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Mathematics · 2023 · 4 marks
CBSE 2023 · Region 4 · Set 1 · Q36
Let f ( x ) be a real valued function. Then its - Left Hand Derivative (L.H.D.): Lf^′( a )= lim_h → 0 (f ( a -h)- f ( a ))/-h - Right Hand Derivative…
Let $\displaystyle \mathrm{f}(\mathrm{x})$ be a real valued function. Then its - Left Hand Derivative (L.H.D.) : $\displaystyle \operatorname{Lf}^{\prime}(\mathrm{a})=\lim _{h \rightarrow 0} \frac{\mathrm{f}(\mathrm{a}-h)-\mathrm{f}(\mathrm{a})}{-h}$ - Right Hand Derivative (R.H.D.) : $\displaystyle R \mathrm{f}^{\prime}(\mathrm{a})=\lim _{h \rightarrow 0} \frac{\mathrm{f}(\mathrm{a}+h)-\mathrm{f}(\mathrm{a})}{h}$ Also, a function $\displaystyle \mathrm{f}(\mathrm{x})$ is said to be differentiable at $\displaystyle \mathrm{x}=\mathrm{a}$ if its L.H.D. and R.H.D. at $\displaystyle \mathrm{x}=\mathrm{a}$ exist and both are equal. For the function $\displaystyle \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}|\mathrm{x}-3|, \mathrm{x} \geq 1 \\ \frac{\mathrm{x}^{2}}{4}-\frac{3 \mathrm{x}}{2}+\frac{13}{4}, \mathrm{x}<1\end{array}\right.$ answer the following questions :
(i)
What is R.H.D. of $\displaystyle \mathrm{f}(\mathrm{x})$ at $\displaystyle \mathrm{x}=1$ ? $\displaystyle 1$
(ii)
What is L.H.D. of $\displaystyle \mathrm{f}(\mathrm{x})$ at $\displaystyle \mathrm{x}=1$ ?
(iii)
(a)
Check if the function $\displaystyle \mathrm{f}(\mathrm{x})$ is differentiable at $\displaystyle \mathrm{x}=1$.
OR
(b)
Find $\displaystyle \mathrm{f}^{\prime}(2)$ and $\displaystyle \mathrm{f}^{\prime}(-1)$. Case Study - $\displaystyle 2$
Marking-scheme solution
$$\begin{aligned}
& \text { (i)R.H.D. of } \mathrm{f}(\mathrm{x}) \text { at } \mathrm{x}=1=\lim _{h \rightarrow 0} \frac{\mathrm{f}(1+h)-\mathrm{f}(1)}{h} \\
& \qquad=\lim _{\boldsymbol{h} \rightarrow \mathbf{0}} \frac{|1+h-3|-|-2|}{h}=\lim _{\boldsymbol{h} \rightarrow \mathbf{0}} \frac{2-h-2}{h}=-1
\end{aligned}
$$
(ii) L.H.D. of $\displaystyle \mathrm{f}(\mathrm{x})$ at $\displaystyle \mathrm{x}=1=\lim _{\boldsymbol{h} \rightarrow \mathbf{0}} \frac{\boldsymbol{\mathrm{f}}(\mathbf{1}-\mathbf{h})-\boldsymbol{\mathrm{f}}(\mathbf{1})}{-\mathbf{h}}$
$$\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{\left[\dfrac{(1-h)^{2}}{4}-\dfrac{3(1-h)}{2}+\dfrac{13}{4}-2\right]}{-h} \\
& =\lim _{h \rightarrow 0}\left[\frac{h^{2}-2 h+1-6+6 h+13-8}{-4 \mathrm{~h}}\right]
\end{aligned}
$$
$$=\lim _{\boldsymbol{h} \rightarrow \mathbf{0}}\left[\frac{\boldsymbol{h}^{2}+\mathbf{4} \boldsymbol{h}}{-4 \mathrm{~h}}\right]=-1 .
$$
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CBSE Class 12 Mathematics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.