Class 12 Continuity and Differentiability Mock Test (CBSE)

Topics in Continuity and Differentiability

• Continuity of a function
• Algebra of continuous functions
• Differentiability
• Chain rule
• Derivatives of inverse trigonometric functions
• Exponential and logarithmic functions
• Logarithmic differentiation
• Derivatives of implicit and parametric functions
• Second order derivatives

Continuity and Differentiability practice questions with answers

1. Q1Multiple choice·1 mark
   The concentration of a drug in the blood, in mg/L, $t$ hours after an injection is $C(t) = 20e^{-0.5t}$. The rate of change of the concentration at $t = 0$, in mg/L per hour, is:

   1. (a)$10$
   2. (b)$-20$
   3. (c)$-0.5$
   4. (d)$-10$
   Show answer and step marks

   1. Correct option: (d) $-10$, since $C'(t) = -10e^{-0.5t}$ and $C'(0) = -10$.+1

   Answer: (d)
2. Q2Assertion and reason·1 mark
   Assertion (A): The derivative of $\sin(x^2)$ with respect to $x$ is $2x\cos(x^2)$. Reason (R): The derivative of $\sin x$ with respect to $x$ is $\cos x$.

   1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
   2. (b)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
   3. (c)Assertion (A) is true but Reason (R) is false.
   4. (d)Assertion (A) is false but Reason (R) is true.
   Show answer and step marks

   1. Correct option: (b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A), since (A) needs the chain rule and (R) only states a basic derivative.+1

   Answer: (b)
3. Q3Multiple choice·1 mark
   If $x = a(\theta - \sin\theta)$ and $y = a(1 + \cos\theta)$, then the value of $\frac{dy}{dx}$ at $\theta = \frac{\pi}{2}$ is:

   1. (a)$1$
   2. (b)$0$
   3. (c)$-1$
   4. (d)$\frac{1}{\sqrt{3}}$
   Show answer and step marks

   1. Correct option: (c) $-1$. Explanation: $\frac{dx}{d\theta} = a(1 - \cos\theta)$ and $\frac{dy}{d\theta} = -a\sin\theta$, so $\frac{dy}{dx} = \frac{-a\sin\theta}{a(1 - \cos\theta)} = -\cot\left(\frac{\theta}{2}\right)$. At $\theta = \frac{\pi}{2}$, $\frac{dy}{dx} = -\cot\left(\frac{\pi}{4}\right) = -1$.+1

   Answer: (c)
4. Q4Multiple choice·1 mark
   If the function $f(x)$ defined by $f(x) = \begin{cases} kx + 1, & x \le \pi \\ \cos x, & x > \pi \end{cases}$ is continuous at $x = \pi$, then the value of $k$ is:

   1. (a)$-\frac{2}{\pi}$
   2. (b)$\frac{2}{\pi}$
   3. (c)$-1$
   4. (d)$0$
   Show answer and step marks

   1. Correct option: (a) $-\frac{2}{\pi}$. For continuity at $x = \pi$, $\lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^+} f(x) = f(\pi)$. Thus, $k\pi + 1 = \cos\pi \implies k\pi + 1 = -1 \implies k\pi = -2 \implies k = -\frac{2}{\pi}$.+1

   Answer: (a)
5. Q5Multiple choice·1 mark
   If $y = x^x$ for $x > 0$, then $\frac{dy}{dx}$ is:

   1. (a)$x^x(1 + \ln x)$
   2. (b)$x^x \ln x$
   3. (c)$x \cdot x^{x-1}$
   4. (d)$x^x(1 - \ln x)$
   Show answer and step marks

   1. Correct option: (a) $x^x(1 + \ln x)$, from taking logarithms: $\ln y = x\ln x$, so $\frac{1}{y}\frac{dy}{dx} = 1 + \ln x$.+1

   Answer: (a)
6. Q6Long answer·5 marks
   Let $y = \sin^{-1}\left(\sqrt{1 - x^2}\right)$, where $0 < x < 1$. (a) Show that $y = \cos^{-1} x$ for $0 < x < 1$. [2] (b) Find $\frac{dy}{dx}$ for $0 < x < 1$. [2] (c) Find the value of $\frac{dy}{dx}$ at $x = \frac{1}{2}$. [1]
   Show answer and step marks

   1. Put $x = \cos\theta$ with $\theta \in (0, \frac{\pi}{2})$, since $0 < x < 1$; then $\sqrt{1 - x^2} = \sin\theta$.+1
   2. So $y = \sin^{-1}(\sin\theta) = \theta = \cos^{-1}x$, as $\theta$ lies in the principal range of $\sin^{-1}$ and $\cos^{-1}x = \theta$.+1
   3. Since $y = \cos^{-1}x$, $\frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}}$ by the standard derivative of $\cos^{-1}x$.+1
   4. Hence $\frac{dy}{dx} = -\frac{1}{\sqrt{1-x^2}}$ holds for $0 < x < 1$.+1
   5. At $x = \frac{1}{2}$, $\sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$, so $\frac{dy}{dx} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$.+1

   Answer: (a) $y = \cos^{-1}x$ is shown. (b) $\frac{dy}{dx} = -\frac{1}{\sqrt{1-x^2}}$. (c) $\frac{dy}{dx}\Big|_{x=1/2} = -\frac{2\sqrt{3}}{3}$.

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