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5. continuity and differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths5. continuity and differentiation1 Mark
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
  • ✓
    $\frac{{x{{\log }_e}\,2x - {{\log }_e}\,2}}{x}$
  • B
    $log_e\, 2x$
  • C
    $\frac{{x{{\log }_e}\,2x + {{\log }_e}\,2}}{x}$
  • D
    $x log_e\, 2x$
✓

Answer

Correct option: A.
$\frac{{x{{\log }_e}\,2x - {{\log }_e}\,2}}{x}$
a
$2y\ell n2x = \ell n4 + 2x - 2y$

$2y\left( {1 + \ell n2x} \right) = \ell n4 + 2x$

$y = \frac{{\ell n2x - \frac{{\ell n2}}{x}}}{{{{\left( {1 + \ell n2x} \right)}^2}}}.{\left( {1 + \ell n2x} \right)^2}$

$ = \ell n2x - \frac{{\ell n2}}{x} = \frac{{x\ell n\left( {2x} \right) - \ell n2}}{x}$

on differentiating equation (1) w.r.t x we get,

$\begin{array}{l}\log _e 2 x \frac{d y}{d x}+\frac{y}{x}=0+1-\frac{d y}{d x} \\ \frac{d y}{d x}\left[1+\log _e 2 x\right]=1-\frac{y}{x} \ldots \ldots \ldots . .\end{array}$

from equation $(1)$

$\begin{array}{l} y\left[1+\log _e 2 x\right]=x+\log _e 2 \\ y=\frac{x+\log _e 2}{1+\log _e 2 x} \end{array}$

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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