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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
If $\text{y}^\frac{1}{\text{n}}+\text{y}-^\frac{1}{\text{n}}=2\text{x}$ then find $(\text{x}^2-1)\text{y}_2+\text{xy}_1=$
  • ✓
    $-n^2 y$
  • B
    $n^2 y$
  • C
    $0$
  • D
    None of these.
✓

Answer

Correct option: A.
$-n^2 y$
$\text{y}^\frac{1}{\text{n}}+\text{y}-^\frac{1}{\text{n}}=2\text{x}$
Differentiating both sides we get
$\frac{\text{y}_1}{\text{n}}\Big(\text{y}^{\frac{1}{\text{n}}-1}-\text{y}^{\frac{1}{\text{n}}-1}\Big)=2$
$\Rightarrow\text{y}_1\Big(\text{y}^{\frac{1}{\text{n}}}-\text{y}^{\frac{-1}{\text{n}}}\Big)=2\text{ny}$
Again differentiating both sides we get
$\text{y}_2\Big(\text{y}^{\frac{1}{\text{n}}}-\text{y}^{\frac{-1}{\text{n}}}\Big)+\frac{\text{y}_1}{\text{n}}\Big(\text{y}^{\frac{1}{\text{n}}}-\text{y}^{\frac{-1}{\text{n}}-1}\Big)=2\text{ny}_1$
$\Rightarrow\text{ny}_2\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)+\frac{\text{y}^2_1}{\text{y}}\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)=2\text{n}^2\text{y}_1$
$\Rightarrow\text{nyy}_2\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)+2\text{xy}_1^2=2\text{n}^2\text{yy}_1$
$\Rightarrow\text{nyy}_2\frac{2\text{ny}}{\text{y}_1}+2\text{xy}_1^2=2\text{n}^2\text{yy}_1$
$\Rightarrow\frac{\text{n}^2\text{y}^2\text{y}_2}{\text{y}_1^2}+\text{xy}_1=\text{n}^2\text{y}$
$\Rightarrow\text{y}_2\frac{\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)^2}{4}+\text{xy}_1=\text{n}^2\text{y}$
$\Rightarrow\text{y}_2\frac{\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)^2-4}{4}+\text{xy}_1=\text{n}^2\text{y}$
$\Rightarrow\text{y}_2\frac{4\text{x}^2-4}{4}+\text{xy}_1=\text{n}^2\text{y}$
$\Rightarrow(\text{x}^2-1)\text{y}_2+\text{xy}_1=\text{n}^2\text{y}$

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