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  • Provide Solution for RD Sharma Class 12 Chapter 11 Higher Order Derivatives Exercise Multiple Choice Question Question 10

Provide Solution for RD Sharma Class 12 Chapter 11 Higher Order Derivatives Exercise Multiple Choice Question Question 10

Answers (1)

Answer:

                (c)

Hint:

We must know about the derivative of logarithm and  \tan x .

Given:

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

Explanation:

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

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

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

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

               y=\tan ^{-1}\left\{\frac{1-8 \log _{e} x+12\left(\log _{e} x\right)^{2}+3+8 \log _{e} x+4\left(\log _{e} x\right)^{2}}{1-4 \log _{e} x-12\left(\log _{e} x\right)^{2}-3+4 \log _{e} x+4\left(\log _{e} x\right)^{2}}\right\}

                \begin{aligned} &y=\tan ^{-1}\left\{\frac{4+16\left(\log _{e} x\right)^{2}}{-2-8\left(\log _{e} x\right)^{2}}\right\} \\ & \end{aligned}

                y=\tan ^{-1}\left\{\frac{4\left(1+4\left(\log _{e} x\right)^{2}\right)}{-2\left(1+4\left(\log _{e} x\right)^{2}\right)}\right\}

                \begin{aligned} &y=\tan ^{-1}[-2] \\ & \end{aligned}

                \frac{d y}{d x}=0, \frac{d^{2} y}{d x^{2}}=0

Posted by

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