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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
If $\text{y}=\sin(\text{m}\sin^{-1}\text{x}),$ then $(1-\text{x}^2)\text{y}_2-\text{xy}_1$ is equal to :
  • A
    $m^2 y$
  • B
    $my$
  • ✓
    $-m^2 y$
  • D
    None of these
✓

Answer

Correct option: C.
$-m^2 y$
Here,$\text{y}=\sin(\text{m}\sin^{-1}\text{x}),$
$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}^2}{(1-\text{x}^2)}+\frac{\text{mx}\cos(\text{m}\sin^{-1}\text{x})}{(1-\text{x}^2)^\frac{3}{2}}$
$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}^2}{(1-\text{x}^2)}+\frac{\text{xm}\cos(\text{m}\sin^{-1}\text{x})}{(1-\text{x}^2)\times\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}^2}{(1-\text{x}^2)}+\frac{\text{xy}_1}{(1-\text{x}^2)}$
$\Rightarrow(1-\text{x}^2)\text{y}_2=-\text{ym}^2+\text{xy}_1$
$\Rightarrow(1-\text{x}^2)\text{y}_2-\text{xy}_1=-\text{m}^2\text{y}$

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