CBSE Class 12 Mathematics Continuity and Differentiability Class Test Set 01

This class test on Chapter 5 Continuity and Differentiability for Class 12 Mathematics (CBSE) has 12 questions carrying 1 to 3 marks each, covering continuity, differentiability and derivatives of implicit, parametric and exponential functions. Complete worked solutions follow every question.

Class Test Questions with Solutions

Question 1. If \( f(x) = \begin{cases} \frac{\log(1 + mx) - \log(1 - nx)}{x} & \text{if } x \neq 0 \\ k & \text{if } x = 0 \end{cases} \) is continuous at x = 0, then the value of k will be:   [1]
Answer:
Given \( f(x) = \begin{cases} \frac{\log(1 + mx) - \log(1 - nx)}{x}, & x \neq 0 \\ k, & x = 0 \end{cases} \)
Since f(x) is continuous at x = 0,
So, RHL = LHL = f(0)
⇒ \( \lim_{x \to 0} \frac{\log(1 + mx) - \log(1 - nx)}{x} = k \)
⇒ \( \lim_{x \to 0} \frac{m\log(1 + mx)}{mx} + \lim_{x \to 0} \frac{n\log(1 - nx)}{-nx} = k \)   \( \left[\lim_{x \to 0} \frac{\log(1 + x)}{x} = 1\right] \)
⇒ m + n = k
∴ k = m + n

Question 2. Determine the value of 'k' for which the following function is continuous at x = 3:   [1]
\( f(x) = \begin{cases} \frac{(x+3)^2 - 36}{x-3}, & x \neq 3 \\ k, & x = 3 \end{cases} \)
Answer:
\( f(x) = \begin{cases} \frac{(x+3)^2 - 36}{x-3}, & x \neq 3 \\ k, & x = 3 \end{cases} \)
Since the function is continuous at x = 3,
So RHL = LHL = f(3) or \( \lim_{x \to 3} f(x) = f(3) \)
⇒ \( \lim_{x \to 3} \frac{(x+3)^2 - 36}{x - 3} = k \)
⇒ \( \lim_{x \to 3} \frac{(x+3)^2 - (6)^2}{x - 3} = k \)
⇒ \( \lim_{x \to 3} \frac{(x+9)(x-3)}{x - 3} = k \)
⇒ 12 = k
∴ k = 12

Question 3. If y = a sin x + b cos x, then prove that \( \frac{d^2y}{dx^2} + y = 0 \).   [2]
Answer:
y = a sin x + b cos x   …..(1)
\( \frac{dy}{dx} = a\cos x - b\sin x \)   (Differentiating w.r.t. x)
\( \frac{d^2y}{dx^2} = -a\sin x - b\cos x \)   (again differentiating w.r.t. x)
\( \frac{d^2y}{dx^2} = -(a\sin x + b\cos x) \)
\( \frac{d^2y}{dx^2} + y = 0 \)   from eq. (1)

Question 4. Consider the function \( f(x) = \begin{cases} 1 + x & \text{if } x \le 2 \\ 5 - x & \text{if } x \gt 2 \end{cases} \). Check the continuity and differentiability of f(x) at x = 2.   [2]
Answer:
Here, f(2) = 1 + 2 = 3.
\( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (1 + x) = 1 + 2 = 3 \) and \( \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (5 - x) = 5 - 2 = 3 \)
Thus, \( f(2) = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = 3 \)
⇒ f is continuous at x = 2.
Also, \( Lf'(2) = \lim_{h \to 0} \frac{f(2 - h) - f(2)}{-h} = \lim_{h \to 0} \frac{1 + (2 - h) - 3}{-h} = \lim_{h \to 0} \frac{-h}{-h} = 1 \)
and \( Rf'(2) = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h} = \lim_{h \to 0} \frac{5 - (2 + h) - 3}{h} = \lim_{h \to 0} \frac{-h}{h} = -1 \)
⇒ Lf′(2) ≠ Rf′(2) ⇒ f is not derivable (differentiable) at x = 2.

