (ex\u2212a<\/sup>) =<\/p>\n\n(A) ex\u2212a<\/sup> \n(B) (x\u2212a)ex\u2212a<\/sup> \n(C) ex<\/sup> \n(D) \u2212ex\u2212a<\/sup>\n<\/p>\nAnswer:<\/strong> (A) ex\u2212a<\/sup><\/p>\n\n<hr>\n\n22.<\/strong> d\/dx \u221a(x\u00b2 + ax + 1) =<\/p>\n\n(A) (x + a) \/ (2\u221a(x\u00b2 + ax + 1)) \n(B) (2x + a) \/ (2\u221a(x\u00b2 + ax + 1)) \n(C) (2x + a) \/ \u221a(x\u00b2 + ax + 1) \n(D) 1 \/ (2\u221a(x\u00b2 + ax + 1))\n<\/p>\nAnswer:<\/strong> (B) (2x + a) \/ (2\u221a(x\u00b2 + ax + 1))<\/p>\n\n<hr>\n\n23.<\/strong> d\/dx (sin x\u00b2) =<\/p>\n\n(A) 2x cos x\u00b2 \n(B) cos x\u00b2 \n(C) x\u00b2 cos x\u00b2 \n(D) x cos x\u00b2\n<\/p>\nAnswer:<\/strong> (A) 2x cos x\u00b2<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Bihar Board Class 12 Maths \u2013 Section A (Q.24\u201337)<\/title>\n<\/head>\n<body>\n\n24.<\/strong> d\/dx (\u221acot x) =<\/p>\n\n(A) 1 \/ (2\u221acot x) \n(B) \u221acosec\u00b2x \n(C) \u2212cosec\u00b2x \/ (2\u221acot x) \n(D) cosec\u00b2x \/ (2\u221acot x)\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n25.<\/strong> d\/dx (tan\u22121<\/sup>\u221ax + cot\u22121<\/sup>\u221ax) =<\/p>\n\n(A) \u03c0\/2 \n(B) 0 \n(C) 1 \n(D) \u03c0\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n26.<\/strong> d\/dx (2 tan\u22121<\/sup> x) =<\/p>\n\n(A) 1 \/ (1 + x\u00b2) \n(B) 2 \/ (1 + x\u00b2) \n(C) 1 \/ (2(1 + x\u00b2)) \n(D) 1 \/ (1 \u2212 x\u00b2)\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n27.<\/strong> d\/dx (cos \u221ax) =<\/p>\n\n(A) sin \u221ax \n(B) \u2212 sin \u221ax \/ (2\u221ax) \n(C) sin \u221ax \/ (2\u221ax) \n(D) 1 \/ (2\u221ax)\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n28.<\/strong> d\/dx { limx\u21920<\/sub> (x5<\/sup> \u2212 a5<\/sup>) \/ (x \u2212 a) } =<\/p>\n\n(A) a \n(B) 0 \n(C) 5a4<\/sup> \n(D) 5\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n29.<\/strong> d\/dx (cot\u22121<\/sup> x) =<\/p>\n\n(A) 1 \/ (1 + x\u00b2) \n(B) \u22121 \/ (1 + x\u00b2) \n(C) 1 \/ x \n(D) \u22121 \/ x\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n30.<\/strong> The equation of the tangent to the curve y = x\u00b2 + 4x + 1 at x = 3 is<\/p>\n\n(A) x + 10y = 8 \n(B) 10x + y = 8 \n(C) 10x \u2212 y = 8 \n(D) x \u2212 10y = 8\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n31.<\/strong> If y = \u221a(sin x + \u221a(sin x + \u221a(sin x + \u2026))), then dy\/dx =<\/p>\n\n(A) 1 \/ (2y \u2212 1) \n(B) cos x \/ (2y \u2212 1) \n(C) sin x \/ (2y \u2212 1) \n(D) (2y \u2212 1) \/ cos x\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n32.<\/strong> If xn<\/sup> + yn<\/sup> = an<\/sup>, then dy\/dx =<\/p>\n\n(A) \u2212 xn\u22121<\/sup> \/ yn\u22121<\/sup> \n(B) xn\u22121<\/sup> \/ yn\u22121<\/sup> \n(C) \u2212 yn\u22121<\/sup> \/ xn\u22121<\/sup> \n(D) nxn\u22121<\/sup>\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n33.<\/strong> If x = a(1 \u2212 cos\u03b8), y = a(\u03b8 + sin\u03b8), then dy\/dx =<\/p>\n\n(A) tan(\u03b8\/2) \n(B) \u2212tan(\u03b8\/2) \n(C) cot(\u03b8\/2) \n(D) \u2212cot(\u03b8\/2)\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n34.<\/strong> If y = xx<\/sup>, then dy\/dx =<\/p>\n\n(A) xx<\/sup>(log x + 1) \n(B) log x \n(C) log x + 1 \n(D) nxn\u22121<\/sup>\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n35.<\/strong> d\/dx [sin\u22121<\/sup>(3x \u2212 4x\u00b3)] =<\/p>\n\n(A) 3 \/ \u221a(1 \u2212 x\u00b2) \n(B) 1 \/ \u221a(1 \u2212 x\u00b2) \n(C) \u22121 \/ \u221a(1 \u2212 x\u00b2) \n(D) 1 \/ (1 \u2212 x\u00b2)\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n36.<\/strong> d\/dx (sin 3x \u00b7 cos 5x) =<\/p>\n\n(A) 4 cos 8x \n(B) 4 cos 8x \u2212 cos 2x \n(C) cos 2x \n(D) cos 2x \u2212 4 cos 8x\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n37.<\/strong> d\/dx (cos x\u00b3) =<\/p>\n\n(A) \u22123x\u00b2 sin x\u00b3 \n(B) sin x\u00b3 \n(C) 3x\u00b2 sin x\u00b3 \n(D) 3x\u00b2\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Bihar Board Class 12 Maths \u2013 Section A (Q.38\u201350)<\/title>\n<\/head>\n<body>\n\n38.<\/strong> If f : R \u2192 R, where f(x) = 3x \u2212 5, then f\u22121<\/sup>(x) =<\/p>\n\n(A) (1\/3)(x + 5) \n(B) (1\/3)(x \u2212 5) \n(C) 3x \u2212 5 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n39.<\/strong> \u222b (x + 2) dx =<\/p>\n\n(A) (x + 2)3<\/sup> + k \n(B) x2<\/sup>\/2 + k \n(C) x2<\/sup>\/2 + 2x + k \n(D) log(x + 2) + k\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n40.<\/strong> \u222b cos 2x \/ (sin x + cos x)2<\/sup> dx =<\/p>\n\n(A) 2 log(sin x + cos x) + k \n(B) log(sin x + cos x) + k \n(C) log(sin x \u2212 cos x) + k \n(D) \u22121\/(sin x + cos x) + k\n<\/p>\nAnswer:<\/strong> (D)<\/p>\n\n<hr>\n\n41.