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If f(x)=\log \left(\sec ^{2} x\right)^{\cot 2} x for x \neq 0 =K for x= is continuous at x=0 then K 1 \mathrm{~S}
(\mathrm{A}) e^{-1}
(B) 1
(C) e
(D) 0
maths | Maths
If f(x)=\log \left(\sec ^{2} x\right)^{\cot 2} x for x \neq 0 =K for x= is continuous at x=0 then K 1 \mathrm{~S}
(\mathrm{A}) e^{-1}
(B) 1
(C) e
(D) 0

Correct option is

(B) 1

Since, \mathrm{f} is continuous at \mathrm{x}=0

\therefore \mathrm{f}(0)=\lim _{\mathrm{x} \rightarrow 0} \log \left(\sec ^{2} \mathrm{x}\right)^{\cot ^{2} \mathrm{x}}

Therefore we have,

\mathrm{K}=\lim _{\mathrm{x} \rightarrow 0} \cot ^{2} \mathrm{x} \cdot \log \left(\sec ^{2} \mathrm{x}\right)

=\lim _{x \rightarrow 0} \cot ^{2} x \cdot \log \left(1+\tan ^{2} x\right)

=\lim _{x \rightarrow 0} \frac{\log \left(1+\tan ^{2} x\right)}{\tan ^{2} x}

\therefore \mathrm{K}=1 \ldots \ldots \lim _{\mathrm{x} \rightarrow 0} \frac{\log (1+\mathrm{x})}{\mathrm{x}}=1

i

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