Question 5. Differentiate sin⁻¹ x with respect to \( \sqrt{x} \).   [2]
Answer:
Let z = sin⁻¹ x   (Differentiating w.r.t. x)
\( \frac{dz}{dx} = \frac{1}{\sqrt{1 - x^2}} \)   ...(i)
Let \( y = \sqrt{x} \)   (Differentiating w.r.t. x)
\( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \)   ...(ii)
From (i) and (ii),
\( \frac{dz}{dy} = \frac{\frac{dz}{dx}}{\frac{dy}{dx}} = \frac{\frac{1}{\sqrt{1 - x^2}}}{\frac{1}{2\sqrt{x}}} \)
⇒ \( \frac{dz}{dy} = \frac{2\sqrt{x}}{\sqrt{1 - x^2}} \)

Question 6. If x = e^{cos 2t} and y = e^{sin 2t}, prove that \( \frac{dy}{dx} = \frac{-y\log x}{x\log y} \).   [2]
Answer:
x = e^{cos 2t}
⇒ log x = cos 2t   .........(1)
⇒ \( \frac{1}{x}\cdot\frac{dx}{dt} = -2\sin 2t \) ⇒ \( \frac{dx}{dt} = -2x\sin 2t \)
and y = e^{sin 2t}
log y = sin 2t   .........(2)
⇒ \( \frac{1}{y}\cdot\frac{dy}{dt} = 2\cos 2t \)
⇒ \( \frac{dy}{dt} = 2y\cos 2t \)
Now \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = -\frac{2y\cos 2t}{2x\sin 2t} = -\frac{y\log x}{x\log y} \)   (from equations (1) and (2))

Question 7. If \( x\sqrt{1 + y} + y\sqrt{1 + x} = 0 \), for −1 < x < 1, prove that \( \frac{dy}{dx} = -\frac{1}{(1 + x)^2} \).   [3]
Answer:
\( x\sqrt{1 + y} + y\sqrt{1 + x} = 0 \)
⇒ \( x\sqrt{(1 + y)} = -y\sqrt{1 + x} \)   [On squaring both sides]
⇒ x²(1 + y) = y²(1 + x)
x² + x²y = y² + xy²
x² − y² + x²y − xy² = 0
(x − y)(x + y) + xy(x − y) = 0
(x − y)[x + y + xy] = 0   [x ≠ y]
x + y + xy = 0
x + y(1 + x) = 0
\( y = \frac{-x}{1 + x} \)   [Differentiating w.r.t. x]
\( \frac{dy}{dx} = -\left[\frac{(1 + x)\cdot 1 - x\cdot 1}{(1 + x)^2}\right] \)
\( = -\left[\frac{1 + x - x}{(1 + x)^2}\right] \)
\( \frac{dy}{dx} = \frac{-1}{(1 + x)^2} \)

Question 8. Let \( f(x) = \begin{cases} \frac{1 - \sin^3 x}{3\cos^2 x} & \text{if } x \lt \frac{\pi}{2} \\ a & \text{if } x = \frac{\pi}{2} \\ \frac{b(1 - \sin x)}{(\pi - 2x)^2} & \text{if } x \gt \frac{\pi}{2} \end{cases} \). If f(x) is continuous at \( x = \frac{\pi}{2} \), find a and b.   [3]
Answer:
Since f(x) is continuous at \( x = \frac{\pi}{2} \),
So, \( \lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{x \to \frac{\pi}{2}^-} f(x) = f\left(\frac{\pi}{2}\right) \)
\( \lim_{x \to \frac{\pi}{2}^+} \frac{b(1 - \sin x)}{(\pi - 2x)^2} = a \)
Put \( x = \frac{\pi}{2} + h \):
\( \lim_{h \to 0} \frac{b\left[1 - \sin\left(\frac{\pi}{2} + h\right)\right]}{\left[\pi - 2\left(\frac{\pi}{2} + h\right)\right]^2} = a \)
\( \lim_{h \to 0} b\left[\frac{1 - \cos h}{(\pi - \pi - 2h)^2}\right] = a \)
\( \lim_{h \to 0} b \times \frac{2\sin^2\frac{h}{2}}{4h^2} = \lim_{h \to 0} \frac{b}{8}\left(\frac{\sin\frac{h}{2}}{\frac{h}{2}}\right)^2 = a \)
⇒ \( \frac{b}{8} = a \)
b = 8a   …..(1)
Now
\( \lim_{x \to \frac{\pi}{2}^-} \frac{1 - \sin^3 x}{3\cos^2 x} = a \)
\( \lim_{x \to \frac{\pi}{2}^-} \frac{(1 - \sin x)(1 + \sin x + \sin^2 x)}{3(1 - \sin^2 x)} = a \)
\( \lim_{x \to \frac{\pi}{2}^-} \frac{1 + \sin x + \sin^2 x}{3(1 + \sin x)} = a \)
⇒ \( \frac{3}{3(2)} = a \)
⇒ \( a = \frac{1}{2} \)
Put the value of a in eq. (1),
\( b = 8 \times \frac{1}{2} \)
b = 4

