If an SHM the velocity of a particle is `u_1` and `u_2` from the mean position `x_1 and x_2` shwo that time period `T = 2pisqrtfrac(x_2^2 - x_1^2)(u_1^2 - u_2^2)`.

Find the derivate of : y = x cos^-1 (frac(1-x^2)(1+x^2))

If the mean of observations x_1.x_2,x_3…..x_n is barx then the mean of x_1+a,x_2+a,…...x_n+a will be

Find dy/dx : x=frac(1-t^2)(1+t^2),y = frac(2t)(1+t^2)

The position of a moving particle is given by x = 6 + 18t + 9t^2 where x is the distance and t is the time. What is the velocity at t = 2sec.

Prove that V - u = 2as from velocity time graph.

Find dy/dx : y = sqrt((x-1)(2-x)(x+1)(2x+1)

1 and 2 are the numerals of both numeral system whose basis are x and 2x+1 . If decimal conversion of (21)_x is y, then show that the decimal conversion of (21)_(2x+1) is 2y+1 .

A partical of mass m1 is travelling with avelocity u1 strikes head-on against partical of mass m2 at rest.After collision the particales move with a velocity v1 and v2 respectively .show that the ratio of decrese in K.E of the partical of mass m1 to its original K.E is given By (4m1m2)/(m1+m2)^2

Solve: frac(x)(2) -frac(1-2x)(3) =2

During n^(th) second of its motion a body covers a distance S_n with uniform acceleration 'a' and initial velocity 'u'. Show that (a= frac(2S_n - 2u)(2n-1))