<\/strong> \u222b (1 \u2212 cos 2x) \/ (1 + cos 2x) dx =<\/p>\n\n(A) tan x + x + k \n(B) tan x \u2212 x + k \n(C) x \u2212 tan\u00b2x + k \n(D) tan(x\/2) + k\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n42.<\/strong> \u222b dx \/ (x \u2212 1) =<\/p>\n\n(A) log|x + 1| + k \n(B) \u2212log|x + 1| + k \n(C) log|x \u2212 1| + k \n(D) log x + k\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n43.<\/strong> \u222b log x dx =<\/p>\n\n(A) 1\/x + k \n(B) x log x + k \n(C) x log x \u2212 x + k \n(D) x log x + x + k\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n44.<\/strong> \u222b cos \u221ax dx =<\/p>\n\n(A) sin\u221ax + cos\u221ax + k \n(B) 1\/2 (\u221ax sin\u221ax \u2212 cos\u221ax) + k \n(C) 2(\u221ax sin\u221ax + cos\u221ax) + k \n(D) sin\u221ax + k\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n45.<\/strong> \u222b ex<\/sup>(tan\u22121<\/sup>x + 1\/(1 + x2<\/sup>)) dx =<\/p>\n\n(A) ex<\/sup> tan\u22121<\/sup>x + k \n(B) ex<\/sup> \/ (1 + x2<\/sup>) + k \n(C) ex<\/sup> + k \n(D) tan\u22121<\/sup>x + k\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n46.<\/strong> \u222b dx \/ (x2<\/sup> \u2212 a2<\/sup>) =<\/p>\n\n(A) (1\/a) tan\u22121<\/sup>(x\/a) + k \n(B) (1\/2a) log|(x \u2212 a)\/(x + a)| + k \n(C) (1\/2a) log|(a + x)\/(a \u2212 x)| + k \n(D) (1\/a) log|(x \u2212 a)\/(x + a)| + k\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n47.<\/strong> \u222b sin2<\/sup>(x\/2) dx =<\/p>\n\n(A) 1\/2 x \u2212 1\/2 sin x + k \n(B) 1\/2 x \u2212 1\/2 cos x + k \n(C) 1\/2 sin x + k \n(D) \u22121\/2 sin x + k\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n48.<\/strong> \u222b tan(sin\u22121<\/sup>x) \/ \u221a(1 \u2212 x2<\/sup>) dx =<\/p>\n\n(A) log|sec(sin\u22121<\/sup>x)| + k \n(B) log|cos(sin\u22121<\/sup>x)| + k \n(C) tan(sin\u22121<\/sup>x) + k \n(D) log|sin\u22121<\/sup>x| + k\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n49.<\/strong> \u222b dx \/ (ex<\/sup> + e\u2212x<\/sup>) =<\/p>\n\n(A) cot\u22121<\/sup>(ex<\/sup>) + k \n(B) tan\u22121<\/sup>(ex<\/sup>) + k \n(C) log|ex<\/sup> + 1| + k \n(D) sin\u22121<\/sup>(ex<\/sup>) + k\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n50.<\/strong> \u222b0<\/sub>1<\/sup> dx \/ (1 + x2<\/sup>) =<\/p>\n\n(A) \u03c0\/2 \n(B) \u03c0\/3 \n(C) \u03c0\/4 \n(D) \u03c0\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Bihar Board Class 12 Maths \u2013 Section A (Q.51\u201363)<\/title>\n<\/head>\n<body>\n\n51.<\/strong> \u222b0<\/sub>1<\/sup> x\u00b3 \/ (1 + x\u2078) dx =<\/p>\n\n(A) \u03c0\/2 \n(B) \u03c0\/4 \n(C) \u03c0\/8 \n(D) \u03c0\/16\n<\/p>\nAnswer:<\/strong> (C) \u03c0\/8<\/p>\n\n<hr>\n\n52.<\/strong> \u222b1<\/sub>e<\/sup> (log x)\u00b2 \/ x dx =<\/p>\n\n(A) 1\/3 \n(B) (1\/3)e\u00b3 \n(C) (1\/3)(e\u00b3 \u2212 1) \n(D) e\u00b3\n<\/p>\nAnswer:<\/strong> (B) (1\/3)e\u00b3<\/p>\n\n<hr>\n\n53.<\/strong> If A = [4 2 3], then A\u2032 =<\/p>\n\n(A) [4 2 3] \n(B) [3 2 4] \n(C) [3 2 4] \n(D) [4 2 3]T<\/sup>\n<\/p>\nAnswer:<\/strong> (D)<\/p>\n\n<hr>\n\n54.<\/strong> \u222b0<\/sub>\u03c0\/4<\/sup> etan x<\/sup> \/ cos\u00b2x dx =<\/p>\n\n(A) e \u2212 1 \n(B) e + 1 \n(C) 1\/e + 1 \n(D) 1\/e \u2212 1\n<\/p>\nAnswer:<\/strong> (A) e \u2212 1<\/p>\n\n<hr>\n\n55.<\/strong> \u222b0<\/sub>\u03c0\/2<\/sup> cos\u00b2x dx =<\/p>\n\n(A) \u03c0\/2 \n(B) \u03c0 \n(C) \u03c0\/4 \n(D) 1\n<\/p>\nAnswer:<\/strong> (C) \u03c0\/4<\/p>\n\n<hr>\n\n56.<\/strong> \u222b0<\/sub>\u03c0\/3<\/sup> cos\u00b3x dx =<\/p>\n\n(A) 3\u221a3\/8 \n(B) \u221a3\/8 \n(C) 3\/8 \n(D) 1\/8\n<\/p>\nAnswer:<\/strong> (B) \u221a3\/8<\/p>\n\n<hr>\n\n57.<\/strong> \u222b0<\/sub>2\u03c0<\/sup> |sin x| dx =<\/p>\n\n(A) 2 \n(B) 4 \n(C) 1 \n(D) 3\n<\/p>\nAnswer:<\/strong> (B) 4<\/p>\n\n<hr>\n\n58.<\/strong> \u222b0<\/sub>\u03c0\/2<\/sup> sin x \/ (sin x + cos x) dx =<\/p>\n\n(A) \u03c0 \n(B) \u03c0\/2 \n(C) 0 \n(D) \u03c0\/4\n<\/p>\nAnswer:<\/strong> (D) \u03c0\/4<\/p>\n\n<hr>\n\n59.<\/strong> \u222b\u2212\u03c0<\/sub>\u03c0<\/sup> sin\u2075x dx =<\/p>\n\n(A) 3\u03c0\/4 \n(B) 2\u03c0 \n(C) 5\u03c0\/6 \n(D) 0\n<\/p>\nAnswer:<\/strong> (D) 0<\/p>\n\n<hr>\n\n60.<\/strong> \u222b0<\/sub>a<\/sup> \u221ax \/ \u221a(a \u2212 x + \u221ax) dx =<\/p>\n\n(A) a \n(B) a\/2 \n(C) 2a \n(D) 3a\n<\/p>\nAnswer:<\/strong> (A) a<\/p>\n\n<hr>\n\n61.<\/strong> \u222b0<\/sub>1<\/sup> e\u221ax<\/sup> \/ \u221ax dx =<\/p>\n\n(A) 2(e \u2212 1) \n(B) e \u2212 1 \n(C) 2(e + 1) \n(D) e + 1\n<\/p>\nAnswer:<\/strong> (A) 2(e \u2212 1)<\/p>\n\n<hr>\n\n62.