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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Let $ f(x) = \int \frac{7x^{10} + 9x^8}{(1 + x^2 + 2x^9)^2} ,dx $, $ x > 0 $,
$ \lim_{x \to 0} f(x) = 0 $ and $ f(1) = \frac{1}{4} $.
If
$ A = \left[ \matrix{ 0 & 0 & 1 \cr \frac{1}{4} & f'(1) & 1 \cr \alpha^2 & 4 & 1 } \right] $
and $ B = \text{adj(adj } A) $ be such that $ |B| = 81 $, then $ \alpha^2 $ is equal to
Solution
$ f(x) = \int \frac{7x^{10} + 9x^8}{(1 + x^2 + 2x^9)^2} ,dx $
$ = \int \left( \frac{7}{x^8} - \frac{9}{x^{10}} \right) \frac{dx}{\left( \frac{1}{x^9} + x^7 + 2 \right)^2} $
Put $ t = \frac{1}{x^9} + x^7 + 2 $
$ \frac{dt}{dx} = -\frac{9}{x^{10}} + \frac{7}{x^8} $
$ f(x) = \int \frac{-dt}{t^2} $
$ = \frac{1}{t} + C $
$ = \frac{1}{\frac{1}{x^9} + x^7 + 2} + C $
$ = \frac{x^9}{1 + x^2 + 2x^9} + C $
Given $ f(1) = \frac{1}{4} $
$ \frac{1}{4} = \frac{1}{4} + C \Rightarrow C = 0 $
$ f(x) = \frac{x^9}{1 + x^2 + 2x^9} $
Differentiate:
$ f'(x) = \frac{(1 + x^2 + 2x^9)(9x^8) - x^9(2x + 18x^8)}{(1 + x^2 + 2x^9)^2} $
At $ x = 1 $:
$ f'(1) = \frac{36 - 20}{16} = 1 $
Now,
$ A = \left[ \matrix{ 0 & 0 & 1 \cr \frac{1}{4} & 1 & 1 \cr \alpha^2 & 4 & 1 } \right] $
Determinant:
$ |A| = 1 - \alpha^2 $
Given:
$ B = \text{adj(adj } A) $
$ |B| = |A|^2 = 81 $
$ |A| = 3 $
$ 1 - \alpha^2 = -3 \Rightarrow \alpha^2 = 4 $
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