CBSE Class 12 Mathematics HOTs Continuity And Differentiability Set 05

Question. Find the values of \( p \) and \( q \), for which \[ f(x) = \begin{cases} \frac{1 - \sin^3 x}{3\cos^2 x}, & \text{if } x < \pi/2 \\ p, & \text{if } x = \pi/2 \\ \frac{q(1 - \sin x)}{(\pi - 2x)^2}, & \text{if } x > \pi/2 \end{cases} \] is continuous at \( x = \pi/2 \).
Answer: Since \( f(x) \) is continuous at \( \pi/2 \), we have: \[ \lim_{x \to \pi/2^-} f(x) = \lim_{x \to \pi/2^+} f(x) = f(\pi/2) \quad \dots(1) \] Now, computing the Left-Hand Limit (LHL): \[ \lim_{x \to \pi/2^-} f(x) = \lim_{h \to 0} f\left(\frac{\pi}{2} - h\right) \] \[ = \lim_{h \to 0} \frac{1 - \sin^3\left(\frac{\pi}{2} - h\right)}{3\cos^2\left(\frac{\pi}{2} - h\right)} = \lim_{h \to 0} \frac{1 - \cos^3 h}{3\sin^2 h} \] \[ = \lim_{h \to 0} \frac{(1 - \cos h)(1 + \cos^2 h + \cos h)}{3(1 - \cos^2 h)} \] \[ = \lim_{h \to 0} \frac{(1 - \cos h)(1 + \cos^2 h + \cos h)}{3(1 - \cos h)(1 + \cos h)} \] \[ = \lim_{h \to 0} \frac{1 + \cos^2 h + \cos h}{3(1 + \cos h)} = \frac{1 + 1^2 + 1}{3(1 + 1)} = \frac{3}{6} = \frac{1}{2} \] Now, computing the Right-Hand Limit (RHL): \[ \lim_{x \to \pi/2^+} f(x) = \lim_{h \to 0} f\left(\frac{\pi}{2} + h\right) \] \[ = \lim_{h \to 0} \frac{q\left[1 - \sin\left(\frac{\pi}{2} + h\right)\right]}{\left[\pi - 2\left(\frac{\pi}{2} + h\right)\right]^2} = \lim_{h \to 0} \frac{q(1 - \cos h)}{4h^2} \] \[ = \lim_{h \to 0} \frac{q \cdot 2\sin^2\left(\frac{h}{2}\right)}{4h^2} = \frac{q}{2} \lim_{h \to 0} \frac{\sin^2\left(\frac{h}{2}\right)}{4 \left(\frac{h}{2}\right)^2} = \frac{q}{8} \lim_{h \to 0} \left(\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right)^2 = \frac{q}{8} \] We also have \( f(\pi/2) = p \). Using equation (1): \[ \frac{1}{2} = p = \frac{q}{8} \] \[ \Rightarrow p = \frac{1}{2} \text{ and } q = 4 \] end p

Question. Find the value of the constant \( k \) so that the function \( f \), defined below, is continuous at \( x = 0 \), where \[ f(x) = \begin{cases} \frac{1 - \cos 4x}{8x^2}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases} \]
Answer: Since \( f(x) \) is continuous at \( x = 0 \), we have: \[ f(0) = \lim_{x \to 0} f(x) \] Here, \( f(0) = k \). Let's evaluate the limit: \[ \lim_{x \to 0} \frac{1 - \cos 4x}{8x^2} = \lim_{x \to 0} \frac{2\sin^2 2x}{8x^2} \] \[ = \lim_{x \to 0} \left(\frac{\sin 2x}{2x}\right)^2 = 1 \] Equating this to \( f(0) \): \[ k = 1 \]

