CBSE Class 12 Mathematics Continuity and Differentiability Important Questions Set 01

Choose and write the correct option in the following questions.

Question. The function \( f : R \to R \) given by \( f(x) = - |x - 1| \) is [C
(a) continuous as well as differentiable at \( x = 1 \)
(b) not continuous but differentiable at \( x = 1 \)
(c) continuous but not differentiable at \( x = 1 \)
(d) neither continuous nor differentiable at \( x = 1 \)
Answer: (c)

Question. The function \( f(x) = e^{|x|} \) is 
(a) continuous everywhere but not differentiable at \( x = 0 \)
(b) continuous and differentiable everywhere
(c) not continuous at \( x = 0 \)
(d) none of these
Answer: (a)

Question. The function \( f(x) = [x] \), where \( [x] \) denotes the greatest integer function, is continuous at
(a) 4
(b) –2
(c) 1
(d) 1.5
Answer: (d)

Question. The number of points at which the function \( f(x) = \frac{1}{x - [x]} \) is not continuous is 
(a) 1
(b) 2
(c) 3
(d) none of these
Answer: (d)

Question. The function \( f(x) = \begin{cases} \frac{\sin x}{x} + \cos x, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases} \) is continuous at \( x = 0 \), then the value of \( k \) is
(a) 3
(b) 2
(c) 1
(d) 1.5
Answer: (b)

Question. The value of \( k \) which makes the function defined by \( f(x) = \begin{cases} \sin \frac{1}{x}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases} \), continuous at \( x = 0 \) is
(a) 8
(b) 1
(c) –1
(d) none of these
Answer: (d)

Question. The function \( f(x) = \cot x \) is discontinuous on the set 
(a) \( \{x = n\pi : n \in Z\} \)
(b) \( \{x = 2n\pi : n \in Z\} \)
(c) \( \{x = (2n + 1)\frac{\pi}{2} ; n \in Z\} \)
(d) \( \{x = \frac{n\pi}{2} ; n \in Z\} \)
Answer: (a)

Question. Let \( f(x) = |\sin x| \). Then 
(a) \( f \) is everywhere differentiable
(b) \( f \) is everywhere continuous but not differentiable at \( x = n\pi : n \in Z \)
(c) \( f \) is everywhere continuous but not differentiable at \( x = (2n + 1)\frac{\pi}{2}, n \in Z \)
(d) none of these
Answer: (b)

Question. The function \( f(x) = \frac{x - 1}{x(x^2 - 1)} \) is discontinuous at
(a) exactly one point
(b) exactly two points
(c) exactly three points
(d) no point
Answer: (c)

Question. If \( f(x) = x^2 \sin \frac{1}{x} \), where \( x \neq 0 \), then the value of the function \( f \) at \( x = 0 \), so that the function is continuous at \( x = 0 \), is 
(a) 0
(b) –1
(c) 1
(d) None of these
Answer: (a)

Question. The function \( f(x) = |x| + |x - 1| \) is
(a) continuous at \( x = 0 \) as well as at \( x = 1 \).
(b) continuous at \( x = 1 \) but not at \( x = 0 \).
(c) discontinuous at \( x = 0 \) as well as at \( x = 1 \).
(d) continuous at \( x = 0 \) but not at \( x = 1 \).
Answer: (a)

Question. The function \( f(x) = \frac{4 - x^2}{4x - x^3} \) is
(a) discontinuous at only one point
(b) discontinuous at exactly two points
(c) discontinuous at exactly three points
(d) none of these
Answer: (c)

Question. The value of \( c \) in Rolle’s Theorem for the function \( f(x) = e^x \sin x \), in \( [0, \pi] \) is 
(a) \( \frac{\pi}{6} \)
(b) \( \frac{\pi}{4} \)
(c) \( \frac{\pi}{2} \)
(d) \( \frac{3\pi}{4} \)
Answer: (d)

