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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
If $\text{y}=\text{x}^{\text{n}-1}\log\text{x}\ \text{x}^2\text{y}_2+(3-2\text{n})\text{xy}_1$ is equals to :
  • ✓
    $-(n-1)^2 y$
  • B
    $(n-1)^2 y$
  • C
    $-n^2 y$
  • D
    $n^2 y$
✓

Answer

Correct option: A.
$-(n-1)^2 y$
Here,
$\text{y}=\text{x}^{\text{n}-1}\log\text{x}$
$\Rightarrow\text{y}_1=(\text{n}-1)\text{x}^{\text{n-2}}\log\text{x}+\frac{\text{x}^{\text{n}-1}}{\text{x}}$
$\Rightarrow\text{y}_1=\frac{(\text{n}-1)\text{x}^{\text{n}-1}\log\text{x}+\text{x}^{\text{n}-1}}{\text {x}}$
$\Rightarrow\text{xy}_1=(\text{n}-1)\text{y}+\text{x}^{\text{n-1}}$
$\Rightarrow\text{xy}_2+\text{y}_1=(\text{n}-1)\text{y}_1+(\text{n}-1)\text{x}^{\text{n-2}}$
$\Rightarrow\text{xy}_2+\text{y}_1=(\text{n}-1)\text{y}_1+\frac{(\text{n}-1)\text{x}^{\text{n}-1}}{\text{x}}$
$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1=\text{x}(\text{n}-1)\text{y}_1+(\text{n}-1)\text{x}^{\text{n}-1}$
$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1=\text{x}(\text{n}-1)\text{y}_1+(\text{n}-1)\{\text{xy}_1-(\text{n}-1)\text{y}\}$
$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1=\text{x}(\text{n}-1)\text{y}_1+(\text{n}-1)\text{xy}_1-(\text{n}-1)^2\text{y}$
$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1=2\text{x}(\text{n}-1)\text{y}_1+(\text{n}-1)^2\text{y}$
$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1(1-2\text{n}+2)=-(\text{n}-1)^2\text{y}$
$\Rightarrow\text{x}^2\text{y}_2+(3-2\text{n})\text{xy}_1=-(\text{n}-1)^2\text{y}$

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