Question. Find the value of \( k \), for which \[ f(x) = \begin{cases} \frac{\sqrt{1 + kx} - \sqrt{1 - kx}}{x}, & \text{if } -1 \le x < 0 \\ \frac{2x + 1}{x - 1}, & \text{if } 0 \le x < 1 \end{cases} \] is continuous at \( x = 0 \).
Answer: Since \( f(x) \) is continuous at \( x = 0 \), we have: \[ \lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x) \quad \dots(1) \] Calculating the value of the function at \( x = 0 \) using the definition for \( 0 \le x < 1 \): \[ f(0) = \frac{2(0) + 1}{0 - 1} = -1 \] Evaluating the RHL: \[ \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h) = \lim_{h \to 0} \frac{2h + 1}{h - 1} = -1 \] Evaluating the LHL: \[ \lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0 - h) = \lim_{h \to 0} \frac{\sqrt{1 - kh} - \sqrt{1 + kh}}{-h} = \lim_{h \to 0} \frac{\sqrt{1 + kh} - \sqrt{1 - kh}}{h} \] Multiplying the numerator and denominator by \( (\sqrt{1 + kh} + \sqrt{1 - kh}) \): \[ = \lim_{h \to 0} \frac{(1 + kh) - (1 - kh)}{h(\sqrt{1 + kh} + \sqrt{1 - kh})} \] \[ = \lim_{h \to 0} \frac{2kh}{h(\sqrt{1 + kh} + \sqrt{1 - kh})} \] \[ = \lim_{h \to 0} \frac{2k}{\sqrt{1 + kh} + \sqrt{1 - kh}} = \frac{2k}{1 + 1} = k \] Using equation (1), we get: \[ k = -1 \]

Question. If \[ f(x) = \begin{cases} \frac{1 - \cos 4x}{x^2}, & \text{when } x < 0 \\ a, & \text{when } x = 0 \\ \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}, & \text{when } x > 0 \end{cases} \] and \( f \) is continuous at \( x = 0 \), find the value of \( a \).
Answer: Since \( f(x) \) is continuous at \( x = 0 \), we have: \[ \lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x) \quad \dots(1) \] Here, \( f(0) = a \). Evaluating the RHL (\( x > 0 \)): \[ \lim_{x \to 0^+} f(x) = \lim_{h \to 0} \frac{\sqrt{h}}{\sqrt{16 + \sqrt{h}} - 4} \] Rationalizing the denominator: \[ = \lim_{h \to 0} \frac{\sqrt{h}(\sqrt{16 + \sqrt{h}} + 4)}{(\sqrt{16 + \sqrt{h}} - 4)(\sqrt{16 + \sqrt{h}} + 4)} \] \[ = \lim_{h \to 0} \frac{\sqrt{h}(\sqrt{16 + \sqrt{h}} + 4)}{(16 + \sqrt{h}) - 16} \] \[ = \lim_{h \to 0} \frac{\sqrt{h}(\sqrt{16 + \sqrt{h}} + 4)}{\sqrt{h}} \] \[ = \lim_{h \to 0} (\sqrt{16 + \sqrt{h}} + 4) = \sqrt{16} + 4 = 8 \] Evaluating the LHL (\( x < 0 \)): \[ \lim_{x \to 0^-} f(x) = \lim_{h \to 0} \frac{1 - \cos 4(-h)}{(-h)^2} = \lim_{h \to 0} \frac{1 - \cos 4h}{h^2} \] \[ = \lim_{h \to 0} \frac{2\sin^2 2h}{h^2} = 8 \lim_{h \to 0} \left(\frac{\sin 2h}{2h}\right)^2 = 8 \] From equation (1), we get: \[ a = 8 \]

Question. If \[ f(x) = \begin{cases} 1, & \text{if } x \le 3 \\ ax + b, & \text{if } 3 < x < 5 \\ 7, & \text{if } x \ge 5 \end{cases} \] find the values of \( a \) and \( b \) so that \( f(x) \) is a continuous function.
Answer: For \( f(x) \) to be continuous everywhere, it must be continuous at \( x = 3 \) and \( x = 5 \). At \( x = 3 \): \[ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3) \] \[ \lim_{x \to 3^-} f(x) = 1 \] \[ \lim_{x \to 3^+} f(x) = \lim_{h \to 0} [a(3 + h) + b] = 3a + b \] \[ f(3) = 1 \] Thus, \[ 3a + b = 1 \quad \dots(1) \] At \( x = 5 \): \[ \lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = f(5) \] \[ \lim_{x \to 5^-} f(x) = \lim_{h \to 0} [a(5 - h) + b] = 5a + b \] \[ \lim_{x \to 5^+} f(x) = 7 \] \[ f(5) = 7 \] Thus, \[ 5a + b = 7 \quad \dots(2) \] Subtracting equation (1) from (2): \[ (5a + b) - (3a + b) = 7 - 1 \] \[ 2a = 6 \Rightarrow a = 3 \] Substituting \( a = 3 \) in equation (1): \[ 3(3) + b = 1 \Rightarrow 9 + b = 1 \Rightarrow b = -8 \] Hence, \( a = 3 \) and \( b = -8 \).

