Latest Posts

sin^-1 [x ?1 – x – ?x ?1 – x²] =

1) sin-1 x + sin-1 √x 2) sin-1 x – sin-1 √x 3) sin-1 √x – sin-1 x 4) None... View Article

If cos^-1 x = ?, (0 < x < 1) and sin^-1 (2x ?1 – x²) + sec^-1 1 / (2x² – 1) = 2? / 3, then tan^-1 2x =

1) π / 6 2) π / 4 3) π / 3 4) π / 2 Solution: (3) π / 3 cos-1 x... View Article

The function f(x) = x/2 + 2/x has a local minimum at

(1) x = 2 (2) x = -2 (3) x = 0 (4) x = -1 Solution: f(x) = x/2... View Article

If u = tan^-1 (y/x), then by Euler’s theorem the value of x ∂u/∂x + y ∂u/∂y =

(1) sin u (2) tan u (3) 0 (4) cos u Solution: Given u = tan-1 (y/x) = x0f(y/x) Here... View Article

[latex]\frac{d}{dx}\left ( e^{\sqrt{1-x^{2}}}\tan x \right )=[/latex]

(1) \(\begin{array}{l}e^{\sqrt{1-x^{2}}} (\sec ^{2}x- \frac{x\tan x}{\sqrt{1-x^{2}}})\end{array} \) (2) \(\begin{array}{l}e^{\sqrt{1-x^{2}}} (\sec ^{2}x+ \frac{x\tan x}{\sqrt{1-x^{2}}})\end{array} \) (3) \(\begin{array}{l}e^{\sqrt{1-x^{2}}} (\sec ^{2}x+ \frac{\tan x}{\sqrt{1-x^{2}}})\end{array}... View Article

If u = (x+y)/(x-y), then ?u/?x + ?u/?y =

(1) 1/(x-y) (2) 2/(x-y) (3) 1/(x-y)2 (4) 2/(x-y)2 Solution: u = (x+y)/(x-y) ∂u/∂x = [(x-y) – (x+y)]/(x-y)2 = -2y/(x-y)2 ∂u/∂y... View Article

If y = x + e^x, then d²y/dx² is

(1) ex (2) -ex/(1 + ex)3 (3) -ex/(1 + ex)2 (4) -1/(1 + ex)3 Solution: Given y = x +... View Article

Let f(x) = tan^-1 x, then f’(x) + f’’(x) is equal to 0, where x is equal to

(1) 0 (2) 1 (3) i (4) -i Solution: Given f(x) = tan–1 x Differentiate w.r.t.x f’(x) = 1/(1+x2) Again... View Article

nth derivative of cos² x =

(1) 2n-1 cos (nπ/2 + 2x) (2) 2n cos (nπ/2 – 2x) (3) 2n+1 cos (nπ/2 + 2x) (4) 2n+2... View Article

sec² (tan^-1 2) + cosec² (cot^-1 3) =

1) 5 2) 13 3) 15 4) 6 Solution: (3) 15 sec2 (tan-1 2) + cosec2 (cot-1 3) = [sec... View Article

If y = (sin^-1x)², then (1 – x²)d²y/dx² – xdy/dx is equal to

(1) 1 (2) 0 (3) -1 (4) 2 Solution: Given y = (sin-1x)2 Differentiate w.r.t.x dy/dx = 2 sin-1x (1/√(1... View Article

[latex]f(x) =\begin{vmatrix} x^{3} & x^{2}&3x^{2} \\ 1 & -6 & 4\\ p&p^{2} & p^{3} \end{vmatrix}[/latex], where p is a constant, then [latex]\frac{d^{3}f(x)}{dx^{3}}[/latex] is

(1) proportional to x (2) proportional to x2 (3) proportional to x3 (4) a constant Solution: Given \(\begin{array}{l}f(x) =\begin{vmatrix} x^{3}... View Article

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Latest Posts

sin^-1 [x ?1 – x – ?x ?1 – x²] =

If cos^-1 x = ?, (0 < x < 1) and sin^-1 (2x ?1 – x²) + sec^-1 1 / (2x² – 1) = 2? / 3, then tan^-1 2x =

The function f(x) = x/2 + 2/x has a local minimum at

If u = tan^-1 (y/x), then by Euler’s theorem the value of x ∂u/∂x + y ∂u/∂y =

[latex]\frac{d}{dx}\left ( e^{\sqrt{1-x^{2}}}\tan x \right )=[/latex]

If u = (x+y)/(x-y), then ?u/?x + ?u/?y =

If y = x + e^x, then d²y/dx² is

Let f(x) = tan^-1 x, then f’(x) + f’’(x) is equal to 0, where x is equal to

nth derivative of cos² x =

sec² (tan^-1 2) + cosec² (cot^-1 3) =

If y = (sin^-1x)², then (1 – x²)d²y/dx² – xdy/dx is equal to

[latex]f(x) =\begin{vmatrix} x^{3} & x^{2}&3x^{2} \\ 1 & -6 & 4\\ p&p^{2} & p^{3} \end{vmatrix}[/latex], where p is a constant, then [latex]\frac{d^{3}f(x)}{dx^{3}}[/latex] is

d²/dx² (2 cos x cos 3x) =

The nth derivative of xe^x vanishes when

If u = x² + y² and x = s + 3t, y = 2s – t , then d²u/ds² =

If y = a cos (log x) + b sin (log x) where a, b are parameters then x²y’’ + xy’ =

If y = ae^x+ be^-x where a, b are parameters then y’’ =

If y = (x + √(1 + x²))ⁿ, then (1 + x²)d²y/dx² + x dy/dx is

If y = x³ log log_e (1+x), then y’’(0) equals

If y = sin x + e^x , then d²x/dy²