# Maths-1997 -Set I

**Q1) Find from first principle the derivative of (x2 + 1)/x w.r.t. x**

Ans1) Let y = (x2 + 1)/x = x + 1/x – (i)

Let x be a small incremental in the value of x and y be the corresponding incremental in the value of y, then

y + y = x + x + 1/(x + x) – (ii)

Now (ii) – (i)

=> y = x + 1/(x + x) – 1/x

y = x – x/(x(x + x)) => y/x = 1 – 1/(x(x + x))

… dy/dx =

y/x

=

(1 – 1/(x(x + x))

= 1 – 1/x2

= (x2 – 1)/x2

**Q2) Evaluate :
1 – cos 5x
1 – cos 6x**

Ans2)

1 – cos 5x =

1 – cos 6x 2sin2 (5x/2)

2sin2 (3x)

(5/2)2[sin (5x/2)/(5x/2)]2

32 [sin 3x/3x]2 = 25/36

**Q3) Using differentials, find the approximate value of 29**

Ans3) To find 29

Let x = 27 and x = 2 then x + x = 27 + 2 = 29

Taking y = x1/3, we get dy/dx = 1/3 x-2/3

Also y + y = (x + x)1/3 = 291/3

where y = x1/3 = 271/3 = 3

… 3 + y = 291/3

Again y = dydx x = 1/3 x-2/3 . 2 = 2/3(27)-2/3 = 2/27

291/3 = 3 + y = 3 + (2/27) = 83/27