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AP EAMCET Engineering 2013 Solved Paper

Section: Mathematics

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Question : 147 of 160

Marks: +1, -0

If

\frac{d}{dx}[(x+1)(x^{2}+1)(x^{4}+1)(x^{8}+1)]

=(15 x^{p}-16 x^{q}+1)(x-1)^{-2}

(p, q)

is equal to

$(12, 11)$
$(15, 14)$
$(16, 15)$
$(16,15)$

Solution:

\frac{d}{dx}[(x+1)(x^{2}+1)(x^{4}+1)(x^{8}+1)]

=\frac{15 x^{p}-16 x^{q}+1}{(x-1)^2}

.......(i)

LHS

=\frac{d}{dx}\left[\frac{(x^{2}-1)(x^{2}+1)(x^{4}+1)(x^{8}+1)}{x-1}\right]

=\frac{d}{dx}\left[\frac{(x^{4}-1)(x^{4}+1)(x^{8}+1)}{x-1}\right]

=\frac{d}{dx}\left[\frac{(x^{8}-1)(x^{8}+1)}{x-1}\right]

=\frac{d}{dx}\left[\frac{x^{16}-1}{x-1}\right]

=\frac{(x-1)(16 x^{15})-(x^{16}-1)}{(x-1)^{2}}

=\frac{16 x^{16}-16 x^{15}-x^{16}+1}{(x-1)^{2}}

=\frac{15 x^{16}-16 x^{15}+1}{(x-1)^{2}}

On comparing LHS = RHS, we get

p = 16 and q = 15

∴

(p, q) = (16, 15)

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