Question. Find the value of \( k \) so that the following function is continuous at \( x = 2 \). \[ f(x) = \begin{cases} \frac{x^3 + x^2 - 16x + 20}{(x - 2)^2}, & \text{if } x \neq 2 \\ k, & \text{if } x = 2 \end{cases} \]
Answer: Since \( f(x) \) is continuous at \( x = 2 \), we must have: \[ \lim_{x \to 2} f(x) = f(2) \] We are given \( f(2) = k \). Let's evaluate the limit: \[ \lim_{x \to 2} \frac{x^3 + x^2 - 16x + 20}{(x - 2)^2} \] Factorizing the numerator \( x^3 + x^2 - 16x + 20 \) using synthetic division or polynomial division by \( (x - 2) \): Since \( x = 2 \) is a root, \( (x - 2) \) is a factor: \[ x^3 + x^2 - 16x + 20 = (x - 2)(x^2 + 3x - 10) = (x - 2)(x - 2)(x + 5) = (x - 2)^2(x + 5) \] Substituting this back into the limit expression: \[ \lim_{x \to 2} \frac{(x - 2)^2(x + 5)}{(x - 2)^2} = \lim_{x \to 2} (x + 5) = 2 + 5 = 7 \] Equating this to \( f(2) \): \[ k = 7 \]

Question. Find the value of \( k \) so that the following function is continuous at \( x = \pi/2 \): \[ f(x) = \begin{cases} \frac{k\cos x}{\pi - 2x}, & \text{if } x \neq \pi/2 \\ 5, & \text{if } x = \pi/2 \end{cases} \]
Answer: Since \( f(x) \) is continuous at \( x = \pi/2 \), we have: \[ f\left(\frac{\pi}{2}\right) = \lim_{x \to \pi/2} f(x) \] Here, \( f\left(\frac{\pi}{2}\right) = 5 \). Evaluating the limit: \[ \lim_{x \to \pi/2} \frac{k\cos x}{\pi - 2x} \] Let \( x = \frac{\pi}{2} + h \). As \( x \to \frac{\pi}{2} \), \( h \to 0 \). \[ = \lim_{h \to 0} \frac{k \cos\left(\frac{\pi}{2} + h\right)}{\pi - 2\left(\frac{\pi}{2} + h\right)} \] \[ = \lim_{h \to 0} \frac{-k\sin h}{-2h} = \frac{k}{2} \lim_{h \to 0} \frac{\sin h}{h} = \frac{k}{2}(1) = \frac{k}{2} \] Equating this to the function value: \[ \frac{k}{2} = 5 \Rightarrow k = 10 \]

Question. If the function \( f(x) \) given by \[ f(x) = \begin{cases} 3ax + b, & \text{if } x > 1 \\ 11, & \text{if } x = 1 \\ 5ax - 2b, & \text{if } x < 1 \end{cases} \] is continuous at \( x = 1 \), find the values of \( a \) and \( b \).
Answer: Since \( f(x) \) is continuous at \( x = 1 \), we have: \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \] Given \( f(1) = 11 \). Evaluating LHL: \[ \lim_{x \to 1^-} f(x) = \lim_{h \to 0} [5a(1 - h) - 2b] = 5a - 2b \] Evaluating RHL: \[ \lim_{x \to 1^+} f(x) = \lim_{h \to 0} [3a(1 + h) + b] = 3a + b \] Therefore: \[ 5a - 2b = 11 \quad \dots(1) \] \[ 3a + b = 11 \quad \dots(2) \] Multiply equation (2) by 2: \[ 6a + 2b = 22 \quad \dots(3) \] Adding equations (1) and (3): \[ (5a - 2b) + (6a + 2b) = 11 + 22 \] \[ 11a = 33 \Rightarrow a = 3 \] Substituting \( a = 3 \) in equation (2): \[ 3(3) + b = 11 \Rightarrow 9 + b = 11 \Rightarrow b = 2 \] Hence, \( a = 3 \) and \( b = 2 \).

Question. Find the values of \( a \) and \( b \) such that the following function \( f(x) \) is a continuous function: \[ f(x) = \begin{cases} 5, & \text{if } x \le 2 \\ ax + b, & \text{if } 2 < x < 10 \\ 21, & \text{if } x \ge 10 \end{cases} \]
Answer: For \( f(x) \) to be continuous everywhere, it must be continuous at \( x = 2 \) and \( x = 10 \). At \( x = 2 \): \[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) \] \[ 5 = 2a + b \quad \dots(1) \] At \( x = 10 \): \[ \lim_{x \to 10^-} f(x) = \lim_{x \to 10^+} f(x) = f(10) \] \[ 10a + b = 21 \quad \dots(2) \] Subtracting equation (1) from (2): \[ (10a + b) - (2a + b) = 21 - 5 \] \[ 8a = 16 \Rightarrow a = 2 \] Substituting \( a = 2 \) into equation (1): \[ 2(2) + b = 5 \Rightarrow 4 + b = 5 \Rightarrow b = 1 \] Hence, \( a = 2 \) and \( b = 1 \).