<\/strong> \u222b0<\/sub>1<\/sup> x ex<\/sup> dx =<\/p>\n\n(A) 1 \n(B) 0 \n(C) 2 \n(D) \u22121\n<\/p>\nAnswer:<\/strong> (B) 0<\/p>\n\n<hr>\n\n63.<\/strong> \u222b1<\/sub>4<\/sup> dx \/ \u221ax =<\/p>\n\n(A) 1 \n(B) \u22122 \n(C) 2 \n(D) \u22121\n<\/p>\nAnswer:<\/strong> (C) 2<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Bihar Board Class 12 Maths \u2013 Section A (Q.64\u201372)<\/title>\n<\/head>\n<body>\n\n64.<\/strong> \u222b0<\/sub>a<\/sup> \u221a(a\u00b2 \u2212 x\u00b2) dx =<\/p>\n\n(A) \u03c0\/4 \n(B) a\u00b2\/4 \n(C) \u03c0a\u00b2\/4 \n(D) \u03c0\n<\/p>\nAnswer:<\/strong> (C) \u03c0a\u00b2\/4<\/p>\n\n<hr>\n\n65.<\/strong> \u222b\u22122<\/sub>2<\/sup> |x| dx =<\/p>\n\n(A) 4 \n(B) 3 \n(C) 2 \n(D) 0\n<\/p>\nAnswer:<\/strong> (A) 4<\/p>\n\n<hr>\n\n66.<\/strong> The order and degree of the differential equation \n(d\u00b2s\/dt\u00b2)\u00b2 + (ds\/dt)\u00b3 + 4 = 0 is<\/p>\n\n(A) order = 2, degree = 1 \n(B) order = 2, degree = 2 \n(C) order = 1, degree = 2 \n(D) order = 1, degree = 1\n<\/p>\nAnswer:<\/strong> (B) order = 2, degree = 2<\/p>\n\n<hr>\n\n67.<\/strong> The integrating factor of the differential equation \n(1 + x\u00b2) dy\/dx + y = etan\u22121<\/sup>x<\/sup> is<\/p>\n\n(A) etan\u22121<\/sup>x<\/sup> \n(B) esin\u22121<\/sup>x<\/sup> \n(C) tan\u22121<\/sup>x \n(D) sin\u22121<\/sup>x\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n68.<\/strong> The solution of the differential equation dy\/dx = ex+y<\/sup> is<\/p>\n\n(A) ex<\/sup> + e\u2212y<\/sup> = k \n(B) ex<\/sup> + ey<\/sup> = k \n(C) e\u2212x<\/sup> + ey<\/sup> = k \n(D) e\u2212x<\/sup> + e\u2212y<\/sup> = k\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n69.<\/strong> The solution of the differential equation \nx (dy\/dx) = cot y is<\/p>\n\n(A) x cos y = k \n(B) x tan y = k \n(C) x sec y = k \n(D) x sin y = k\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n70.<\/strong> If the operation * is defined as a * b = a + 2b, then (2 * 3) * 4 is<\/p>\n\n(A) 30 \n(B) 20 \n(C) 16 \n(D) 15\n<\/p>\nAnswer:<\/strong> (B) 20<\/p>\n\n<hr>\n\n71.<\/strong> If A = \n[\n1 \u22121 \n\u22121 1\n], then A\u00b3 =<\/p>\n\n(A) 3A \n(B) 4A \n(C) 2A \n(D) None of these\n<\/p>\nAnswer:<\/strong> (D)<\/p>\n\n<hr>\n\n72.<\/strong> If A = \n[\ncos\u03b1 sin\u03b1 \n\u2212sin\u03b1 cos\u03b1\n] and A + A\u2032 = I, then \u03b1 =<\/p>\n\n(A) \u03c0 \n(B) \u03c0\/3 \n(C) 3\u03c0\/2 \n(D) \u03c0\/6\n<\/p>\nAnswer:<\/strong> (D) \u03c0\/6<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Bihar Board Class 12 Maths \u2013 Section A (Q.73\u201380)<\/title>\n<\/head>\n<body>\n\n73.<\/strong> If A =\n[\n3 \u22125 \n\u22121 2\n], then adj A =<\/p>\n\n(A)\n[\n2 5 \n1 3\n] \n(B)\n[\n2 3 \n1 5\n] \n(C)\n[\n1 3 \n2 5\n] \n(D) None of these\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n74.<\/strong> If A = [1 2 3 4] and B =\n[\n1 \n2 \n3 \n4\n], then AB =<\/p>\n\n(A) [30] \n(B) [10] \n(C) [20] \n(D) [40]\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n75.<\/strong> A = [aij<\/sub>]m\u00d7n<\/sub> is a square matrix if<\/p>\n\n(A) m = n \n(B) m < n \n(C) m > n \n(D) None of these\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n76.<\/strong> If A =\n[\n3 \u22124 \n1 \u22121\n], then A + A\u2032 =<\/p>\n\n(A)\n[\n6 \u22123 \n\u22123 \u22122\n] \n(B)\n[\n6 3 \n3 2\n] \n(C)\n[\n6 3 \n\u22123 \u22122\n] \n(D)\n[\n6 \u22123 \n\u22123 2\n]\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n77.<\/strong> f : A \u2192 B will be an onto function if<\/p>\n\n(A) f(A) \u2282 B \n(B) f(A) = B \n(C) f(A) \u2283 B \n(D) None of these\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n78.<\/strong> What type of relation is \u201cless than\u201d in the set of real numbers?<\/p>\n\n(A) Only symmetric \n(B) Only transitive \n(C) Only reflexive \n(D) Equivalence\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n79.<\/strong> | 3i + 4j + 7k | =<\/p>\n\n(A) \u221a14 \n(B) \u221a74 \n(C) \u221a61 \n(D) \u221a94\n<\/p>\nAnswer:<\/strong> (D)<\/p>\n\n<hr>\n\n80.<\/strong> i \u00b7 ( j \u00d7 k ) =<\/p>\n\n(A) 1 \n(B) 0 \n(C) \u22121 \n(D) i\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Bihar Board Class 12 Maths \u2013 Section A (Q.81\u201389)<\/title>\n<\/head>\n<body>\n\n81.<\/strong> If | a\u20d7 + b\u20d7 | = | a\u20d7 \u2212 b\u20d7 |, then<\/p>\n\n(A) |a\u20d7| = |b\u20d7| \n(B) a\u20d7 \u2225 b\u20d7 \n(C) a\u20d7 \u27c2 b\u20d7 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n82.<\/strong> a\u20d7 \u00b7 (a\u20d7 \u00d7 a\u20d7) =<\/p>\n\n(A) 1 \n(B) 0 \n(C) a\u20d7 \n(D) \u22121\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n83.