Question 9. Discuss the continuity and differentiability of the function f(x) = |x| + |x − 1| in the interval (−1, 2).   [3]
Answer:
f(x) = |x| + |x − 1|
\( f(x) = \begin{cases} 1 - 2x, & -1 \lt x \le 0 \\ 1, & 0 \lt x \le 1 \\ 2x - 1, & 1 \lt x \lt 2 \end{cases} \)
The possible points of discontinuity and non-differentiability are x = 0 and x = 1.
(i) We have to check continuity at x = 0
\( \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 1 = 1 \)
\( \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-2x + 1) = 1 \)
f(0) = 1
So RHL = LHL = f(0)
∴ the function is continuous at x = 0
(ii) We have to check continuity at x = 1
\( \text{RHL} = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x - 1) = 1 \)
\( \text{LHL} = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 1 = 1 \)
f(1) = 1
RHL = LHL = f(1)
So the function is continuous at x = 1
(iii) We have to check differentiability at x = 0
\( \text{RHD} = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h};\quad h \gt 0 \)
\( = \lim_{h \to 0} \frac{1 - 1}{h} = 0 \)
\( \text{LHD} = \lim_{h \to 0} \frac{f(0-h) - f(0)}{-h};\quad h \gt 0 \)
\( = \lim_{h \to 0} \frac{-2(-h) + 1 - 1}{-h} = -2 \)
RHD ≠ LHD
So f(x) is not differentiable at x = 0.
(iv) We have to check differentiability at x = 1
\( \text{RHD} = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h};\quad h \gt 0 \)
\( = \lim_{h \to 0} \frac{[2(1+h) - 1] - 1}{h} = 2 \)
\( \text{LHD} = \lim_{h \to 0} \frac{f(1-h) - f(1)}{-h};\quad h \gt 0 \)
\( = \lim_{h \to 0} \frac{1 - 1}{-h} = 0 \)
RHD ≠ LHD
So f(x) is not differentiable at x = 1
Hence, f(x) is continuous at every point of the interval (−1, 2), but it is not differentiable at x = 0 and x = 1.

