Board Exam-Pattern Question Bank

Continuity and Differentiability is among the heaviest-weighted chapters in the CBSE paper — expect a 2-mark continuity-constant problem, a 3-mark implicit or logarithmic differentiation, and a 5-mark second-order relation proof nearly every year. The bank below mirrors that structure. Present answers the way the marking scheme rewards: state the LHL/RHL/value comparison explicitly in continuity questions, show the logarithm step before differentiating any variable power, and end relation proofs with the substituted line that visibly cancels to zero.

The questions below follow the pattern of recent board papers; they are original items written for practice.

2-mark questions

Q1. Examine the continuity of $f(x) = 2x^2 - 1$ at $x = 3$.

Solution. $\lim_{x \to 3}(2x^2 - 1) = 2(9) - 1 = 17$ and $f(3) = 17$. Limit $=$ value, so $f$ is continuous at $x = 3$.

Q2. Find the value of $k$ for which $f(x) = \begin{cases} kx + 1, & x \leq \pi \\ \cos x, & x > \pi \end{cases}$ is continuous at $x = \pi$.

Solution. LHL $= k\pi + 1$; RHL $= \cos\pi = -1$. Continuity needs $k\pi + 1 = -1$, so $k = -\dfrac{2}{\pi}$.

Q3. Find the value of $k$ for which $f(x) = \begin{cases} kx^2, & x \leq 2 \\ 3, & x > 2 \end{cases}$ is continuous at $x = 2$.

Solution. LHL $= 4k$ must equal RHL $= 3$ (and $f(2) = 4k$), so $k = \dfrac{3}{4}$.

Q4. Differentiate $\sqrt{e^{\sqrt{x}}}$, $x > 0$, w.r.t. $x$.

Solution. Write $y = e^{\sqrt{x}/2}$. Then $\frac{dy}{dx} = e^{\sqrt{x}/2}\cdot\frac{1}{2}\cdot\frac{1}{2\sqrt{x}} = \dfrac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}$.

3-mark questions

Q5. Show that $f(x) = \vert x - 5 \vert$ is continuous but not differentiable at $x = 5$.

Solution. Continuity: LHL $=$ RHL $= 0 = f(5)$. Differentiability: LHD $= \lim_{h \to 0^-}\frac{\vert h \vert}{h} = -1$ while RHD $= 1$; unequal one-sided derivatives, so $f'(5)$ does not exist. Hence continuous but not differentiable at $x = 5$. ∎

Q6. If $xy = e^{x - y}$, find $\dfrac{dy}{dx}$.

Solution. Differentiating: $y + xy' = e^{x-y}(1 - y')$, and $e^{x-y} = xy$, so $y + xy' = xy - xy\,y'$. Collecting: $y'(x + xy) = xy - y = y(x - 1)$, giving $$\frac{dy}{dx} = \frac{y(x - 1)}{x(1 + y)}$$

Q7. Differentiate $y = x^x - 2^{\sin x}$, $x > 0$, w.r.t. $x$.

Solution. For $x^x$: logarithmic differentiation gives $x^x(1 + \log x)$. For $2^{\sin x}$: chain rule on the constant-base formula gives $2^{\sin x}\cos x\log 2$. Hence $$\frac{dy}{dx} = x^x(1 + \log x) - 2^{\sin x}\cos x\log 2$$

Q8. If $x = a(\theta - \sin\theta)$ and $y = a(1 + \cos\theta)$, find $\dfrac{dy}{dx}$ and its value at $\theta = \dfrac{\pi}{2}$.

Solution. $\frac{dx}{d\theta} = a(1 - \cos\theta)$, $\frac{dy}{d\theta} = -a\sin\theta$. So $$\frac{dy}{dx} = \frac{-\sin\theta}{1 - \cos\theta} = \frac{-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}} = -\cot\frac{\theta}{2}$$ At $\theta = \frac{\pi}{2}$: $-\cot\frac{\pi}{4} = -1$.

5-mark questions

Q9. If $y = \left(\tan^{-1} x\right)^2$, show that $(x^2 + 1)^2 y_2 + 2x(x^2 + 1)y_1 = 2$.

Solution. $y_1 = \dfrac{2\tan^{-1}x}{1 + x^2}$, so $(1 + x^2)y_1 = 2\tan^{-1}x$. Differentiating this cleaned equation: $(1 + x^2)y_2 + 2x\,y_1 = \dfrac{2}{1 + x^2}$. Multiplying through by $(1 + x^2)$: $$(x^2 + 1)^2 y_2 + 2x(x^2 + 1)y_1 = 2$$ ∎

Q10. If $y = e^{a\cos^{-1}x}$, $-1 \leq x \leq 1$, show that $(1 - x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} - a^2y = 0$.

Solution. $y_1 = e^{a\cos^{-1}x}\cdot\dfrac{-a}{\sqrt{1 - x^2}} = \dfrac{-ay}{\sqrt{1 - x^2}}$. Clearing the radical: $\sqrt{1 - x^2}\,y_1 = -ay$. Differentiating: $\sqrt{1 - x^2}\,y_2 - \dfrac{x}{\sqrt{1 - x^2}}y_1 = -ay_1$. Multiplying by $\sqrt{1 - x^2}$ and using $\sqrt{1 - x^2}\,y_1 = -ay$ on the right: $$(1 - x^2)y_2 - xy_1 = -a(-ay) = a^2y \qquad\Rightarrow\qquad (1 - x^2)y_2 - xy_1 - a^2y = 0$$ ∎

Q11. If $x = a\left(\cos t + \log\tan\dfrac{t}{2}\right)$ and $y = a\sin t$, find $\dfrac{dy}{dx}$.

Solution. $\frac{dy}{dt} = a\cos t$, and $$\frac{dx}{dt} = a\left(-\sin t + \frac{1}{\tan\frac{t}{2}}\cdot\sec^2\frac{t}{2}\cdot\frac{1}{2}\right) = a\left(-\sin t + \frac{1}{\sin t}\right) = \frac{a\cos^2 t}{\sin t}$$ (using $2\sin\frac{t}{2}\cos\frac{t}{2} = \sin t$). Hence $$\frac{dy}{dx} = a\cos t \cdot \frac{\sin t}{a\cos^2 t} = \tan t$$

Q12. If $y = 3\cos(\log x) + 4\sin(\log x)$, show that $x^2y_2 + xy_1 + y = 0$.

Solution. $y_1 = \dfrac{-3\sin(\log x) + 4\cos(\log x)}{x}$, so $xy_1 = -3\sin(\log x) + 4\cos(\log x)$. Differentiating this cleaned equation: $$y_1 + xy_2 = \frac{-3\cos(\log x) - 4\sin(\log x)}{x} = \frac{-y}{x}$$ Multiplying by $x$: $xy_1 + x^2y_2 = -y$, i.e. $x^2y_2 + xy_1 + y = 0$. ∎ (Both 5-mark proofs used the clear-then-differentiate trick — it is the intended method and saves a page of quotient rules.)