The differential coefficient of [latex]tan^{-1}left ( frac{2x}{1-x^{2}} right )[/latex] w.r.t [latex]sin^{-1}left ( frac{2x}{1+x^{2}} right )[/latex] is (1) 1 (2) -1 (3) 0 (4) none of these Solution: Let u = tan-1 2x/(1 - x2) Put x = tan θ So u = tan-1 (2 tan θ/(1 - tan2 θ)) =... View Article
Differential coefficient of sin-1 x w.r.t.cos-1 √(1 – x2) is (1) 1 (2) 2 (3) 1/(1 + x2) (4) none of these Solution: Let u = sin-1 x Differentiate w.r.t.x du/dx = 1/√(1 - x2) Let v = cos-1... View Article
Differential coefficient of sin-1 (1 – x)/(1 + x) w.r.t. √x is (1) 1/2√x (2) √x/√(1-x) (3) 1 (4) -2/(1+x) Solution: Let u = sin-1(1 - x)/(1 + x) v = √x Differentiate w.r.t.x du/dx = [1/√(1 -... View Article
[latex]\frac{d}{dx}\left ( \tan^{-1}\frac{\sqrt{1+x^{2}}-1}{x} \right )[/latex] (1) 1/(1 + x2) (2) 1/2(1 + x2) (3) -2/(1 + x2) (4) 2/(1 + x2) Solution: Let y = (tan-1 (√(1 + x2) - 1)/x) Put x = tan θ So y =... View Article
If [latex]y=\sin^{-1}\left ( \frac{1-x^{2}}{{1+x^{2}}} \right )[/latex], then dy/dx equals (1) 2/(1 - x2) (2) 1/(1 + x2) (3) -2/(1 + x2) (4) 2/(1 - x2) Solution: Given y = sin-1(1-x2)/(1+x2) Put x = tan θ θ = tan-1x So... View Article
If [latex]y=\tan^{-1}\left ( \frac{x}{\sqrt{1-x^{2}}} \right )[/latex], then dy/dx = (1) -1/√(1 - x2) (2) x/√(1 - x2) (3) 1/√(1 - x2) (4) √(1 - x2)/x Solution: Given y = tan-1(x/√(1 - x2)) Put x = sin θ Then y =... View Article
Differential coefficient of cos-1 √x with respect to √(1 – x) is (1) √x (2) -√x (3) 1/√x (4) -1/√x Solution: Let u = cos-1√x v = √(1 - x) Differentiate w.r.t.x du/dx = [-1/√(1 - x)] × 1/2√x =... View Article
If [latex]f(x)=\cot^{-1}\left ( \frac{x^{x}-x^{-x}}{2} \right )[/latex], then f’(1) is equal to (1) -1 (2) 1 (3) log 2 (4) -log 2 Solution: Given f(x) = cot-1 (xx - x-x)/2 = cot-1 (xx - 1/xx)/2 = cot-1 ((xx)2 - 1)/2xx) =... View Article
If f(x) = x + 2, then f’(f(x)) at x = 4 is (1) 8 (2) 1 (3) 4 (4) 5 Solution: Given f(x) = x + 2 f(f(x)) = x + 2 + 2 = x + 4 Differentiate w.r.t.x f’(f(x)) = 1 f’(f(x))... View Article
If y2 =ax2 + bx + c, then y3 d2y/dx2 is (1) a constant (2) a function of x only (3) a function of y only (4) a function of x and y Solution: Given y2 = ax2 + bx + c... View Article
If [latex]\Delta _{1}=\begin{vmatrix} x & a& b\\ b & x & a\\ a& b & x \end{vmatrix}[/latex] and [latex]\Delta _{2}=\begin{vmatrix} x & b\\ a& x \end{vmatrix}[/latex] are the given determinants, then (1) ∆1 = 3(∆2)2 (2) d/dx ∆1 = 3(∆2) (3) d/dx ∆1 = 3(∆2)2 (4) ∆1 = 3(∆2)3/2 Solution: Given and ∆1 = x(x2 -ab) - a(xb - a2)... View Article
If y = sin-1 √(1 – x2), then dy/dx = (1) 1/√(1 - x2) (2) -1/√(1 - x2) (3) 1/√(1 + x2) (4) -1/√(x2 -1) Solution: Given y = sin-1 √(1 - x2) Put x = cos θ θ = cos-1 x... View Article
If [latex]y=\tan^{-1}\sqrt{\frac{a-x}{a+x}}[/latex], then dy/dx = (1) cos-1(x/a) (2) -cos-1(x/a) (3) ½ cos-1(x/a) (4) none of these Solution: Given y = tan-1 √(a - x)/(a + x) Put x = a cos 2θ θ... View Article
[latex]\frac{d}{dx}\sin^{-1}\left ( 2ax\sqrt{1-a^{2}x^{2}} \right )=[/latex] (1) 2a/√(a2 - x2) (2) a/√(a2 - x2) (3) 2a/√(1 - a2x2) (4) a/√(1 - a2x2) Solution: Let y = sin-1 (2ax √(1 - a2x2)) Put ax = sin θ... View Article
If √(1 – x2) + √(1 – y2) = a(x – y), then dy/dx = (1) √(1 - x2)/√(1 - y2) (2) √(1 - y2)/√(1 - x2) (3) √(x2 - 1)/√(1 - y2) (4) none of these Solution: Given √(1 - x2) + √(1 - y2) =... View Article
The solution of the equation sin-1 (tan ? / 4) – sin-1 (?3 / x) – (x / 6) = 0 is given by x = 1) 1 2) 2 3) 3 4) 4 Solution: (4) 4 The given equation is sin-1 (tan Ï€ / 4) - sin-1 (√3 / x) - (x / 6) = 0 It becomes sin-1 1 - sin-1 √3... View Article
If tan-1 1 / [1 + 2] + tan-1 1 / [1 + 2 – 3] + tan-1 1 / [1 + 3 – 4] + …. + tan-1 1 / [1 + n (n + 1)] = tan-1 ?, then ? = 1) n 2) n / (n + 1) 3) n / (n + 2) 4) n + 1 Solution: (3) n / (n + 2) LHS = tan-1 1 / [1 + 2] + tan-1 1 / [1 + 2 - 3] + tan-1 1 /... View Article
If ?C = 90o in triangle ABC, then tan-1 b / (b + c) + tan-1 b / (c + a) = 1) π / 2 2) π / 4 3) π / 3 4) π Solution: (2) π / 4 tan-1 b / (b + c) + tan-1 b / (c + a) = tan-1 [a / (b + c)] + [b / (c + a)] /... View Article
tan [(1 / 2) sin-1 2x / [1 + x2] + (1 / 2) cos-1 [1 – y2] / [1 + y2] where (xy ? 1) = 1) x + y 2) [x + y] / [1 - xy] 3) [x + y] / [1 + xy] 4) [x - y] / [1 + xy] Solution: (2) [x + y] / [1 - xy] tan [(1 / 2) sin-1 2x... View Article
tan-1 x + tan-1 y + tan-1 [1 – x – y + xy] / [1 + x + y – xy] = 1) Ï€ / 2 2) Ï€ / 4 3) Ï€ 4) Ï€ / 3 Solution: (2) Ï€ / 4 LHS = tan-1 x + tan-1 y + tan-1 [1 - x - y + xy] / [1 + x + y - xy] = tan-1... View Article