Question. The value of \( c \) in Mean Value Theorem for the function \( f(x) = x(x - 2), x \in [1, 2] \) is
(a) \( \frac{3}{2} \)
(b) \( \frac{2}{3} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{7}{4} \)
Answer: (a)

Question. The value of \( c \) in Rolle’s theorem for the function \( f(x) = x^3 - 3x \) in the interval \( [0, \sqrt{3}] \) is
(a) 1
(b) –1
(c) \( \frac{3}{2} \)
(d) \( \frac{1}{3} \)
Answer: (a)

Question. The set of points where the functions \( f \) given by \( f(x) = |x - 3| \cos x \) is differentiable is
(a) \( R \)
(b) \( R - \{3\} \)
(c) \( (0, \infty) \)
(d) none of these
Answer: (b)

Question. Differential coefficient of \( \sec (\tan^{-1}x) \) w.r.t. \( x \) is 
(a) \( \frac{x}{\sqrt{1 + x^2}} \)
(b) \( \frac{x}{1 + x^2} \)
(c) \( x\sqrt{1 + x^2} \)
(d) \( \frac{1}{\sqrt{1 + x^2}} \)
Answer: (a)

Question. If \( u = \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \) and \( v = \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \), then \( \frac{du}{dv} \) is
(a) \( \frac{1}{2} \)
(b) \( x \)
(c) \( \frac{1 - x^2}{1 + x^2} \)
(d) 1
Answer: (d)

Question. If \( y = \log \sqrt{\tan x} \), then the value of \( \frac{dy}{dx} \) at \( x = \frac{\pi}{4} \) is
(a) 0
(b) 1
(c) \( \frac{1}{2} \)
(d) \( \infty \)
Answer: (b)

Question. If \( y = \sqrt{\sin x + y} \), then \( \frac{dy}{dx} \) is equal to
(a) \( \frac{\cos x}{2y - 1} \)
(b) \( \frac{\cos x}{1 - 2y} \)
(c) \( \frac{\sin x}{1 - 2y} \)
(d) \( \frac{\sin x}{2y - 1} \)
Answer: (a)