<\/strong> If 3i\u20d7 + j\u20d7 \u2212 2k\u20d7 and i\u20d7 + \u03bbj\u20d7 \u2212 3k\u20d7 are perpendicular to each other, then the value of \u03bb is<\/p>\n\n(A) \u22123 \n(B) \u22126 \n(C) \u22129 \n(D) \u22121\n<\/p>\nAnswer:<\/strong> (D)<\/p>\n\n<hr>\n\n84.<\/strong> The projection of the vector i\u20d7 + 3j\u20d7 + 7k\u20d7 on the vector 2i\u20d7 \u2212 3j\u20d7 + 6k\u20d7 is<\/p>\n\n(A) 5 \n(B) 25 \n(C) 6 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n85.<\/strong> If |a\u20d7| = 2, |b\u20d7| = 3 and a\u20d7 \u00b7 b\u20d7 = 4, then |a\u20d7 \u2212 b\u20d7| =<\/p>\n\n(A) 5 \n(B) \u221a5 \n(C) 4 \n(D) 2\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n86.<\/strong> If a\u20d7 = i\u20d7 \u2212 j\u20d7 + 2k\u20d7 and b\u20d7 = 2i\u20d7 + 3j\u20d7 \u2212 4k\u20d7, then |a\u20d7 \u00d7 b\u20d7| =<\/p>\n\n(A) \u221a174 \n(B) \u221a87 \n(C) \u221a93 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (B)<\/p>\n\n<hr>\n\n87.<\/strong> If |a\u20d7 + b\u20d7| = |a\u20d7 \u2212 b\u20d7|, then<\/p>\n\n(A) |a\u20d7| = |b\u20d7| \n(B) a\u20d7 \u2225 b\u20d7 \n(C) a\u20d7 \u27c2 b\u20d7 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n88.<\/strong> 2j\u20d7 \u00b7 (\u22123k\u20d7) =<\/p>\n\n(A) 6 \n(B) \u22126 \n(C) 0 \n(D) \u22126i\u20d7\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n89.<\/strong> The direction ratios of a straight line are 2, 6, \u22123. Then its direction cosines are<\/p>\n\n(A) 1\/7, 2\/7, 3\/7 \n(B) 2\/7, \u22126\/7, 3\/7 \n(C) 2\/7, 6\/7, \u22123\/7 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Bihar Board Class 12 Maths \u2013 Section A (Q.90\u201395)<\/title>\n<\/head>\n<body>\n\n90.<\/strong> If a line makes angles \u03b1, \u03b2 and \u03b3 with the positive directions of x, y and z axes respectively, then<\/p>\n\n(A) cos\u00b2\u03b1 + cos\u00b2\u03b2 + cos\u00b2\u03b3 = 1 \n(B) sin\u00b2\u03b1 + sin\u00b2\u03b2 + sin\u00b2\u03b3 = 4 \n(C) cos\u00b2\u03b1 + cos\u00b2\u03b2 + cos\u00b2\u03b3 = 2 \n(D) sin\u00b2\u03b1 + sin\u00b2\u03b2 + sin\u00b2\u03b3 = 1\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n91.<\/strong> The angle between the straight lines \n(x \u2212 2)\/2 = (y \u2212 1)\/7 = (z + 3)\/(\u22123) \nand \n(x + 2)\/(\u22121) = (y \u2212 4)\/2 = (z \u2212 5)\/4 \nis<\/p>\n\n(A) \u03c0\/2 \n(B) 0 \n(C) \u03c0\/6 \n(D) \u03c0\/4\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n92.<\/strong> The distance of the plane x + 2y \u2212 2z = 9 from the point (2, 3, \u22125) is<\/p>\n\n(A) 1 \n(B) 2 \n(C) 3 \n(D) 4\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n93.<\/strong> If two planes x \u2212 4y + \u03bbz + 3 = 0 and 2x + 2y + 3z = 5 are perpendicular to each other, then \u03bb =<\/p>\n\n(A) 1 \n(B) 2 \n(C) 3 \n(D) 4\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n94.<\/strong> If the line \n(x \u2212 x\u2081)\/a\u2081 = (y \u2212 y\u2081)\/b\u2081 = (z \u2212 z\u2081)\/c\u2081 \nis parallel to the plane a\u2082x + b\u2082y + c\u2082z + d = 0, then<\/p>\n\n(A) a\u2081\/a\u2082 = b\u2081\/b\u2082 = c\u2081\/c\u2082 \n(B) a\u2081x + b\u2081y + c\u2081z + d = 0 \n(C) a\u2081a\u2082 + b\u2081b\u2082 + c\u2081c\u2082 = 0 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n95.<\/strong> The equation of the plane whose intercepts on the x, y and z axes are \u22122, 3 and 4 respectively is<\/p>\n\n(A) 6x \u2212 4y \u2212 3z + 12 = 0 \n(B) 6x + 4y + 3z + 12 = 0 \n(C) 6x \u2212 4y \u2212 3z = 0 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Bihar Board Class 12 Maths \u2013 Section A (Q.96\u2013100)<\/title>\n<\/head>\n<body>\n\n96.<\/strong> If n(A) = 4 and n(B) = 2, then n(A \u00d7 B) =<\/p>\n\n(A) 6 \n(B) 8 \n(C) 16 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (B) 8<\/p>\n\n<hr>\n\n97.<\/strong> The solution of the differential equation \nx (dy\/dx) = cot y is<\/p>\n\n(A) x cos y = k \n(B) x tan y = k \n(C) x sec y = k \n(D) x sin y = k\n<\/p>\nAnswer:<\/strong> (C)<\/p>\n\n<hr>\n\n98.<\/strong> | a + ib\u2003\u2003c + id | \n | \u2212c + id\u2003a \u2212 ib | =<\/p>\n\n(A) a\u00b2 + b\u00b2 + c\u00b2 + d\u00b2 \n(B) a\u00b2 \u2212 b\u00b2 \u2212 c\u00b2 \u2212 d\u00b2 \n(C) a\u00b2 \u2212 b\u00b2 + c\u00b2 + d\u00b2 \n(D) a\u00b2 + b\u00b2 + c\u00b2 \u2212 d\u00b2\n<\/p>\nAnswer:<\/strong> (A)<\/p>\n\n<hr>\n\n99.<\/strong> The maximum value of Z = 4x + y subject to the constraints \nx + y \u2264 50, x \u2265 0, y \u2265 0 is<\/p>\n\n(A) 50 \n(B) 250 \n(C) 0 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (B) 250<\/p>\n\n<hr>\n\n100.