Question. For what value of \( a \) is the function \( f \) defined by \[ f(x) = \begin{cases} a\sin \frac{\pi}{2}(x + 1), & \text{if } x \le 0 \\ \frac{\tan x - \sin x}{x^3}, & \text{if } x > 0 \end{cases} \] continuous at \( x = 0 \)?
Answer: Since \( f(x) \) is continuous at \( x = 0 \), we have: \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \] Evaluating \( f(0) \): \[ f(0) = a \sin \left(\frac{\pi}{2}(0 + 1)\right) = a \sin \left(\frac{\pi}{2}\right) = a \] Thus, the LHL is: \[ \lim_{x \to 0^-} f(x) = a \] Evaluating the RHL: \[ \lim_{x \to 0^+} f(x) = \lim_{h \to 0} \frac{\tan h - \sin h}{h^3} \] \[ = \lim_{h \to 0} \frac{\frac{\sin h}{\cos h} - \sin h}{h^3} = \lim_{h \to 0} \frac{\sin h(1 - \cos h)}{h^3 \cos h} \] \[ = \lim_{h \to 0} \left[ \frac{\sin h}{h} \cdot \frac{1 - \cos h}{h^2} \cdot \frac{1}{\cos h} \right] \] \[ = \lim_{h \to 0} \left[ \frac{\sin h}{h} \cdot \frac{2\sin^2(h/2)}{h^2} \cdot \frac{1}{\cos h} \right] \] \[ = \lim_{h \to 0} \left[ \frac{\sin h}{h} \cdot \frac{2\sin^2(h/2)}{4(h/2)^2} \cdot \frac{1}{\cos h} \right] \] \[ = 1 \cdot \frac{2}{4} \cdot \lim_{h \to 0} \left(\frac{\sin(h/2)}{h/2}\right)^2 \cdot 1 \] \[ = \frac{1}{2} \cdot 1 = \frac{1}{2} \] Equating LHL and RHL: \[ a = \frac{1}{2} \]

Question. Find the relationship between \( a \) and \( b \) so that the function \( f \) defined by \[ f(x) = \begin{cases} ax + 1, & \text{if } x \le 3 \\ bx + 3, & \text{if } x > 3 \end{cases} \] is continuous at \( x = 3 \).
Answer: Since \( f(x) \) is continuous at \( x = 3 \), we have: \[ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3) \] Evaluating LHL and \( f(3) \): \[ \lim_{x \to 3^-} f(x) = a(3) + 1 = 3a + 1 \] Evaluating RHL: \[ \lim_{x \to 3^+} f(x) = b(3) + 3 = 3b + 3 \] Since LHL = RHL: \[ 3a + 1 = 3b + 3 \] \[ \Rightarrow 3a - 3b = 2 \] \[ \Rightarrow a - b = \frac{2}{3} \] This is the required relation between \( a \) and \( b \).

Question. Discuss the continuity of the function \( f(x) \) at \( x = 1/2 \), when \( f(x) \) is defined as follows: \[ f(x) = \begin{cases} \frac{1}{2} - x, & 0 \le x < 1/2 \\ 1, & x = 1/2 \\ \frac{3}{2} - x, & 1/2 < x \le 1 \end{cases} \]
Answer: Let us check the LHL and RHL at \( x = 1/2 \). Evaluating LHL: \[ \lim_{x \to 1/2^-} f(x) = \lim_{h \to 0} f\left(\frac{1}{2} - h\right) \] \[ = \lim_{h \to 0} \left[ \frac{1}{2} - \left(\frac{1}{2} - h\right) \right] = \lim_{h \to 0} h = 0 \] Evaluating RHL: \[ \lim_{x \to 1/2^+} f(x) = \lim_{h \to 0} f\left(\frac{1}{2} + h\right) \] \[ = \lim_{h \to 0} \left[ \frac{3}{2} - \left(\frac{1}{2} + h\right) \right] = \lim_{h \to 0} (1 - h) = 1 \] Since \( \lim_{x \to 1/2^-} f(x) \neq \lim_{x \to 1/2^+} f(x) \), the limit of \( f(x) \) as \( x \to 1/2 \) does not exist. Hence, the function \( f(x) \) is discontinuous at \( x = 1/2 \).

