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JEE Main 2019 · 12 Jan · Shift I

JEE Main 2019 Mathematics question on Limits Continuity and Differentiability

For x>1x > 1, if (2x)2y=4e2x−2y(2x)^{2y} = 4e^{2x-2y}, then (1+log⁡e2x)2dydx(1 + \log_e 2x)^2 \dfrac{dy}{dx} is equal to
  1. A.
    xlog⁡e2x+log⁡e2x\dfrac{x \log_e 2x + \log_e 2}{x}
  2. B.
    xlog⁡e2x−log⁡e2x\dfrac{x \log_e 2x - \log_e 2}{x}
  3. C.
    xlog⁡e2xx \log_e 2x
  4. D.
    log⁡e2x\log_e 2x
Show answer and solution

Correct option: B

Solution

Taking log of both sides: 2y·ln(2x) = ln4 + (2x-2y) = 2ln2 + 2x - 2y.

So 2y·ln(2x) + 2y = 2ln2 + 2x, giving y(ln(2x) + 1) = ln2 + x.

Thus y = (x + ln2)/(1 + ln(2x)).

Differentiate using quotient rule. Let u = x + ln2, v = 1 + ln(2x):
dy/dx = (v·u' - u·v')/v²

u' = 1, v' = 1/x, so:
dy/dx = [(1 + ln(2x))·1 - (x + ln2)·(1/x)] / (1 + ln(2x))²

Numerator = 1 + ln(2x) - 1 - (ln2)/x = ln(2x) - (ln2)/x = (x·ln(2x) - ln2)/x.

Therefore (1 + ln(2x))²·dy/dx = (x·ln(2x) - ln2)/x, which is option 2.

[JEE Main 2019 · 12 Jan · Shift I]

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