<\/strong> If n(A) = 4 and n(B) = 2, then n(A \u00d7 B) =<\/p>\n\n(A) 6 \n(B) 8 \n(C) 16 \n(D) None of these\n<\/p>\nAnswer:<\/strong> (B) 8<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<h2 class=\"wp-block-heading\"><\/span>Section B Bihar Board 12th Math 2026 Question<\/span><\/h2>\n\n\n\n<h3 class=\"wp-block-heading\"><\/span>Short Answer Type Question<\/span><\/h3>\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Section B \u2013 Short Answer Solutions<\/title>\n<\/head>\n<body>\n\n1. Prove that<\/strong> \ntan\u22121<\/sup>1 + tan\u22121<\/sup>2 + tan\u22121<\/sup>3 = \u03c0<\/p>\n\nSolution:<\/strong> \nUsing the identity \ntan\u22121<\/sup>a + tan\u22121<\/sup>b = tan\u22121<\/sup>((a+b)\/(1\u2212ab))<\/p>\n\n\ntan\u22121<\/sup>1 + tan\u22121<\/sup>2 = tan\u22121<\/sup>(3 \/ \u22121) = tan\u22121<\/sup>(\u22123)\n<\/p>\n\n\ntan\u22121<\/sup>(\u22123) + tan\u22121<\/sup>3 = \u03c0\n<\/p>\n\nHence proved.<\/p>\n\n<hr>\n\n2. Find the value of<\/strong> \n2tan\u22121<\/sup>(1\/3) + tan\u22121<\/sup>(1\/7)<\/p>\n\nSolution:<\/strong> \n2tan\u22121<\/sup>x = tan\u22121<\/sup>(2x \/ (1 \u2212 x\u00b2))<\/p>\n\n\n2tan\u22121<\/sup>(1\/3) = tan\u22121<\/sup>(3\/4)\n<\/p>\n\n\ntan\u22121<\/sup>(3\/4) + tan\u22121<\/sup>(1\/7) = tan\u22121<\/sup>(1)\n<\/p>\n\n\n= \u03c0\/4\n<\/p>\n\n<hr>\n\n3. Prove that<\/strong> \nsec2<\/sup>(tan\u22121<\/sup>2) + cosec2<\/sup>(cot\u22121<\/sup>3) = 15<\/p>\n\nSolution:<\/strong> \nLet tan\u03b8 = 2 \u21d2 sec2<\/sup>\u03b8 = 1 + tan2<\/sup>\u03b8 = 5<\/p>\n\n\nLet cot\u03c6 = 3 \u21d2 cosec2<\/sup>\u03c6 = 1 + cot2<\/sup>\u03c6 = 10\n<\/p>\n\n\n\u2234 Required sum = 5 + 10 = 15\n<\/p>\n\nHence proved.<\/p>\n\n<hr>\n\n4. Prove that the value of the determinant is zero:<\/strong><\/p>\n\n\n| 102 18 36 | \n| 1 3 4 | \n| 17 3 6 |\n<\/p>\n\nSolution:<\/strong> \nFirst row = 6 \u00d7 Third row<\/p>\n\n\nSince two rows are proportional, the determinant is zero.\n<\/p>\n\nHence proved.<\/p>\n\n<hr>\n\n5. Prove that<\/strong><\/p>\n\n\n| 0 a b | \n| c 0 b | = 2abc \n| c a 0 |\n<\/p>\n\nSolution:<\/strong> \nExpanding along the first row:<\/p>\n\n\n= \u2212a(\u2212bc) + b(ca) = abc + abc = 2abc\n<\/p>\n\nHence proved.<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Section B \u2013 Short Answer Solutions (Q.6\u201311)<\/title>\n<\/head>\n<body>\n\n6. Find the value of x, when<\/strong><\/p>\n\n| 2x 8 | \n| 4 x | = 0\n<\/p>\n\nSolution:<\/strong> \nDeterminant = (2x)(x) \u2212 (8)(4)<\/p>\n\n\n2x\u00b2 \u2212 32 = 0 \nx\u00b2 = 16 \nx = \u00b14\n<\/p>\n\n<hr>\n\n7. Find the slope at the point (1, \u221a2) of the curve<\/strong> \nx\u00b2 + y\u00b2 = 3<\/p>\n\nSolution:<\/strong> \nDifferentiate implicitly:<\/p>\n\n\n2x + 2y (dy\/dx) = 0 \ndy\/dx = \u2212x \/ y\n<\/p>\n\n\nAt (1, \u221a2): \ndy\/dx = \u22121 \/ \u221a2\n<\/p>\n\n<hr>\n\n8. Find dy\/dx when<\/strong> \nx = at\u00b2, y = 2at<\/p>\n\nSolution:<\/strong> \ndx\/dt = 2at \ndy\/dt = 2a<\/p>\n\n\ndy\/dx = (dy\/dt)\/(dx\/dt) = 2a \/ (2at) = 1\/t\n<\/p>\n\n<hr>\n\n9. If<\/strong> y = sec(tan\u221ax), find dy\/dx<\/strong><\/p>\n\nSolution:<\/strong> \nLet u = tan\u221ax<\/p>\n\n\ndy\/dx = sec(u)tan(u) \u00b7 du\/dx\n<\/p>\n\n\ndu\/dx = sec\u00b2\u221ax \u00b7 (1 \/ 2\u221ax)\n<\/p>\n\n\n\u2234 dy\/dx = sec(tan\u221ax) \u00b7 tan(tan\u221ax) \u00b7 sec\u00b2\u221ax \/ (2\u221ax)\n<\/p>\n\n<hr>\n\n10. Find dy\/dx when<\/strong> \nxsin y<\/sup> = ysin x<\/sup><\/p>\n\nSolution:<\/strong> \nTaking log on both sides:<\/p>\n\n\nsin y \u00b7 log x = sin x \u00b7 log y\n<\/p>\n\n\nDifferentiate implicitly:<\/p>\n\n\ncos y \u00b7 dy\/dx \u00b7 log x + sin y \u00b7 (1\/x) \n= cos x \u00b7 log y + sin x \u00b7 (1\/y) \u00b7 dy\/dx\n<\/p>\n\n\nCollecting dy\/dx terms and simplifying gives the required dy\/dx.\n<\/p>\n\n<hr>\n\n11. Find the inverse matrix of<\/strong><\/p>\n\n[ 4 1 ] \n[ 2 3 ]\n<\/p>\n\nSolution:<\/strong> \nDeterminant = (4)(3) \u2212 (1)(2) = 10<\/p>\n\n\nInverse matrix = (1\/10)\n<\/p>\n\n\n[ 3 \u22121 ] \n[ \u22122 4 ]\n<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Section B \u2013 Short Answer Solutions (Q.12\u201316)<\/title>\n<\/head>\n<body>\n\n12. If<\/strong> \nA = \n[\n1 2 \n\u22123 4\n], \nB =\n[\n0 7 \n2 \u22124\n], \nfind the value of (4A \u2212 B).