For $x > 1$, if $(2x)^{2y} = 4e^{2x - 2y}$, then $\left( 1 +\log_{e} 2x\right)^{2} \frac{dy}{dx} $ is equal to :
- $\log_e 2x$
- $\frac{x \log_{e} 2x +\log_{e} 2}{x} $
- $x \log_e 2x$
- $\frac{x \log_{e} 2x - \log_{e} 2}{x} $
The Correct Option is D
Solution and Explanation
$2y \ell n2x = \ell n4 + 2x - 2y$
$y = \frac{x + \ell n 2}{1 + \ell n 2 x}$
$y' = \frac{(1 + \ell n 2 x) - (x + \ell n 2 ) \frac{1}{x}}{( 1+ \ell n 2x )^2}$
$y ' = 1 + \ell n 2 x)^2 = [\frac{x \ell n 2x - \ell n 2}{x} ] $
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JEE Main Notification
Concepts Used:
Continuity
A function is said to be continuous at a point x = a, if
limx→a
f(x) Exists, and
limx→a
f(x) = f(a)
It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is undefined or does not exist, then we say that the function is discontinuous.
Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:
- The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
- The limit of the function as x approaches a, exists.
- The limit of the function as x approaches a, must be equal to the function value at x = a.