Science (English Medium) कक्षा १२ - CBSE Question Bank Solutions

The general solution of the differential equation y dx – x dy = 0 is ______.

Solve the differential equation: y dx + (x – y²)dy = 0

Evaluate: `int_0^(pi/4) (dx)/(1 + tanx)`

Find: `int e^x.sin2xdx`

Find: `int (2x)/((x^2 + 1)(x^2 + 2)) dx`

Find the general solution of the differential equation: `e^((dy)/(dx)) = x^2`.

Find `int e^(cot^-1x) ((1 - x + x^2)/(1 + x^2))dx`.

Find `int (sin^-1x)/(1 - x^2)^(3//2) dx`.

Find `int e^x ((1 - sinx)/(1 - cosx))dx`.

The derivative of x^2x w.r.t. x is ______.

Two vectors `veca = a_1 hati + a_2 hatj + a_3 hatk` and `vecb = b_1 hati + b_2 hatj + b_3 hatk` are collinear if ______.

Find: `int e^(x^2) (x^5 + 2x^3)dx`.

If `veca = 4hati + 6hatj` and `vecb = 3hatj + 4hatk`, then the vector form of the component of `veca` along `vecb` is ______.

The general solution of the differential equation ydx – xdy = 0; (Given x, y > 0), is of the form

(Where 'c' is an arbitrary positive constant of integration)

Find the equations of the tangent and normal to the curve x = a sin³θ and y = a cos³θ at θ=π/4.

Find the equation of the normal at a point on the curve x² = 4y which passes through the point (1, 2). Also find the equation of the corresponding tangent.

Find the equation of the normal at a point on the curve x² = 4y which passes through the point (1, 2). Also find the equation of the corresponding tangent.

The equation of tangent at (2, 3) on the curve y² = ax³ + b is y = 4x – 5. Find the values of a and b.

 If x=a sin 2t(1+cos 2t) and y=b cos 2t(1−cos 2t), find `dy/dx `

Evaluate : `∫_0^(π/2)(sin^2 x)/(sinx+cosx)dx`