<\/p>\n\nSolution:<\/strong> \n4A =\n[\n4 8 \n\u221212 16\n]<\/p>\n\n\n4A \u2212 B =\n[\n4\u22120 8\u22127 \n\u221212\u22122 16\u2212(\u22124)\n]\n<\/p>\n\n\n=\n[\n4 1 \n\u221214 20\n]\n<\/p>\n\n<hr>\n\n13. Integrate:<\/strong> \n\u222b dx \/ (1 + cos x)<\/p>\n\nSolution:<\/strong> \nMultiply numerator and denominator by (1 \u2212 cos x):<\/p>\n\n\n\u222b (1 \u2212 cos x) \/ (1 \u2212 cos\u00b2x) dx \n= \u222b (1 \u2212 cos x) \/ sin\u00b2x dx\n<\/p>\n\n\n= \u222b (cosec\u00b2x \u2212 cot x cosec x) dx\n<\/p>\n\n\n= \u2212cot x + cosec x + C\n<\/p>\n\n<hr>\n\n14. Integrate:<\/strong> \n\u222b sec\u2074x dx<\/p>\n\nSolution:<\/strong> \nsec\u2074x = sec\u00b2x \u00b7 sec\u00b2x<\/p>\n\n\n\u222b sec\u00b2x (1 + tan\u00b2x) dx\n<\/p>\n\n\n= \u222b (1 + tan\u00b2x) d(tan x)\n<\/p>\n\n\n= tan x + (tan\u00b3x)\/3 + C\n<\/p>\n\n<hr>\n\n15. Integrate:<\/strong> \n\u222b sin\u22121<\/sup>x \/ \u221a(1 \u2212 x\u00b2) dx<\/p>\n\nSolution:<\/strong> \nLet t = sin\u22121<\/sup>x<\/p>\n\n\ndt = dx \/ \u221a(1 \u2212 x\u00b2)\n<\/p>\n\n\n\u2234 Integral = \u222b t dt = t\u00b2\/2 + C\n<\/p>\n\n\n= (sin\u22121<\/sup>x)\u00b2 \/ 2 + C\n<\/p>\n\n<hr>\n\n16. Integrate:<\/strong> \n\u222b dx \/ (x\u00b2 + 6x + 13)<\/p>\n\nSolution:<\/strong> \nx\u00b2 + 6x + 13 = (x + 3)\u00b2 + 4<\/p>\n\n\n\u222b dx \/ [(x + 3)\u00b2 + 2\u00b2]\n<\/p>\n\n\n= (1\/2) tan\u22121<\/sup>((x + 3)\/2) + C\n<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Section B \u2013 Short Answer Solutions (Q.17\u201321)<\/title>\n<\/head>\n<body>\n\n17. Find the value of<\/strong> \n\u222b0<\/sub>a<\/sup> x \/ \u221a(a\u00b2 + x\u00b2) dx<\/p>\n\nSolution:<\/strong> \nLet u = a\u00b2 + x\u00b2 \u21d2 du = 2x dx<\/p>\n\n\n\u222b x \/ \u221a(a\u00b2 + x\u00b2) dx = \u00bd \u222b du \/ \u221au = \u221au\n<\/p>\n\n\n= \u221a(a\u00b2 + x\u00b2)\n<\/p>\n\n\nApplying limits 0 to a:\n<\/p>\n\n\n= \u221a(a\u00b2 + a\u00b2) \u2212 \u221a(a\u00b2) = a(\u221a2 \u2212 1)\n<\/p>\n\n<hr>\n\n18. Solve:<\/strong> \nloge<\/sub>(dy\/dx) = ax + by<\/p>\n\nSolution:<\/strong> \ndy\/dx = eax+by<\/sup><\/p>\n\n\ne\u2212by<\/sup> dy = eax<\/sup> dx\n<\/p>\n\n\n\u222b e\u2212by<\/sup> dy = \u222b eax<\/sup> dx\n<\/p>\n\n\n(\u22121\/b)e\u2212by<\/sup> = (1\/a)eax<\/sup> + C\n<\/p>\n\n\nThis is the required solution.\n<\/p>\n\n<hr>\n\n19. Solve:<\/strong> \n\u222b1<\/sub>2<\/sup> (log x)\/x dx<\/p>\n\nSolution:<\/strong> \nLet t = log x \u21d2 dt = dx\/x\n<\/p>\n\n\nIntegral = \u222b t dt = t\u00b2\/2\n<\/p>\n\n\nApplying limits:\n<\/p>\n\n\n= (log 2)\u00b2\/2 \u2212 (log 1)\u00b2\/2 = (log 2)\u00b2 \/ 2\n<\/p>\n\n<hr>\n\n20. Find the value of<\/strong> \n\u222b0<\/sub>\u03c0\/2<\/sup> \u221acos x \/ (\u221acos x + \u221asin x) dx<\/p>\n\nSolution:<\/strong> \nLet I = \u222b \u221acos x \/ (\u221acos x + \u221asin x) dx<\/p>\n\n\nReplace x by (\u03c0\/2 \u2212 x):\n<\/p>\n\n\nI = \u222b \u221asin x \/ (\u221asin x + \u221acos x) dx\n<\/p>\n\n\nAdding both:\n<\/p>\n\n\n2I = \u222b dx = \u03c0\/2\n<\/p>\n\n\n\u2234 I = \u03c0\/4\n<\/p>\n\n<hr>\n\n21. Solve:<\/strong> \ndy\/dx + 2y tan x = sin x<\/p>\n\nSolution:<\/strong> \nThis is a linear differential equation<\/p>\n\n\nIF = e\u222b2 tan x dx<\/sup> = sec\u00b2x\n<\/p>\n\n\nMultiplying throughout by IF:\n<\/p>\n\n\nd\/dx (y sec\u00b2x) = sin x sec\u00b2x\n<\/p>\n\n\nIntegrating:\n<\/p>\n\n\ny sec\u00b2x = sec x + C\n<\/p>\n\n\ny = cos x + C cos\u00b2x\n<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Section B \u2013 Short Answer Solutions (Q.22\u201326)<\/title>\n<\/head>\n<body>\n\n22. If<\/strong> a<\/b> + b<\/b> + c<\/b> = 0, prove that \na \u00d7 b = b \u00d7 c = c \u00d7 a<\/b>.<\/p>\n\nSolution:<\/strong> \nGiven: a + b + c = 0<\/b> \u21d2 a = \u2212(b + c)<\/b><\/p>\n\n\na \u00d7 b = \u2212(b + c) \u00d7 b \n= \u2212(b \u00d7 b + c \u00d7 b) \n= \u2212(0 + c \u00d7 b) \n= b \u00d7 c\n<\/p>\n\n\nSimilarly, b \u00d7 c = c \u00d7 a\n<\/p>\n\n\n\u2234 a \u00d7 b = b \u00d7 c = c \u00d7 a. \u2714\n<\/p>\n\n<hr>\n\n23. Find the area of the parallelogram whose adjacent sides are<\/strong> \n(i + 2j + 3k)<\/b> and (\u22123i \u2212 2j + k)<\/b>.<\/p>\n\nSolution:<\/strong> \nArea = |a \u00d7 b|<\/p>\n\n\na \u00d7 b =\n| i j k | \n| 1 2 3 | \n| \u22123 \u22122 1 |\n<\/p>\n\n\n= i(2\u00d71 \u2212 3\u00d7(\u22122)) \u2212 j(1\u00d71 \u2212 3\u00d7(\u22123)) + k(1\u00d7(\u22122) \u2212 2\u00d7(\u22123))\n<\/p>\n\n\n= 8i \u2212 10j + 4k\n<\/p>\n\n\n|a \u00d7 b| = \u221a(8\u00b2 + (\u221210)\u00b2 + 4\u00b2) = \u221a180 = 6\u221a5\n<\/p>\n\n\n\u2234 Area = 6\u221a5 square units<\/b>.\n<\/p>\n\n<hr>\n\n24. Find the sine of the angle between the vectors<\/strong> \n3i + j + 2k<\/b> and 2i \u2212 2j + 4k<\/b>.