CBSE 2023 · Region 5 · Set 3 · Q21 · 2 marks
Find the value of k for which the function f given as \[\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cl} \frac{1-\cos \mathrm{x}}{2 \mathrm{x}^{2}}, & \text { if } \mathrm{x} \neq 0 \\ \mathrm{k}, & \text { if } \mathrm{x}=0 \end{array} \text { is continuous at } \mathrm{x}=0 .\right. \]If $\displaystyle \mathrm{x}=\mathrm{a} \cos \mathrm{t}$ and $\displaystyle \mathrm{y}=\mathrm{b} \sin \mathrm{t}$, then find $\displaystyle \frac{d^{2} \mathrm{y}}{d \mathrm{x}^{2}}$.
Find the value of k for which the function f given as \[\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cl} \frac{1-\cos \mathrm{x}}{2 \mathrm{x}^{2}}, & \text { if } \mathrm{x} \neq 0 \\ \mathrm{k}, & \text { if } \mathrm{x}=0 \end{array} \text { is continuous at } \mathrm{x}=0 .\right. \]
If $\displaystyle \mathrm{x}=\mathrm{a} \cos \mathrm{t}$ and $\displaystyle \mathrm{y}=\mathrm{b} \sin \mathrm{t}$, then find $\displaystyle \frac{d^{2} \mathrm{y}}{d \mathrm{x}^{2}}$.
Marking-scheme solution
Here,
$$\mathrm{f}(\mathrm{x})=\frac{1-\cos \mathrm{x}}{2 \mathrm{x}^{2}}=\frac{2 \sin ^{2} \dfrac{\mathrm{x}}{2}}{2 \mathrm{x}^{2}}=\left(\frac{\sin \dfrac{\mathrm{x}}{2}}{2 \dfrac{\mathrm{x}}{2}}\right)^{2}
\Rightarrow \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 0} \frac{1}{4}\left(\frac{\sin \dfrac{\mathrm{x}}{2}}{\dfrac{\mathrm{x}}{2}}\right)=\frac{1}{4}
Given $\displaystyle \mathrm{x}=\mathrm{a} \cos \mathrm{t}$ and $\displaystyle \mathrm{y}=\mathrm{b} \sin \mathrm{t}$, we have
\frac{d \mathrm{x}}{d \mathrm{t}}=-\mathrm{a} \sin \mathrm{t} \text { and } \frac{d \mathrm{y}}{d \mathrm{t}}=\mathrm{b} \cos \mathrm{t}
\therefore \frac{d \mathrm{y}}{d \mathrm{x}}=\frac{d \mathrm{y}}{d \mathrm{t}} \cdot \frac{d \mathrm{t}}{d \mathrm{x}}=\frac{\mathrm{b} \cos \mathrm{t}}{-\mathrm{a} \sin \mathrm{t}}=-\frac{\mathrm{b}}{\mathrm{a}} \cot \mathrm{t}
\frac{d^{2} \mathrm{y}}{d \mathrm{x}^{2}}=\frac{d}{d \mathrm{t}}\left(-\frac{\mathrm{b}}{\mathrm{a}} \cot \mathrm{t}\right) \cdot \frac{d \mathrm{t}}{d \mathrm{x}}
\begin{aligned}
& =\frac{\mathrm{b}}{\mathrm{a}} \operatorname{cosec}^{2} \mathrm{t} \cdot \frac{1}{-\mathrm{a} \sin \mathrm{t}} \\
& =-\frac{\mathrm{b}}{\mathrm{a}^{2}} \cdot \frac{1}{\sin ^{3} \mathrm{t}} \text { or }-\frac{\mathrm{b}}{\mathrm{a}^{2}} \operatorname{cosec}^{3} \mathrm{t}
\end{aligned}
$$
Continuity and DifferentiabilityContinuityApplyvery_short_answermedium
More from Continuity and Differentiability
- If f(x)=. is continuous at x=0, then the value of a is:2025 · asked 4×
- The value of k for which function f(x)=. is differentiable at x=0 is:2023 · asked 3×
- If y=( cos x- sin x)/( cos x+ sin x), then (d y)/(d x) is:2023 · asked 3×
- If f(x)=. is continuous at x=0, then the value of k is:2025 · asked 3×
- Differentiate sec^-1( 1√1-x^2) w.r.t. sin^-1(2 x √1-x^2). OR If y= tan x+ sec x, then prove that (d^2 y)/(d…2023 · asked 3×
- If f(x)=. is continuous at x=-1, then the value of k is:2026 · asked 3×
- The value of k for which the function f(x)= casesx^2 sin 1/x, & x ≠ 0 k(x+1), & x=0 cases is a continuous…2026 · asked 3×
- If tan ((x+y)/(x-y))=k, then (dy)/(dx) is equal to2023 · asked 3×
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.