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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
For $x > 1$, if ${\left( {2x} \right)^{2y}} = 4{e^{2x - 2y}}$, then ${\left( {1 + {{\log }_e}\,2x} \right)^2}\frac{{dy}}{{dx}}$ is equal to
  • A
    $\frac{{x{{\log }_e}\,2x - {{\log }_e}\,2}}{x}$
  • B
    $log_e\, 2x$
  • C
    $\frac{{x{{\log }_e}\,2x + {{\log }_e}\,2}}{x}$
  • D
    $x log_e\, 2x$
✓

Answer

$2y\ \ell n\ 2x = \ell n4 + 2x - 2y$$2y\left( {1 + \ell n2x} \right) = \ell n4 + 2x$
$y = \frac{{\ell n2x - \frac{{\ell n2}}{x}}}{{{{\left( {1 + \ell n2x} \right)}^2}}}.{\left( {1 + \ell n2x} \right)^2}$
$ = \ell n2x - \frac{{\ell n2}}{x} = \frac{{x\ell n\left( {2x} \right) - \ell n2}}{x}$
on differentiating equation $(1)\ \text{w.r.t  x}$ we get,
$\log _e 2 x \frac{d y}{d x}+\frac{y}{x}=0+1-\frac{d y}{d x}$
$\frac{d y}{d x}\left[1+\log _e 2 x\right]=1-\frac{y}{x} \ldots \ldots \ldots . .$
from equation $(1)$
$ y\left[1+\log _e 2 x\right]=x+\log _e 2$
$y=\frac{x+\log _e 2}{1+\log _e 2 x} $
from equation $(2)$
$\left(1+\log _e 2 x\right) \frac{d y}{d x}=1-\frac{x+\log _e 2}{x\left(1+\log _e 2 x\right)}$
$\left(1+\log _e 2 x\right)^2 \frac{d y}{d x}=\frac{x+x \log _e 2 x-x-\log _e 2}{x}$
$ \left(1+\log _e 2 x\right)^2 \frac{d y}{d x}=\frac{x \log _e 2 x-\log _e 2}{x}$

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