# Maths-2000-Set I

**Q1) Evaluate
{(1 – Cos x)/x2}**

Ans1)

{(1 – Cos x)/x2} = (2Sin2x/2)/x2

= 2/4 (Sin(x/2)/(x/2))2 =2/4 . 12 = 1/2 (as Sinx/x=1)

**Q2) A particle moves along a straight line so that S = t . Show that the acceleration is negative and is proportional to the cube of velocity.**

Ans2) Here the equation of motion is

S = t

differentiating both sides w.r.t. ‘t’

v = dS/dt = 1/2t = 1/2S ( S = t )

or v = 1/2S

differentiating again w.r.t. ‘t’

accn = dv/dt = -1/2s2 ds/dt

= -2/4s2 ds/dt

= -(2/(1/v2)). v = -2v3

accn is negative and is proportional to cube of velocity.

**Q3) Evaluate
Cos4xdx**

Ans3) let

I = Cos4xdx = (Cos2x)2dx

= ((1 + Cos 2x)/2)2dx = 1/4 (1 + Cos22x + 2Cos 2x)dx

=1/4 [1 + (1 + Cos 4x)/2 + 2Cos 2x]dx

= 1/4[x + x/2 + (Sin 4x)/8 + (2Sin 2x)/2] + C

= 3x/8 + (Sin 4x)/32 + (1/4)Sin 2x + C

**Q4) Evaluate
(1 + tan x)/(x + log sec x) dx**

Ans4) Let

I = (1 + tan x)/(x + log sec x) dx

= log | x + log sec x | + C

[ as f'(x)/f(x) dx = log | f(x) | + C ]