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Derivative of sec^-1 (1/1-2x²) with respect to sin^-1 (3x – 4x³) is

(1) 1/4 (2) 3/2 (3) 2/3 (4) 1 Solution: Let u = sec-1(1/1-2x2) Put x = sin θ θ = sin-1x Then u = sec-1(1/1-2 sin2 θ) =... View Article

If [latex]y=\tan^{-1}\left ( \frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}} \right )[/latex] , then dy/dx is equal to

(1) x2/√(1-x4) (2) x2/√(1+x4) (3) x/√(1+x4) (4) x/√(1-x4) Solution: Given y = tan-1 (√(1+x2) - √(1-x2) )/(√(1+x2) + √(1-x2))) Put... View Article

If x = cos θ, y = sin 5θ, then (1-x²)d²y/dx² – xdy/dx is equal to

(1) -5y (2) 5y (3) 25y (4) -25y Solution: Given x = cos θ y = sin 5θ dx/dθ = -sin θ dy/dθ = 5 cos 5θ dy/dx = (dy/dθ)/(dx/dθ)... View Article

d²x/dy² is equal to

(1) (d2y/dx2)-1 (2) -(d2y/dx2)-1(dy/dx)-3 (3) -(d2y/dx2)(dy/dx)-3 (4) none of these Solution: dx/dy = 1/(dy/dx) Use... View Article

If y = sin (log_ex) then x² d²y/dx² + xdy/dx is equal to

(1) sin (logex) (2) cos (logex) (3) y2 (4) -y Solution: Given y = sin (logex) Differentiate w.r.t.x dy/dx = cos loge x ×(1/x) =... View Article

If x = log_e t, t > 0 and y+1 = t². Then d²y/dx² is equal to

(1) 4e2x (2) (-½)e-4x (3) (-¾)e-5x (4) 4ex Solution: Given x = loge t So ex = t Given y+1 = t2 (ex)2 = y+1 y+1 = e2x... View Article

If [latex]f(x)=1+nx+\frac{n(n-1)}{2!}x^{2}+\frac{n(n-1)(n-2)}{3!}x^{3}+…x^{n}[/latex], then f’’(1) is equal to

(1) n(n-1)2n-1 (2) (n-1)2n-1 (3) n(n-1)2n-2 (4) n(n-1)2n Solution: f(x) = 1 + nx + n(n-1)x2/2! + n(n-1)(n-2)x3/3! +....+xn =... View Article

[latex]y=e^{a\sin^{-1}x}[/latex], then [latex](1-x^{2})y_{n+2}-(2n+1)xy_{n+1}[/latex] is equal to

1) – (n2 + a2)yn 2) (n2 - a2)yn 3) (n2 + a2)yn 4) -(n2 - a2)yn Solution: Given Differentiate w.r.t.x Squaring y12 (1-x2) =... View Article

If f (x) = tan^–1 x, Then, f ‘ (x) + f ”(x) = 0, where x is equal to

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

If y = sec (tan^–1 x), then dy/dx at x = 1 is equal to

1) 1/√2 2) 1/2 3) 1 4) √2 Solution: y = sec (tan–1 x) dy/dx = sec (tan–1 x) tan (tan–1 x) ×1/(1+x2) = sec (tan–1 x) x/(1+x2) At... View Article

If x = a cos³ θ and y = a sin³ θ, 1 + (dy/dx)² is equal to

1) tan θ 2) tan2 θ 3) 1 4) sec2 θ Solution: Given x = a cos3 θ dx/dθ = -3a cos2 θ sin θ y = a sin3 θ dy/dθ = 3a sin2 θ cos θ... View Article

If x = sin t and y = tan t, then dy/dx is equal to

(1) cos3 t (2) 1/cos3 t (3) 1/cos2 t (4) 1/sin2 t Solution: Given x = sin t y = tan t dx/dt = cos t dy/dt = sec2 t dy/dx =... View Article

If y = (tan^–1 x)², then (x² +1)² y₂ + 2x(x² + 1)y₁is equal to

1) 4 2) 0 3) 2 4) 1 Solution: y = (tan–1 x)2 Differentiate w.r.t.x dy/dx = 2 tan-1x (1/1+x2) dy/dx = (2 tan-1x)/(1+x2)...(i)... View Article

If x √(1+y) + y√(1+x) = 0, then dy/dx is equal to

(1) 1/(1+x)2 (2) -1/(1+x)2 (3) 1/(1+x2) (4) 1/(1-x2) Solution: Given x √(1+x) + y√(1+x) = 0 x √(1+x) = - y√(1+x) Squaring both... View Article

If (1 / x) + (1 / y) = 1, the function a³x² + b³y² is minimum, then x and y are

1) a + b, a - b 2) a, b 3) (a / (a + b)), b / (a + b) 4) (a + b) / a, (a + b) / b Solution: (4) (a + b) / a, (a + b) / b (1 / x) + (1 / y)... View Article

If two sides of a triangle are given, then the area of the triangle will be maximum if the angle between the given sides be

1) π / 3 2) π / 4 3) π / 6 4) π / 2 Solution: (4) π / 2 Let a and b are given, then area A = (1 / 2) ab sin C dA / dC = (1 / 2)... View Article

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