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One of the values of x satisfying tan (sec^-1 x) – sin cos^-1 (1 / ?5) is

1) √5 / 3 2) 3 / √5 3) 2 / √5 4) √5 / 2 Solution: (2) 3 / √5 tan (sec-1 x) = √sec2 (sec-1 x - 1) = sin (sin-1 [2 / √5]) cos-1 1 /... View Article

If 2 tan^-1 (cos ?) = tan^-1 (2 cosec ?), then ? =

1) nπ - (π / 4), n ∈ I 2) nπ + (π / 4), n ∈ I 3) 2nπ - (π / 4), n ∈ I 4) 2nπ + (π / 4), n ∈ I Solution: (2) nπ + (π / 4), n ∈ I 2... View Article

tan^-1 5 + tan^-1 3 – cot^-1 (4 / 7) =

1) π / 3 2) π / 4 3) π / 2 4) π / 6 Solution: (3) π / 2 LHS = tan-1 [5 + 3] / [1 - 5 * 3] - cot-1 (4 / 7) = tan-1 (8 - 14) -... View Article

sin^-1 (- ?2 / 2) + cos^-1 (- 1 / 2) – tan^-1 (- ?3) + cot^-1 (- 1 / ?3) =

1) 7π / 12 2) 17π / 12 3) 5π / 12 4) π / 12 Solution: (2) 17π / 12 sin-1 (- √2 / 2) + cos-1 (- 1 / 2) - tan-1 (- √3) + cot-1 (- 1... View Article

The value of sin^-1 {cot (sin^-1 ?(2 – ?3) / 4 + cos^-1 (?12 / 4) + sec^-1 ?2)} =

1) 1 2) 0 3) -1 4) 2 Solution: (2) 0 √(2 - √3) / 4 = √4 - 2√3) / 8 = √(√3 - 1)2 / 8 = [√3 - 1] / 2√2 sin-1 √(2 - √3) / 4 = sin-1... View Article

If 1 < x < ?2, the number of solutions of the equation tan^-1 (x – 1) + tan^-1 x + tan^-1 (x + 1) = tan^-1 3x is

1) 0 2) 1 3) 2 4) 3 Solution: (3) 2 tan-1 (x - 1) + tan-1 x + tan-1 (x + 1) = tan-1 3x LHS = tan-1 (x - 1) + tan-1 (x + 1) = tan-1 3x -... View Article

The inequality sin^-1 (sin 5) > x² – 4x holds if

1) x = 2 - √9 - 2π 2) x = 2 + √9 - 2π 3) x ∈ (2 - √9 - 2π, 2 + √9 - 2π) 4) x > 2 + √9 - 2π Solution: (3) x ∈ (2 - √9 - 2π, 2 + √9... View Article

tan^-1 1 / (2x + 1) + tan^-1 1 / (4x + 1) = tan^-1 2 / x² has

1) one solution 2) two solutions 3) three solutions 4) no solution Solution: (2) two solutions tan-1 1 / (2x + 1) + tan-1 1 / (4x... View Article

tan [(π / 4) + (1 / 2) cos^-1 x) + tan [(π / 4) – (1 / 2) cos^-1 x)] = 1 has

1) one solution 2) two solutions 3) three solutions 4) no solution Solution: (4) no solution Put (1 / 2) cos-1 x = ɑ tan [(π / 4)... View Article

tan^-1 [x + (2 / x)] – tan^-1 (4 / x) – tan^-1 [x – (2 / x)] = 0 has real roots

1) 1 2) 2 3) 3 4) 4 Solution: (2) 2 tan-1 [x + (2 / x)] - tan-1 (4 / x) - tan-1 [x - (2 / x)] = 0 tan-1 [x + (2 / x)] - tan-1 [x... View Article

y = a^x , then yⁿ= ..; a> 0, x∈R

(1) anx (log a)n (2) nax log a (3) (ax)n log a (4) ax (log a)n Solution: Given y = ax Differentiate w.r.t.x dy/dx = ax log a... View Article

dⁿ/dxⁿ (e^2x + e^-2x) =

(1) (e2x + (-1)n e-2x) (2) 2n (e2x + e-2x) (3) 2n (e2x + (-1)n e-2x) (4) none of these Solution: Let y = (e2x + e-2x)... View Article

If x = log p and y = 1/p, then

(1) d2y/dx2 - 2p = 0 (2) d2y/dx2 + y = 0 (3) d2y/dx2 + dy/dx = 0 (4) d2y/dx2 - dy/dx = 0 Solution: Given x = log p y = 1/p So x... View Article

If y = A cos nx + B sin nx, then d²y/dx² =

(1) n2y (2) -y (3) -n2y (4) none of these Solution: Given y =A cos nx + B sin nx Differentiate w.r.t.x dy/dx = -An sin nx + Bn... View Article

The second derivative of a sin³ t with respect to a cos³ t at t = π/4 is

(1) 4√2/3a (2) 2 (3) 1/12a (4) none of these Solution: Let u = a sin3 t du/dt = 3a sin2 t cos t v = a cos3 t dv/dt = -3a cos2 t... View Article

The derivative of [latex]e^{x^{3}}[/latex] with respect to log x is

(1) (2) (3) (4) Solution: Let v = log x dv/dx = 1/x du/dv = (du/dx)/(dv/dx) Hence option (1) is the answer.... View Article

The derivative of [latex]cos^{-1}left ( frac{1-x^{2}}{1+x^{2}} right )[/latex] w.r.t. [latex]cot^{-1}left ( frac{1-3x^{2}}{3x-x^{3}} right )[/latex] is

(1) 1 (2) 3/2 (3) 2/3 (4) 1/2 Solution: Let u = cos-1 (1-x2)/(1+x2) Put x = tan θ So u = cos-1 ((1 - tan2 θ)/(1 + tan2 θ)) =... View Article

The derivative of sin²x with respect to cos² x is

(1) tan x (2) tan2 x (3) -tan x (4) none of these Solution: Let u = sin2x Differentiate w.r.t.x du/dx = 2 sin x cos x v = cos2... View Article

The derivative of [latex]sin^{-1}left ( frac{2x}{1+x^{2}} right )[/latex] w.r.t [latex]cos^{-1}left ( frac{1-x^{2}}{1+x^{2}} right )[/latex] is

(1) 1 (2) 2 (3) -1 (4) 4 Solution: Let u = sin-1 2x/(1 + x2) v = cos-1 ((1 - x2)/(1 + x2)) Put x = tan θ So u = sin-1 (2 tan θ/(1 + tan2... View Article

If [latex]u=\tan^{-1}\left ( \frac{\sqrt{1+x^{2}}-1}{x} \right )[/latex] and [latex]v=\tan^{-1}x[/latex], then du/dv is equal to

(1) 4 (2) 1 (3) 1/4 (4) -1/4 Solution: Given u = tan-1 (√(1+x2)-1)/x Put x = tan θ u = tan-1 (√(1 + tan2 θ) - 1)/tan θ = tan-1... View Article

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