CONTINUITY AND DIFFERENTIABILITY

Question. Find the values of \( p \) and \( q \), for which \[ f(x) = \begin{cases} \frac{1 - \sin^3 x}{3 \cos^2 x} , & \text{if } x < \frac{\pi}{2} \\ p , & \text{if } x = \frac{\pi}{2} \\ \frac{q(1 - \sin x)}{(\pi - 2x)^2} , & \text{if } x > \frac{\pi}{2} \end{cases} \] is continuous at \( x = \frac{\pi}{2} \). 
Answer: We have, \[ f(x) = \begin{cases} \frac{1 - \sin^3 x}{3 \cos^2 x} , & \text{if } x < \frac{\pi}{2} \\ p , & \text{if } x = \frac{\pi}{2} \\ \frac{q(1 - \sin x)}{(\pi - 2x)^2} , & \text{if } x > \frac{\pi}{2} \end{cases} \] is continuous at \( x = \frac{\pi}{2} \).
Now, \( \lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{h \to 0} f\left(\frac{\pi}{2} + h\right) \) \( \left[ \text{Let } x = \frac{\pi}{2} + h, x \to \frac{\pi}{2}^+ \Rightarrow h \to 0 \right] \)
\( = \lim_{h \to 0} \frac{q\left\{1 - \sin\left(\frac{\pi}{2} + h\right)\right\}}{\left\{\pi - 2\left(\frac{\pi}{2} + h\right)\right\}^2} = \lim_{h \to 0} \frac{q\{1 - \cos h\}}{\{\pi - \pi - 2h\}^2} = \lim_{h \to 0} \frac{q(1 - \cos h)}{4h^2} \)
\( = \lim_{h \to 0} \frac{q. 2 \sin^2 \frac{h}{2}}{4h^2} = \lim_{h \to 0} \frac{q. \sin^2 \frac{h}{2}}{2h^2} = q. \lim_{h \to 0} \left[ \frac{\sin \frac{h}{2}}{\frac{h}{2}} \right]^2 \times \frac{1}{8} = \frac{q}{8} \)
Again \( \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{h \to 0} f\left(\frac{\pi}{2} - h\right) \) \( \left[ \text{Let } x = \frac{\pi}{2} - h, x \to \frac{\pi}{2}^- \Rightarrow h \to 0 \right] \)
\( = \lim_{h \to 0} \frac{1 - \sin^3 \left(\frac{\pi}{2} - h\right)}{3 \cos^2 \left(\frac{\pi}{2} - h\right)} = \lim_{h \to 0} \frac{1 - \cos^3 h}{3 \sin^2 h} = \lim_{h \to 0} \frac{(1 - \cos h)(1 + \cos h + \cos^2 h)}{3 \sin^2 h} \)
\( = \lim_{h \to 0} \frac{2 \sin^2 \frac{h}{2} \cdot (1 + 1 + 1)}{3 \sin^2 h} = \lim_{h \to 0} \frac{2 \sin^2 \frac{h}{2} \cdot 3}{3 \sin^2 h} = \lim_{h \to 0} \frac{2 \sin^2 \frac{h}{2}}{\sin^2 h} \)
Dividing \( N^r \) and \( D^r \) by \( h^2 \), we get
\( = \lim_{h \to 0} \frac{2. \frac{\sin^2 \frac{h}{2}}{h^2}}{\frac{\sin^2 h}{h^2}} = \lim_{h \to 0} \frac{2. \frac{\sin^2 \frac{h}{2}}{\frac{h^2}{4} \times 4}}{\frac{\sin^2 h}{h^2}} = \frac{1}{2} \frac{\left[ \lim_{\frac{h}{2} \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \right]^2}{\left[ \lim_{h \to 0} \frac{\sin h}{h} \right]^2} = \frac{1}{2} \)
Also \( f\left(\frac{\pi}{2}\right) = p \)
\( \because f(x) \) is continuous at \( x = \frac{\pi}{2} \Rightarrow \lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{x \to \frac{\pi}{2}^-} f(x) = f\left(\frac{\pi}{2}\right) \)
\( \Rightarrow \frac{q}{8} = \frac{1}{2} = p \Rightarrow p = \frac{1}{2} \text{ and } q = 4 \)

Question. Show that the function \( f(x) = 2x - |x| \) is continuous but not differentiable at \( x = 0 \). 
Answer: Here \( f(x) = 2x - |x| \)
For continuity at \( x = 0 \)
\( \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h) = \lim_{h \to 0} f(h) \)
\( = \lim_{h \to 0} \{2h - |h|\} = \lim_{h \to 0} (2h - h) \)
\( = \lim_{h \to 0} h \)
\( = 0 \) ...(i)
\( \lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0 - h) = \lim_{h \to 0} f(-h) \)
\( = \lim_{h \to 0} \{2(-h) - |-h|\} = \lim_{h \to 0} \{-2h - h\} \)
\( = \lim_{h \to 0} (-3h) \)
\( = 0 \) ...(ii)
Also, \( f(0) = 2 \times 0 - |0| = 0 \) ...(iii)
(i), (ii) and (iii) \( \Rightarrow \lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = f(0) \)
Hence, \( f(x) \) is continuous at \( x = 0 \).
For differentiability at \( x = 0 \)
LHD \( = \lim_{h \to 0} \frac{f(0 - h) - f(0)}{-h} = \lim_{h \to 0} \frac{f(-h) - f(0)}{-h} \)
\( = \lim_{h \to 0} \frac{\{2(-h) - |-h|\} - \{2 \times 0 - |0|\}}{-h} = \lim_{h \to 0} \frac{-2h - h - 0}{-h} \)
\( = \lim_{h \to 0} \frac{-3h}{-h} = \lim_{h \to 0} 3 \)
LHD \( = 3 \) ...(iv)
Again RHD \( = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} \)
\( = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{2h - |h| - \{2 \times 0 - |0|\}}{h} = \lim_{h \to 0} \frac{2h - h}{h} = \lim_{h \to 0} \frac{h}{h} \)
\( = \lim_{h \to 0} 1 \)
RHD \( = 1 \) ...(v)
From (iv) and (v), we get
LHD \(\neq\) RHD i.e., function \( f(x) = 2x - |x| \) is not differentiable at \( x = 0 \).
Hence, \( f(x) \) is continuous but not differentiable at \( x = 0 \).

