If sec-1 (?1 + x2) + cosec-1 (?1 + y2) / y + cot-1 (1 / z) = 3?, then x + y + z =
1) xyz 2) 2xyz 3) xyz2 4) x2yz Solution: (1) xyz sec-1 (√1 + x2) + cosec-1 (√1 + y2)... View Article
1) xyz 2) 2xyz 3) xyz2 4) x2yz Solution: (1) xyz sec-1 (√1 + x2) + cosec-1 (√1 + y2)... View Article
1) π/6 2) π/4 3) π/3 4) π/2 Solution: (3) π / 3 cos-1 x = ɑ, (0 < x... View Article
1) 23 / 25 2) 25 / 23 3) 23 / 24 4) 24 / 23 Solution: (2) 25 /... View Article
1) π / 4 2) – (π / 4) 3) π / 2 4) – (π / 2) Solution: (4)... View Article
1) 0 2) 2 3) 1 4) – 1 5) 2√2 Solution: (4) – 1 (tan-1 x)2 + (cot-1 x)2... View Article
1) 3 2) -√3 3) √2 4) 2 5) √3 Solution: (5) √3 5 cos-1 [1 – x2] / [1... View Article
1) 2√126 / 65 2) 4√65 / 65 3) 8√63 / 65 4) √63 / 65 Solution: (1) 2√126 /... View Article
1) 7/3 2) 3 3) 11/3 4) 13/3 Solution: (4) 13 / 3 tan-1 (1 / 3) + tan-1 (3/4)... View Article
1) – (1/√3) 2) 1/√3 3) -√3 4) √3 Solution: (4) √3 tan-1 x + 2 cot-1 x = 2π... View Article
1) α > β 2) α = β 3) α < β 4) α + β = 2π Solution: (3)... View Article
1) 1 / √1 + x2 2) x / √1 + x2 3) – 1 / √1 + x2 4)... View Article
1) 3 2) 5 3) 7 4) 11 Solution: (2) 5 sin-1 (3 / x) + sin-1 (4 / x)... View Article
1) 5/17 2) 6/17 3) 3/17 4) 4/17 Solution: (2) 6 / 17 Let (cosec-1 (5 / 3) = θ,... View Article
1) x / √1 + x2 2) x 3) x √1 + x2 4) √1 + x2 Solution: (3) x... View Article
(1) F’[f{φ(x)}] (2) F’[f{φ(x)}] f’{φ(x)} (3) F’[f{φ(x)}] f’{φ(x)} φ’(x) (4) none of these Solution: Let y = F[f{φ(x)}] Differentiate w.r.t.x... View Article
(1) x log x (2) x/log x (3) 1/x log x (4) log x/x Solution: Given f(x) = log x... View Article
(1) 1/2 (2) 2/3 (3) 2 (4) none of these Solution: Given fog = I(identity function) So f(g(x)) = x... View Article
1) 1/(1 + {f(x)}3) 2) (1 + {f(x)}3) 3) 1/(1 + {g(x)}3) 4) 1 + {g(x)}3 Solution: Given g(x) is... View Article
(1) f’(c) (2) 1/f’(c) (3) f(c) (4) None of these Solution: gof(x)) = I(x) (I(x) is the identity function since... View Article
(1) -(3x2 + 8y)/(8x + 3y2) (2) (3x2 + 8y)/(8x + 3y2) (3) (3x + 8y)/(8x2 + 3y) (4) none... View Article