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  •  If the function  <img alt="\mathrm{f(x)=\left\{\begin{array}{cc} \frac{\log _{\mathrm{e}}\left(1-x+x^{2}\right)+\log _{\mathrm{e}}\left(1+x+x^{2}\right)}{\sec x-\cos x} & , \quad x \

 If the function  \mathrm{f(x)=\left\{\begin{array}{cc} \frac{\log _{\mathrm{e}}\left(1-x+x^{2}\right)+\log _{\mathrm{e}}\left(1+x+x^{2}\right)}{\sec x-\cos x} & , \quad x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)-\{0\} \\ \mathrm{k} & , \quad x=0 \end{array}\right.}    is continuous at \mathrm{x=0}, then \mathrm{k} is equal to:

Option: 1

1


Option: 2

-1


Option: 3

\mathrm{e}


Option: 4

0


Answers (1)

best_answer

\mathrm{\lim _{x \rightarrow 0} \frac{\left(\ln \left(1+x^{2}+x^{4}\right)\right) \cos x}{1-\cos ^{2} x}}

\mathrm{\lim _{x \rightarrow 0} \frac{\left(\frac{\ln \left(1+x^{2}+x^{4}\right)}{x^{2}+x^{4}}\right) x^{2}\left(1+x^{2}\right) \cos x}{\left(\frac{\sin ^{2}-x}{x^{2}}\right) x^{2}}=1}

\mathrm{\therefore k=1}

Hence correct option is 1

Posted by

shivangi.shekhar

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