Question 10. Find the value of \( \frac{dy}{dx} \) if x^y + y^x = 1.   [3]
Answer:
Given x^y + y^x = 1
Let u = x^y and v = y^x
⇒ u + v = 1
\( \frac{du}{dx} + \frac{dv}{dx} = \frac{d(1)}{dx} \)   [Differentiating w.r.t. x]
⇒ \( \frac{du}{dx} + \frac{dv}{dx} = 0 \)   ....(i)
Now, u = x^y
⇒ log u = y log x   [taking log on both sides]
⇒ \( \frac{d}{dx}(\log u) = \frac{d}{dx}(y\log x) \)
⇒ \( \frac{1}{u}\frac{du}{dx} = y\frac{1}{x} + \log x\frac{dy}{dx} \)
⇒ \( \frac{du}{dx} = u\left(y\frac{1}{x} + \log x\frac{dy}{dx}\right) \)
\( \frac{du}{dx} = x^y\left(\frac{y}{x} + \log x\frac{dy}{dx}\right) \)   …..(ii)
Again, v = y^x
log v = x log y   [taking log on both sides]
\( \frac{d}{dx}\log v = \frac{d}{dx}(x\log y) \)
\( \frac{1}{v}\frac{dv}{dx} = x\frac{1}{y}\frac{dy}{dx} + \log y\cdot 1 \)
\( \frac{dv}{dx} = v\left[\frac{x}{y}\frac{dy}{dx} + \log y\right] \)
\( \frac{dv}{dx} = y^x\left[\frac{x}{y}\frac{dy}{dx} + \log y\right] \)   …..(iii)
Putting values from eq. (ii) and (iii) in eq. (i),
⇒ \( x^y\left(\frac{y}{x} + \log x\frac{dy}{dx}\right) + y^x\left(\frac{x}{y}\frac{dy}{dx} + \log y\right) = 0 \)
⇒ \( yx^{y-1} + x^y\log x\frac{dy}{dx} + xy^{x-1}\frac{dy}{dx} + y^x\log y = 0 \)
⇒ \( \frac{dy}{dx}\left(x^y\log x + xy^{x-1}\right) = -x^{y-1}y - y^x\log y \)
\( \frac{dy}{dx} = -\left(\frac{x^{y-1}y + y^x\log y}{x^y\log x + xy^{x-1}}\right) \)

Question 11. If (ax + b)e^y/x = x, then show that \( x^3\left(\frac{d^2y}{dx^2}\right) = \left(x\frac{dy}{dx} - y\right)^2 \).   [3]
Answer:
(ax + b)e^y/x = x
\( e^{y/x} = \frac{x}{ax + b} \)   .......(1)
\( \frac{y}{x} = \left[\log x - \log(ax + b)\right] \)   taking log on both sides
Differentiating w.r.t. x,
\( \frac{x\cdot\frac{dy}{dx} - y}{x^2} = \frac{1}{x} - \frac{a}{ax + b} \)
\( \frac{x\cdot\frac{dy}{dx} - y}{x^2} = \frac{ax + b - ax}{x(ax + b)} \)
\( x\frac{dy}{dx} - y = \frac{bx}{(ax + b)} \)
From equation (1),
\( x\frac{dy}{dx} - y = be^{y/x} \)   .........(2)
Again differentiating w.r.t. x,
\( x\cdot\frac{d^2y}{dx^2} + \frac{dy}{dx} - \frac{dy}{dx} = \frac{be^{y/x}\left(x\cdot\frac{dy}{dx} - y\right)}{x^2} \)
From equation (2),
\( x^3\left(\frac{d^2y}{dx^2}\right) = \left(x\frac{dy}{dx} - y\right)^2 \)

Question 12. Differentiate \( \tan^{-1}\left(\frac{a\cos x - b\sin x}{b\cos x + a\sin x}\right) \) with respect to x.   [3]
Answer:
Let \( z = \tan^{-1}\left(\frac{a\cos x - b\sin x}{b\cos x + a\sin x}\right) \)
Put a = r sin α and b = r cos α, so that \( \tan\alpha = \frac{a}{b} \), i.e. \( \alpha = \tan^{-1}\frac{a}{b} \) (a constant).
⇒ \( z = \tan^{-1}\left(\frac{r\sin\alpha\cos x - r\cos\alpha\sin x}{r\cos\alpha\cos x + r\sin\alpha\sin x}\right) \)
⇒ \( z = \tan^{-1}\left(\frac{\sin(\alpha - x)}{\cos(\alpha - x)}\right) = \tan^{-1}\left[\tan(\alpha - x)\right] \)
⇒ \( z = \alpha - x = \tan^{-1}\frac{a}{b} - x \)   [Differentiating w.r.t. x]
⇒ \( \frac{dz}{dx} = 0 - 1 = -1 \)