Question. Find the value of ‘\( a \)’ for which the function \( f \) defined as \[ f(x) = \begin{cases} a \sin \frac{\pi}{2}(x + 1) , & x \le 0 \\ \frac{\tan x - \sin x}{x^3} , & x > 0 \end{cases} \] is continuous at \( x = 0 \).
Answer: \( \because f(x) \) is continuous at \( x = 0 \).
\( \Rightarrow \) (LHL of \( f(x) \) at \( x = 0 \)) = (RHL of \( f(x) \) at \( x = 0 \)) = \( f(0) \) ...(i)
Now, \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} a \sin \frac{\pi}{2}(x + 1) \) \( \left[ \because f(x) = a \sin \frac{\pi}{2}(x + 1), \text{if } x \le 0 \right] \)
\( = \lim_{x \to 0^-} a \sin \left( \frac{\pi x}{2} + \frac{\pi}{2} \right) = \lim_{x \to 0^-} a \cos \frac{\pi x}{2} = a. \cos 0 = a \) ...(ii)
Again, \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} \) \( \left[ \because f(x) = \frac{\tan x - \sin x}{x^3}, \text{if } x > 0 \right] \)
\( = \lim_{x \to 0} \frac{\frac{\sin x}{\cos x} - \sin x}{x^3} = \lim_{x \to 0} \frac{\sin x - \sin x \cdot \cos x}{\cos x \cdot x^3} = \lim_{x \to 0} \frac{\sin x(1 - \cos x)}{\cos x \cdot x^3} \)
\( = \lim_{x \to 0} \frac{1}{\cos x} \cdot \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{2 \sin^2 \frac{x}{2}}{\frac{x^2}{4} \times 4} \) \( \left[ \because 1 - \cos x = 2 \sin^2 \frac{x}{2} \right] \)
\( = \frac{1}{1} \cdot 1 \cdot \frac{1}{2} \lim_{\frac{x}{2} \to 0} \left[ \frac{\sin \frac{x}{2}}{\frac{x}{2}} \right]^2 = \frac{1}{2} \cdot \left[ \lim_{\frac{x}{2} \to 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}} \right]^2 = \frac{1}{2} \times 1 = \frac{1}{2} \) ...(iii)
Also, \( f(0) = a \sin \frac{\pi}{2}(0 + 1) = a \sin \frac{\pi}{2} = a \) ...(iv)
\( \because f \) is continuous at \( x = 0 \).
\( \therefore \) (i), (ii), (iii) and (iv) \( \Rightarrow a = \frac{1}{2} \)

