TS ICET 2015 Question Paper with Solutions for online practice

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Question : 111 of 200

Marks: +1, -0

\lim↙{x→π/2}

{1- \sin^3 x}/{\cos^2 x}

=

$3/2$
1
$2/3$
- 1

Solution:

\lim↙{x→{π}/{2}}({1-\sin^{3}x}/{\cos^{2}x})=\lim↙{x→{π}/{2}}={1^{3}-\sin^{3}x}/{1-\sin^{2}x}

=\lim↙{x→{π}/{2}}{(1-\sinx)(1+\sin^{2}x+\sinx)}/{(1-\sinx)(1+\sinx)}

[Using

a^{3}-b^{3}=(a-b)(a^{2}+b^{2}+ab)

and

\cos^{2}x=1-\sin^{2}x=(1-\sinx)(1+\sinx)]

=\lim↙{x→{π}/{2}}{1+\sin^{2}x+\sinx}/{1+\sinx}

={1+\sin^{2}{π}/{2}+\sin{π}/{2}}/{1+\sin{π}/{2}}={1+1+1}/{1+1}={3}/{2}

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