MCQ
If $\text{y}=\frac{\text{ax}+\text{b}}{\text{x}^2+\text{c}},$ then $(2\text{xy}_1+\text{y})\text{y}_3=$
- ✓$3(xy_2 + y_1) y_2$
- B$3(xy_1 + y_2) y_2$
- C$3(xy_1 + y_2) y_1$
- DNone of these
$\text{y}=\frac{\text{ax}+\text{b}}{\text{x}^2+\text{c}}$
$\Rightarrow(\text{x}^2+\text{c})\text{y}=\text{ax}+\text{b}$
Differentiating w.r.t.x, we get
$2\text{xy}+(\text{x}^2+\text{c})\frac{\text{dy}}{\text{dx}}=\text{a}$
Differentiating 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 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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