Question. If \( f(x) = \begin{cases} \frac{\sin (a + 1)x + 2\sin x}{x} , & x < 0 \\ 2 , & x = 0 \\ \frac{\sqrt{1 + bx} - 1}{x} , & x > 0 \end{cases} \) is continuous at \( x = 0 \), then find the values of \( a \) and \( b \). 
Answer: We have, \( f(x) = \begin{cases} \frac{\sin (a + 1)x + 2\sin x}{x} , & x < 0 \\ 2 , & x = 0 \\ \frac{\sqrt{1 + bx} - 1}{x} , & x > 0 \end{cases} \) is continuous at \( x = 0 \)
Since, \( f(x) \) is continuous at \( x = 0 \)
\( \Rightarrow \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \) ... (i)
Now, \( \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h) \) [Let \( x = 0 + h \), \( h \) is +ve small quantity \( x \to 0^+ \Rightarrow h \to 0 \)]
\( = \lim_{h \to 0} f(h) = \lim_{h \to 0} \frac{\sqrt{1 + bh} - 1}{h} = \lim_{h \to 0} \frac{\sqrt{1 + bh} - 1}{h} \times \frac{\sqrt{1 + bh} + 1}{\sqrt{1 + bh} + 1} \)
\( = \lim_{h \to 0} \frac{1 + bh - 1}{h(\sqrt{1 + bh} + 1)} = \lim_{h \to 0} \frac{bh}{h(\sqrt{1 + bh} + 1)} = \lim_{h \to 0} \frac{b}{\sqrt{1 + bh} + 1} = \frac{b}{2} \)
Again \( \lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0 - h) \) [Let \( x = 0 - h \), \( h \) is +ve small quantity \( x \to 0^- \Rightarrow h \to 0 \)]
\( = \lim_{h \to 0} f(-h) = \lim_{h \to 0} \frac{\sin(a + 1)(-h) + 2\sin(-h)}{-h} \)
\( = \lim_{h \to 0} \frac{-\sin(a + 1)h - 2 \sin h}{-h} = \lim_{h \to 0} \left[ \frac{\sin(a + 1)h}{h} + \frac{2 \sin h}{h} \right] \)
\( = \lim_{h \to 0} \left[ \frac{\sin(a + 1)h}{(a + 1)h} \times (a + 1) + 2 \frac{\sin h}{h} \right] \)
\( = 1 \times (a + 1) + 2 = a + 3 \)
Also \( f(0) = 2 \)
Now from (i) \( \frac{b}{2} = a + 3 = 2 \Rightarrow b = 4, a = -1 \)

Question. Show that the function \( f(x) = |x - 3|, x \in \mathbb{R} \), is continuous but not differentiable at \( x = 3 \).
Answer: Here, \( f(x) = |x - 3| \Rightarrow f(x) = \begin{cases} -(x - 3) , & x < 3 \\ 0 , & x = 3 \\ (x - 3) , & x > 3 \end{cases} \)
For Continuity:
Now, \( \lim_{x \to 3^+} f(x) = \lim_{h \to 0} f(3 + h) \) [Let \( x = 3 + h \) and \( x \to 3^+ \Rightarrow h \to 0 \)]
\( = \lim_{h \to 0} (3 + h - 3) = \lim_{h \to 0} h = 0 \) ...(i)
\( \lim_{x \to 3^-} f(x) = \lim_{h \to 0} f(3 - h) \) [Let \( x = 3 - h \) and \( x \to 3^- \Rightarrow h \to 0 \)]
\( = \lim_{h \to 0} - (3 - h - 3) = \lim_{h \to 0} h = 0 \) ...(ii)
Also, \( f(3) = 0 \) ...(iii)
From equation (i), (ii) and (iii), we get \( \lim_{x \to 3^+} f(x) = \lim_{x \to 3^-} f(x) = f(3) \)
Hence, \( f(x) \) is continuous at \( x = 3 \).
For Differentiability:
RHD \( = \lim_{h \to 0} \frac{f(3 + h) - f(3)}{h} = \lim_{h \to 0} \frac{(3 + h - 3) - 0}{h} = \lim_{h \to 0} \frac{h}{h} = \lim_{h \to 0} 1 = 1 \) ...(iv)
LHD \( = \lim_{h \to 0} \frac{f(3 - h) - f(3)}{-h} = \lim_{h \to 0} \frac{-(3 - h - 3) - 0}{-h} = \lim_{h \to 0} \frac{h}{-h} = \lim_{h \to 0} (-1) = -1 \) ...(v)
Equation (iv) and (v) \( \Rightarrow \) RHD \(\neq\) LHD at \( x = 3 \).
Hence, \( f(x) \) is not differentiable at \( x = 3 \).
Therefore, \( f(x) = |x - 3|, x \in \mathbb{R} \) is continuous but not differentiable at \( x = 3 \).

