JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If $y(x) = x^x, x > 0$ , then $y''(2) - 2y'(2)$ is equal to:

Choose the correct answer:

Explanation

Solution Given $y = x^x$ .

Taking log: $\log y = x \log x$ .

Differentiating: $\frac{1}{y} y' = 1 + \log x \implies y' = x^x(1 + \log x)$ .

At $x=2$ : $y'(2) = 2^2(1 + \log 2) = 4(1 + \log 2)$ .

Differentiating again:

$y'' = y'(1 + \log x) + y(\frac{1}{x})$

$y'' = x^x(1 + \log x)^2 + x^x(\frac{1}{x})$

At $x=2$ : $y''(2) = 4(1 + \log 2)^2 + 4(\frac{1}{2}) = 4(1 + \log 2)^2 + 2$ .

Now, $y''(2) - 2y'(2)$ :

$= [4(1 + \log 2)^2 + 2] - [2(4(1 + \log 2))]$

$= 4[1 + (\log 2)^2 + 2 \log 2] + 2 - 8 - 8 \log 2$

$= 4 + 4(\log 2)^2 + 8 \log 2 + 2 - 8 - 8 \log 2$

$= 4(\log 2)^2 - 2$ .

Correct Option: (4)

Choose the correct answer:

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