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If f\left ( x \right )=\begin{bmatrix} \cos x &x &1 \\ 2\sin x &x^{2} &2x \\ \tan x &x & 1 \end{bmatrix}    then    \lim_{x\rightarrow 0}\frac{{f}'\left ( x \right )}{x}

  • Option 1)

    does not exist.

     

  • Option 2)

    exists and is equal to 2.

  • Option 3)

    exists and is equal to 0.

  • Option 4)

    exists and is equal to −2.

 

Answers (2)

best_answer

 

As we learned 

Value of determinants of order 3 -

-

 

 f\left ( x \right )=\begin{vmatrix} \cos x & x&1 \\ 2\sin x & x^{2} &2x \\ \tan x & x & 1 \end{vmatrix}

On expansion we get 

f\left ( x \right )=x^{2}\left ( \tan x-\cos x \right )

f\left{}' ( x \right )=2x\left ( \tan x-\cos x \right )+x^{2}\left ( \sec x+\sin x \right )

\lim_{x\rightarrow 0}\frac{{f}'\left ( x \right )}{x}=\frac{2x\left ( \tan x-\cos x \right )+x^{2}\left ( \sec x+\sin x \right )}{x}

=2\left ( \tan x-\cos x \right )+x\left ( \sec x+\sin x \right )

Put x=0

we get answer = -2


Option 1)

does not exist.

 

Option 2)

exists and is equal to 2.

Option 3)

exists and is equal to 0.

Option 4)

exists and is equal to −2.

Posted by

Himanshu

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