Question. Discuss the continuity and differentiability of the function \( f(x) = |x| + |x - 1| \) in the interval \((-1, 2)\). 
Answer: Given function is \( f(x) = |x| + |x - 1| \)
Function is also written as
\( f(x) = \begin{cases} - x - (x - 1), & \text{if } -1 < x < 0 \\ x - (x - 1), & \text{if } 0 \le x < 1 \\ x + (x - 1), & \text{if } x \ge 1 \end{cases} \Rightarrow f(x) = \begin{cases} -2x + 1, & \text{if } x < 0 \\ 1, & \text{if } 0 \le x < 1 \\ 2x - 1, & \text{if } x \ge 1 \end{cases} \)
Obviously, in given function we need to discuss the continuity and differentiability of the function \( f(x) \) at \( x = 0 \) or 1 only.
For continuity at \( x = 0 \)
\( \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h) \) [Let \( x = 0 + h \) and \( x \to 0^+ \Rightarrow h \to 0 \)]
\( = \lim_{h \to 0} f(h) = \lim_{h \to 0} 1 = 1 \) [\( \because h \) is very small positive quantity]
\( = 1 \) ...(i)
\( \lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0 - h) \) [Let \( x = 0 - h \) and \( x \to 0^- \Rightarrow h \to 0 \)]
\( = \lim_{h \to 0} f(-h) = \lim_{h \to 0} \{-2(-h) + 1\} = \lim_{h \to 0} (2h + 1) = 1 \) ...(ii)
Also, \( f(0) = 1 \) ...(iii)
(i), (ii) and (iii) \( \Rightarrow \lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = f(0) \)
Hence, \( f(x) \) is continuous at \( x = 0 \).
For differentiability at \( x = 0 \)
RHD \( = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0} \frac{1 - 1}{h} = \lim_{h \to 0} \frac{0}{h} = 0 \) [\( \because |h|= h, |0|= 0 \)]
RHD \( = 0 \) ...(iv)
LHD \( = \lim_{h \to 0} \frac{f(0 - h) - f(0)}{-h} = \lim_{h \to 0} \frac{-2(-h) + 1 - 1}{-h} = \lim_{h \to 0} \frac{2h}{-h} = \lim_{h \to 0} (-2) \)
LHD \( = - 2 \) ...(v)
(iv) and (v) \( \Rightarrow \) RHD \(\neq\) LHD at \( x = 0 \).
Hence, \( f(x) \) is not differentiable at \( x = 0 \) but continuous at \( x = 0 \).
Similarly, we can prove \( f(x) \) is not differentiable at \( x = 1 \) but continuous at \( x = 1 \). (Do yourself)

Question. Find ‘\( a \)’ and ‘\( b \)’ if the function given by \( f(x) = \begin{cases} ax^2 + b, & \text{if } x < 1 \\ 2x + 1, & \text{if } x \ge 1 \end{cases} \) is differentiable at \( x = 1 \). 
Answer: Since, \( f \) is differentiable at 1 \( \Rightarrow \) \( f \) is also continuous at 1.
Now \( \lim_{x \to 1^+} f(x) = \lim_{h \to 0} f(1 + h) \) [Here \( h \) is +ve and very small quantity]
\( = \lim_{h \to 0} \{2(1 + h) + 1\} = 2 + 1 = 3 \)
\( \lim_{x \to 1^-} f(x) = \lim_{h \to 0} f(1 - h) = \lim_{h \to 0} \{a(1 - h)^2 + b\} = a + b \)
Since \( f(x) \) is continuous at \( x = 1 \)
\( \Rightarrow a + b = 3 \) ...(i)
Again, since \( f \) is differentiable
\( \Rightarrow \) LHD (at \( x = 1 \)) = RHD (at \( x = 1 \))
\( \Rightarrow \lim_{h \to 0} \frac{f(1 - h) - f(1)}{-h} = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} \)
\( \Rightarrow \lim_{h \to 0} \frac{a(1 - h)^2 + b - 3}{-h} = \lim_{h \to 0} \frac{2(1 + h) + 1 - 3}{h} \)
\( \Rightarrow \lim_{h \to 0} \frac{a - 2ah + ah^2 + b - 3}{-h} = \lim_{h \to 0} \frac{2 + 2h + 1 - 3}{h} \)
\( \Rightarrow \lim_{h \to 0} \frac{(a + b - 3) - 2ah + ah^2}{-h} = \lim_{h \to 0} \frac{2h}{h} \)
\( \Rightarrow \lim_{h \to 0} \frac{0 - 2ah + ah^2}{-h} = 2 \) [From equation (i)]
\( \Rightarrow \lim_{h \to 0} \frac{ah(h - 2)}{-h} = 2 \Rightarrow 2a = 2 \Rightarrow a = 1 \Rightarrow b = 2 \) [From equation (i)]

