Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

Write the correct alternative in the following:

Question:

Write the correct alternative in the following:

If $y=\tan ^{-1}\left\{\frac{\log _{e}\left(e / x^{2}\right)}{\log _{e}\left(e x^{2}\right)}\right\}+\tan ^{-1}\left(\frac{3+2 \log _{e} x}{1-6 \log _{e} x}\right)$, then $\frac{d^{2} y}{d x^{2}}=$

A. 2

B. 1

C. 0

D. $-1$

Solution:

Given:

$y=\tan ^{-1}\left\{\frac{\log _{e}\left(\frac{e}{x^{2}}\right)}{\log _{e}\left(e x^{2}\right)}\right\}+\tan ^{-1}\left\{\frac{3+2 \log _{e} x}{1-6 \log _{e} x}\right\}$

$y=\tan ^{-1}\left\{\frac{\log _{e} e-\log _{e} x^{2}}{\log _{e} e+\log _{e} x^{2}}\right\}+\tan ^{-1}\left\{\frac{3 \log _{e} e+2 \log _{e} x}{1-3 \log _{e} e \times 2 \log _{e} x}\right\}$

$y=\tan ^{-1}\left\{\frac{1-\log _{e} x^{2}}{1+\log _{e} x^{2}}\right\}+\tan ^{-1}\left(3 \log _{e} e\right)+\tan ^{-1}\left(2 \log _{e} x\right)$

$y=\tan ^{-1}\left\{\frac{\log _{e} e-2 \log _{e} x}{1+\log _{e} e \times 2 \log _{e} x}\right\}+\tan ^{-1}\left(3 \log _{e} e\right)+\tan ^{-1}\left(2 \log _{e} x\right)$

$y=\tan ^{-1}\left(\log _{e} e\right)-\tan ^{-1}\left(2 \log _{e} x\right)+\tan ^{-1}\left(3 \log _{e} e\right)+\tan ^{-1}\left(2 \log _{e} x\right)$

$y=\tan ^{-1}(1)+\tan ^{-1}(3)$

$y=\tan ^{-1}\left(\frac{1+3}{1-3}\right)=\tan ^{-1}(-2)$

$\frac{d y}{d x}=0$

Leave a comment

None
Free Study Material