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5. continuity and differentiation — ગણિત ધોરણ 12 સાયન્સ — Question

Gujarat Boardગુજરાતી માધ્યમધોરણ 12 સાયન્સગણિત5. continuity and differentiation1 Mark
MCQ
જો $\cos ^{-1}\left(\frac{y}{2}\right)=\log _{e}\left(\frac{x}{5}\right)^{5},|y|<2$, તો
  • A
    $x^{2} y^{\prime \prime}+x y^{\prime}-25 y=0$
  • B
    $x^{2} y^{\prime \prime}+x y^{\prime}-25 y=0$
  • C
    $x^{2} y^{\prime \prime}-x y^{\prime}+25 y=0$
  • D
    $x^{2} y^{\prime \prime}+x y^{\prime \prime}+25 y=0$
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Answer

$\cos ^{-1}\left(\frac{y}{2}\right)=\log _{e}\left(\frac{x}{5}\right)^{5}$

$\cos ^{-1}\left(\frac{y}{2}\right)=5 \log _{e}\left(\frac{x}{5}\right)$

$\frac{-1}{\sqrt{1-\frac{y^{2}}{4}}} \cdot \frac{y^{\prime}}{2}=5 \cdot \frac{1}{\frac{x}{5}} \times \frac{1}{5}$

$\Rightarrow \frac{-y^{\prime}}{\sqrt{4-y^{2}}}=\frac{5}{x}$

$-x y^{\prime}=5 \sqrt{4-y^{2}}$

$-x y^{\prime \prime}-y^{\prime}=5 \cdot \frac{1}{2 \sqrt{4-y^{2}}}\left(-2 y y^{\prime}\right)$

$\Rightarrow x y^{\prime \prime}+y^{\prime}=\frac{5 y^{\prime} \cdot y}{\sqrt{4-y^{2}}}$

$x y^{\prime \prime}+y^{\prime}=5 \cdot\left(\frac{-5}{x}\right) y$

$x^{2} y^{\prime \prime}+x y^{\prime}=-25 y$

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