<\/p>\n\nSolution:<\/strong> \nsin\u03b8 = |a \u00d7 b| \/ (|a||b|)<\/p>\n\n\na \u00d7 b =\n| i j k | \n| 3 1 2 | \n| 2 \u22122 4 |\n<\/p>\n\n\n= 8i \u2212 8j \u2212 8k\n<\/p>\n\n\n|a \u00d7 b| = 8\u221a3 \n|a| = \u221a14, |b| = \u221a24 = 2\u221a6\n<\/p>\n\n\nsin\u03b8 = (8\u221a3) \/ (\u221a14 \u00d7 2\u221a6) = 2 \/ \u221a7<\/b>\n<\/p>\n\n<hr>\n\n25. If<\/strong> P(A)=1\/4, P(B)=1\/3 and P(A\u222aB)=1\/2, prove that A and B are independent.<\/p>\n\nSolution:<\/strong> \nP(A\u222aB) = P(A) + P(B) \u2212 P(A\u2229B)<\/p>\n\n\n1\/2 = 1\/4 + 1\/3 \u2212 P(A\u2229B)\n<\/p>\n\n\n\u21d2 P(A\u2229B) = 1\/12\n<\/p>\n\n\nP(A)\u00b7P(B) = (1\/4)(1\/3) = 1\/12\n<\/p>\n\n\n\u2234 P(A\u2229B) = P(A)\u00b7P(B) \nHence, A and B are independent events. \u2714\n<\/p>\n\n<hr>\n\n26. Two dice are thrown. Find the probability that the sum is 8 if the second die always shows 4.<\/strong><\/p>\n\nSolution:<\/strong> \nGiven: second die = 4<\/p>\n\n\nSum = 8 \u21d2 first die must be 4\n<\/p>\n\n\nTotal possible outcomes = 6 \nFavourable outcomes = 1\n<\/p>\n\n\nProbability = 1\/6\n<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Section B \u2013 Short Answer Solutions (Q.27\u201330)<\/title>\n<\/head>\n<body>\n\n27. Maximize<\/strong> Z = 3x + 2y \nsubject to: \n3x + y \u2264 15, \nx \u2265 0, y \u2265 0<\/p>\n\nSolution:<\/strong> \nThe feasible region is bounded by the axes and the line 3x + y = 15.<\/p>\n\n\nCorner points are: \n(0,0), (5,0), (0,15)\n<\/p>\n\n\nEvaluate Z at these points:\n<\/p>\n\n\nZ(0,0) = 0 \nZ(5,0) = 15 \nZ(0,15) = 30\n<\/p>\n\n\n\u2234 Maximum value of Z is 30<\/strong> at point (0,15)<\/strong>.\n<\/p>\n\n<hr>\n\n28. Find the distance between the planes<\/strong> \n2x \u2212 y + 3z + 4 = 0 and 6x \u2212 3y + 9z \u2212 3 = 0<\/p>\n\nSolution:<\/strong> \nDivide the second equation by 3:<\/p>\n\n\n2x \u2212 y + 3z \u2212 1 = 0\n<\/p>\n\n\nDistance between parallel planes is:\n<\/p>\n\n\n|d2<\/sub> \u2212 d1<\/sub>| \/ \u221a(a\u00b2 + b\u00b2 + c\u00b2)\n<\/p>\n\n\n= |\u22121 \u2212 4| \/ \u221a(2\u00b2 + (\u22121)\u00b2 + 3\u00b2)\n<\/p>\n\n\n= 5 \/ \u221a14\n<\/p>\n\n\n\u2234 Required distance = 5\/\u221a14<\/strong>.\n<\/p>\n\n<hr>\n\n29. Find the value of p so that the straight lines<\/strong> \n(x \u2212 1)\/(-1) = (y + 3)\/p = (z \u2212 7)\/2 \nand \n(x + 1)\/p = y\/2 = (z + 6)\/(-3) \nare perpendicular.<\/p>\n\nSolution:<\/strong> \nDirection ratios of first line: (\u22121, p, 2)<\/p>\n\n\nDirection ratios of second line: (p, 2, \u22123)\n<\/p>\n\n\nFor perpendicular lines, dot product = 0:\n<\/p>\n\n\n(\u22121)(p) + (p)(2) + (2)(\u22123) = 0\n<\/p>\n\n\n\u2212p + 2p \u2212 6 = 0 \u21d2 p = 6\n<\/p>\n\n\n\u2234 Value of p = 6<\/strong>.\n<\/p>\n\n<hr>\n\n30. If<\/strong> f : R \u2192 R, f(x) = (3 \u2212 x\u00b3)1\/3<\/sup>, prove that (f \u2218 f)(x) = x.<\/p>\n\nSolution:<\/strong> \nGiven f(x) = (3 \u2212 x\u00b3)1\/3<\/sup><\/p>\n\n\n(f \u2218 f)(x) = f(f(x))\n<\/p>\n\n\n= [3 \u2212 {(3 \u2212 x\u00b3)1\/3<\/sup>}\u00b3]1\/3<\/sup>\n<\/p>\n\n\n= (3 \u2212 (3 \u2212 x\u00b3))1\/3<\/sup>\n<\/p>\n\n\n= (x\u00b3)1\/3<\/sup> = x\n<\/p>\n\n\n\u2234 (f \u2218 f)(x) = x. \u2714\n<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<h3 class=\"wp-block-heading\"><\/span>Long Answer Type Question<\/span><\/h3>\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Section C \u2013 Long Answer Type Questions<\/title>\n<\/head>\n<body>\n\n31. Prove that<\/strong><\/p>\n\n\n| x y z | \n| x\u00b2 y\u00b2 z\u00b2 | = (x \u2212 y)(y \u2212 z)(z \u2212 x)(xy + yz + zx) \n| yz zx xy |\n<\/p>\n\nSolution:<\/strong><\/p>\n\n\nLet \u0394 =\n<\/p>\n\n\n| x y z | \n| x\u00b2 y\u00b2 z\u00b2 | \n| yz zx xy |\n<\/p>\n\n\nApply the operation: \nR2<\/sub> \u2192 R2<\/sub> \u2212 xR1<\/sub> \nR3<\/sub> \u2192 R3<\/sub> \u2212 yzR1<\/sub>\n<\/p>\n\n\nThen the determinant simplifies and common factors (x\u2212y), (y\u2212z), (z\u2212x) appear.\n<\/p>\n\n\nAfter simplification, we get:\n<\/p>\n\n\n\u0394 = (x \u2212 y)(y \u2212 z)(z \u2212 x)(xy + yz + zx)\n<\/p>\n\n\nHence proved.\n<\/p>\n\n<hr>\n\n32. If<\/strong> sin\u22121<\/sup>x + sin\u22121<\/sup>y + sin\u22121<\/sup>z = \u03c0, prove that<\/strong><\/p>\n\n\nx\u221a(1 \u2212 x\u00b2) + y\u221a(1 \u2212 y\u00b2) + z\u221a(1 \u2212 z\u00b2) = 2xyz\n<\/p>\n\nSolution:<\/strong><\/p>\n\n\nLet sin\u22121<\/sup>x = A, sin\u22121<\/sup>y = B, sin\u22121<\/sup>z = C\n<\/p>\n\n\nThen A + B + C = \u03c0\n<\/p>\n\n\nSo, sin(A + B) = sin(\u03c0 \u2212 C) = sin C\n<\/p>\n\n\nUsing identity:\n<\/p>\n\n\nsin(A + B) = sinA cosB + cosA sinB\n<\/p>\n\n\n\u21d2 x\u221a(1 \u2212 y\u00b2) + y\u221a(1 \u2212 x\u00b2) = z\n<\/p>\n\n\nMultiplying both sides appropriately and adding cyclic terms, we get:\n<\/p>\n\n\nx\u221a(1 \u2212 x\u00b2) + y\u221a(1 \u2212 y\u00b2) + z\u221a(1 \u2212 z\u00b2) = 2xyz\n<\/p>\n\n\nHence proved.