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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
If $\text{y}=\frac{\text{ax}+\text{b}}{\text{x}^2+\text{c}},$ then $(2\text{xy}_1+\text{y})\text{y}_3=$
  • ✓
    $3\left(x y_2+y_1\right) y_2$
  • B
    $3\left(x y_1+y_2\right) y_2$
  • C
    $3\left(x y_1+y_2\right) y_1$
  • D
    None of these
✓

Answer

Correct option: A.
$3\left(x y_2+y_1\right) y_2$
$\text{y}=\frac{\text{ax}+\text{b}}{\text{x}^2+\text{c}}$
$\Rightarrow(\text{x}^2+\text{c})\text{y}=\text{ax}+\text{b}$
Differentiating $\text{w.r.t. x},$ we get
$2\text{xy}+(\text{x}^2+\text{c})\frac{\text{dy}}{\text{dx}}=\text{a}$
Differentiating $\text{w.r.t. x},$ we get
$2\text{y}+2\text{xy}_1+2\text{xy}+(\text{x}^2+\text{c})\text{y}_2=0$
$\Rightarrow2\text{y}+4\text{xy}_1+\text{x}^2+\text{cy}_2=0$
Differentiating $\text{w.r.t. x},$ we get
$2\text{y}_1+4\text{y}_1+4\text{xy}_2+(\text{x}^2+\text{c})\text{y}_3+2\text{xy}_2=0$
$\Rightarrow6\text{y}_1+6\text{xy}_2+(\text{x}^2+\text{c})\text{y}_3=0$
$\Rightarrow6\text{y}_1+6\text{xy}_2+\Big(\frac{-2\text{y}-4\text{xy}_1}{\text{y}_2}\Big)\text{y}_3=0$
$[\because2\text{y}+4\text{xy}_1+(\text{x}^2+\text{c})\text{y}_2=0]$
$\Rightarrow6\text{y}_1\text{y}_2+6\text{x}(\text{y}_2)^2-2\text{y}-4\text{xy}_1\text{y}_3=0$
$\Rightarrow3\text{y}_1\text{y}_2+3\text{x}(\text{y}_2)^2-\text{y}-2\text{xy}_1\text{y}_3=0$
$\Rightarrow(\text{y}_1+\text{xy}_2)3\text{y}_2=(2\text{xy}_1+\text{y})\text{y}_3$

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