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

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