DERIVATIVES

Question. If \( \tan^{-1}\left(\frac{y}{x}\right) = \log\sqrt{x^2 + y^2} \), prove that \( \frac{dy}{dx} = \frac{x + y}{x - y} \). 
Answer: Given, \( \tan^{-1}\left(\frac{y}{x}\right) = \log\sqrt{x^2 + y^2} \)
\( \Rightarrow \tan^{-1}\left(\frac{y}{x}\right) = \frac{1}{2} \log(x^2 + y^2) \)
Differentiating w.r.t \( x \), we have
\( \Rightarrow \frac{1}{1 + \left(\frac{y}{x}\right)^2} \times \left\{ \frac{x \frac{dy}{dx} - y \times 1}{x^2} \right\} = \frac{1}{2} \times \frac{1}{x^2 + y^2} \times \left(2x + 2y \frac{dy}{dx}\right) \)
\( \Rightarrow \frac{x^2}{x^2 + y^2} \times \frac{x \frac{dy}{dx} - y}{x^2} = \frac{1}{x^2 + y^2} \left(x + y \frac{dy}{dx}\right) \)
\( \Rightarrow x \frac{dy}{dx} - y = x + y \frac{dy}{dx} \)
\( \Rightarrow (x - y) \frac{dy}{dx} = x + y \Rightarrow \frac{dy}{dx} = \frac{x + y}{x - y} \)

Question. Differentiate \( \tan^{-1} \frac{3x - x^3}{1 - 3x^2}, |x| < \frac{1}{\sqrt{3}} \) w.r.t. \( \tan^{-1} \frac{x}{\sqrt{1 - x^2}} \). 
Answer: Let \( y = \tan^{-1} \frac{3x - x^3}{1 - 3x^2} = \tan^{-1} \left( \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \right) \), put \( x = \tan \theta \)
\( y = \tan^{-1}(\tan 3\theta) = 3\theta = 3 \tan^{-1} x \)
\( \therefore \frac{dy}{dx} = \frac{3}{1 + x^2} \)
And let \( t = \tan^{-1} \frac{x}{\sqrt{1 - x^2}} = \tan^{-1} \left( \frac{\sin \phi}{\sqrt{1 - \sin^2 \phi}} \right) \), put \( x = \sin \phi \)
\( t = \tan^{-1} \left( \frac{\sin \phi}{\cos \phi} \right) = \tan^{-1}(\tan \phi) = \phi \)
\( t = \sin^{-1} x \Rightarrow \frac{dt}{dx} = \frac{1}{\sqrt{1 - x^2}} \)
\( \therefore \frac{dy}{dt} = \frac{\frac{dy}{dx}}{\frac{dt}{dx}} = \frac{\frac{3}{1 + x^2}}{\frac{1}{\sqrt{1 - x^2}}} = \frac{3\sqrt{1 - x^2}}{1 + x^2} \)