\n<\/p>\n\n<hr>\n\n33. Solve:<\/strong><\/p>\n\n\n(x\u00b2 \u2212 y\u00b2) dy\/dx = 2xy\n<\/p>\n\nSolution:<\/strong><\/p>\n\n\nRewriting the equation:\n<\/p>\n\n\n(x\u00b2 \u2212 y\u00b2) dy = 2xy dx\n<\/p>\n\n\nSeparate variables:\n<\/p>\n\n\n(x\u00b2 \u2212 y\u00b2)\/y dy = 2x dx\n<\/p>\n\n\n\u21d2 (x\u00b2\/y \u2212 y) dy = 2x dx\n<\/p>\n\n\nIntegrate both sides:\n<\/p>\n\n\n\u222b(x\u00b2\/y \u2212 y) dy = \u222b2x dx\n<\/p>\n\n\n\u21d2 x\u00b2 ln|y| \u2212 y\u00b2\/2 = x\u00b2 + C\n<\/p>\n\n\nThis is the required solution.\n<\/p>\n\n<\/body>\n<\/html>\n\n\n\n\n<!DOCTYPE html>\n<html lang=\"en\">\n<head>\n<meta charset=\"UTF-8\">\n<title>Section C \u2013 Long Answer Solutions (Q.34\u201338)<\/title>\n<\/head>\n<body>\n\n34. Prove that<\/strong><\/p>\n\n\n\u222b0<\/sub>\u03c0\/2<\/sup> ( \u221atan x + \u221acot x ) dx = \u03c0\u221a2\n<\/p>\n\nSolution:<\/strong><\/p>\n\n\nLet I = \u222b0<\/sub>\u03c0\/2<\/sup> ( \u221atan x + \u221acot x ) dx\n<\/p>\n\n\nUsing the property: \n\u222b0<\/sub>\u03c0\/2<\/sup> f(x) dx = \u222b0<\/sub>\u03c0\/2<\/sup> f(\u03c0\/2 \u2212 x) dx\n<\/p>\n\n\n\u21d2 I = \u222b0<\/sub>\u03c0\/2<\/sup> ( \u221acot x + \u221atan x ) dx\n<\/p>\n\n\nAdding both expressions:\n<\/p>\n\n\n2I = 2 \u222b0<\/sub>\u03c0\/2<\/sup> \u221atan x dx\n<\/p>\n\n\n\u21d2 I = \u222b0<\/sub>\u03c0\/2<\/sup> \u221atan x dx\n<\/p>\n\n\nUsing standard result:\n<\/p>\n\n\n\u222b0<\/sub>\u03c0\/2<\/sup> \u221atan x dx = \u03c0\/\u221a2\n<\/p>\n\n\n\u2234 I = \u03c0\u221a2\n<\/p>\n\n\nHence proved.\n<\/p>\n\n<hr>\n\n35. Find<\/strong> dy\/dx, when<\/strong> xy<\/sup> + yx<\/sup> = 1.<\/p>\n\nSolution:<\/strong><\/p>\n\n\nDifferentiate both sides w.r.t. x:\n<\/p>\n\n\nd\/dx (xy<\/sup>) + d\/dx (yx<\/sup>) = 0\n<\/p>\n\n\nxy<\/sup>(y’\/ln x + y\/x) + yx<\/sup>(ln y + x\u00b7y’\/y) = 0\n<\/p>\n\n\nCollect y’:\n<\/p>\n\n\ny’ [ xy<\/sup> ln x + yx<\/sup> x\/y ] = \u2212 [ xy<\/sup> y\/x + yx<\/sup> ln y ]\n<\/p>\n\n\n\u2234 dy\/dx = \u2212 ( xy<\/sup> y\/x + yx<\/sup> ln y ) \/ ( xy<\/sup> ln x + yx<\/sup> x\/y )\n<\/p>\n\n<hr>\n\n36. Prove that<\/strong><\/p>\n\n\n| a \u00d7 b |2<\/sup> + (a \u00b7 b)2<\/sup> = a2<\/sup>b2<\/sup>\n<\/p>\n\nSolution:<\/strong><\/p>\n\n\nWe know:\n<\/p>\n\n\n|a \u00d7 b| = ab sin\u03b8 \na \u00b7 b = ab cos\u03b8\n<\/p>\n\n\nSquaring and adding:\n<\/p>\n\n\n|a \u00d7 b|2<\/sup> + (a \u00b7 b)2<\/sup> \n= a2<\/sup>b2<\/sup>(sin\u00b2\u03b8 + cos\u00b2\u03b8)\n<\/p>\n\n\n= a2<\/sup>b2<\/sup>\n<\/p>\n\n\nHence proved.\n<\/p>\n\n<hr>\n\n37. Maximize<\/strong> Z = x + y<\/p>\n\n\nsubject to: \nx \u2212 y \u2264 \u22121 \n\u2212x + y \u2264 0 \nx \u2265 0, y \u2265 0\n<\/p>\n\nSolution:<\/strong><\/p>\n\n\nThe feasible region is bounded by the lines: \nx \u2212 y = \u22121, y = x, x = 0\n<\/p>\n\n\nCorner points are:\n<\/p>\n\n\n(0,1), (0,0), (1,1)\n<\/p>\n\n\nEvaluate Z:\n<\/p>\n\n\nZ(0,1) = 1 \nZ(0,0) = 0 \nZ(1,1) = 2\n<\/p>\n\n\n\u2234 Maximum value of Z is 2<\/strong> at point (1,1)<\/strong>.\n<\/p>\n\n<hr>\n\n38. 10 coins are tossed. Find the probability that exactly 5 heads appear.<\/strong><\/p>\n\nSolution:<\/strong><\/p>\n\n\nTotal number of outcomes = 210<\/sup>\n<\/p>\n\n\nNumber of favourable outcomes = 10<\/sup>C5<\/sub>\n<\/p>\n\n\nProbability = 10<\/sup>C5<\/sub> \/ 210<\/sup>\n<\/p>\n\n\n= 252 \/ 1024 = 63 \/ 256<\/strong>\n<\/p>\n\n<\/body>\n<\/html>\n\n","protected":false},"excerpt":{"rendered":"","protected":false},"author":1,"featured_media":2426,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[196,311,193],"tags":[],"class_list":["post-2413","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-12th","category-answer-key","category-previous-year-papers"],"_links":{"self":[{"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/posts\/2413","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/comments?post=2413"}],"version-history":[{"count":12,"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/posts\/2413\/revisions"}],"predecessor-version":[{"id":2428,"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/posts\/2413\/revisions\/2428"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/media\/2426"}],"wp:attachment":[{"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/media?parent=2413"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/categories?post=2413"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/examdhara.com\/school\/wp-json\/wp\/v2\/tags?post=2413"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}