CBSE 2023 · Region 2 · Set 1 · Q25 · 2 marks
If $\displaystyle \mathrm{y}=\sqrt{\mathrm{a} x+\mathrm{b}}$, prove that $\displaystyle \mathrm{y}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}\right)+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}=0$.If $\displaystyle \mathrm{f}(x)=\left\{\begin{array}{ll}\mathrm{ax}+\mathrm{b} & ; 0<x \leq 1 \\ 2 x^{2}-x & ; 1<x<2\end{array}\right.$ is a differentiable function in $\displaystyle (0, 2)$, then find the values of a and b .
If $\displaystyle \mathrm{y}=\sqrt{\mathrm{a} x+\mathrm{b}}$, prove that $\displaystyle \mathrm{y}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}\right)+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}=0$.
If $\displaystyle \mathrm{f}(x)=\left\{\begin{array}{ll}\mathrm{ax}+\mathrm{b} & ; 0<x \leq 1 \\ 2 x^{2}-x & ; 1<x<2\end{array}\right.$ is a differentiable function in $\displaystyle (0, 2)$, then find the values of a and b .
Marking-scheme solution
(a)
$\displaystyle \mathbf{y}=\sqrt{\mathbf{a x}+\mathbf{b}} \Rightarrow \mathbf{y}^{2}=\mathbf{a x}+\mathbf{b}$
Differentiate with respect to ' $\displaystyle x$ ', $\displaystyle 2 \mathrm{y} \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}=\mathrm{a}$
Differentiate with respect to ' $\displaystyle x, 2 \mathrm{y} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}+2\left(\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}\right)^{2}=0 \Rightarrow \mathrm{y} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}+\left(\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}\right)^{2}=0$
Or
(b) $\displaystyle \mathbf{f}(\mathbf{x})$ is differentiable in $\displaystyle (\mathbf{0 , 2}) \Rightarrow \mathbf{f}(\mathbf{x})$ is continuous on $\displaystyle (\mathbf{0 , 2}) \Rightarrow \mathrm{f}(x)$ is continuous at $\displaystyle x=1 \therefore \lim _{x \rightarrow 1^{-}}(\mathrm{a} x+\mathrm{b})=\lim _{x \rightarrow 1^{+}}\left(2 x^{2}-x\right) \Rightarrow \mathrm{a}+\mathrm{b}=1$
Also, $\displaystyle \mathrm{f}(x)$ is differentiable at $\displaystyle x=1, \therefore$ L.H.D. $\displaystyle (x=1)=$ R.H.D. $\displaystyle (x=1)$
$$\Rightarrow \mathrm{a}=$\displaystyle 4$($\displaystyle 1$)-$\displaystyle 1$ \therefore \mathrm{a}=$\displaystyle 3$ & \mathrm{b}=$\displaystyle 1$-\mathrm{a}=-$\displaystyle 2$
Continuity and DifferentiabilitySecond Order DerivativeApplyvery_short_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.