Question. Find the value of \( a \) if the function \( f(x) \) defined by \[ f(x) = \begin{cases} 2x - 1, & \text{if } x < 2 \\ a, & \text{if } x = 2 \\ x + 1, & \text{if } x > 2 \end{cases} \] is continuous at \( x = 2 \).
Answer: For \( f(x) \) to be continuous at \( x = 2 \), we must have: \[ \lim_{x \to 2^-} f(x) = f(2) = \lim_{x \to 2^+} f(x) \quad \dots(1) \] Here, \( f(2) = a \). Evaluating the LHL: \[ \lim_{x \to 2^-} f(x) = \lim_{h \to 0} [2(2 - h) - 1] = 3 \] Evaluating the RHL: \[ \lim_{x \to 2^+} f(x) = \lim_{h \to 0} [(2 + h) + 1] = 3 \] Using equation (1), we get: \[ a = 3 \]

Question. Find all points of discontinuity of \( f \), where \( f \) is defined as follows: \[ f(x) = \begin{cases} |x| + 3, & x \le -3 \\ -2x, & -3 < x < 3 \\ 6x + 2, & x \ge 3 \end{cases} \]
Answer: The function is defined in intervals. The only potential points of discontinuity are the boundary points, \( x = -3 \) and \( x = 3 \). At \( x = -3 \): \[ \text{LHL} = \lim_{x \to -3^-} f(x) = \lim_{x \to -3^-} (|x| + 3) = |-3| + 3 = 6 \] \[ \text{RHL} = \lim_{x \to -3^+} f(x) = \lim_{x \to -3^+} (-2x) = -2(-3) = 6 \] \[ f(-3) = |-3| + 3 = 6 \] Since \( \text{LHL} = \text{RHL} = f(-3) \), the function \( f(x) \) is continuous at \( x = -3 \). At \( x = 3 \): \[ \text{LHL} = \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (-2x) = -2(3) = -6 \] \[ \text{RHL} = \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (6x + 2) = 6(3) + 2 = 20 \] Since \( \text{LHL} \neq \text{RHL} \), the limit does not exist at \( x = 3 \). Thus, \( f(x) \) is discontinuous at \( x = 3 \). The only point of discontinuity of the function \( f \) is \( x = 3 \).

Question. For what value of \( k \) is the function defined by \[ f(x) = \begin{cases} k(x^2 + 2), & \text{if } x \le 0 \\ 3x + 1, & \text{if } x > 0 \end{cases} \] continuous at \( x = 0 \)? Also, write whether the function is continuous at \( x = 1 \).
Answer: For \( f(x) \) to be continuous at \( x = 0 \), we must have: \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \] Evaluating LHL and \( f(0) \): \[ \lim_{x \to 0^-} f(x) = k(0^2 + 2) = 2k \] Evaluating RHL: \[ \lim_{x \to 0^+} f(x) = 3(0) + 1 = 1 \] Equating LHL and RHL: \[ 2k = 1 \Rightarrow k = \frac{1}{2} \] Next, let's discuss continuity at \( x = 1 \): For \( x > 0 \), \( f(x) = 3x + 1 \), which is a polynomial function. Since polynomial functions are continuous everywhere in their domain, \( f(x) \) is continuous at \( x = 1 \).

Question. Find the values of \( a \) and \( b \) such that the function defined as follows is continuous: \[ f(x) = \begin{cases} x + 2, & x \le 2 \\ ax + b, & 2 < x < 5 \\ 3x - 2, & x \ge 5 \end{cases} \]
Answer: Since the function is continuous everywhere, it must be continuous at \( x = 2 \) and \( x = 5 \). At \( x = 2 \): \[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) \] \[ 2 + 2 = a(2) + b \] \[ 2a + b = 4 \quad \dots(1) \] At \( x = 5 \): \[ \lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) \] \[ a(5) + b = 3(5) - 2 \] \[ 5a + b = 13 \quad \dots(2) \] Subtracting equation (1) from (2): \[ (5a + b) - (2a + b) = 13 - 4 \] \[ 3a = 9 \Rightarrow a = 3 \] Substituting \( a = 3 \) in equation (1): \[ 2(3) + b = 4 \Rightarrow b = -2 \] Hence, \( a = 3 \) and \( b = -2 \).

Question. Show that the function \( f(x) \) defined by \[ f(x) = \begin{cases} \frac{\sin x}{x} + \cos x, & x > 0 \\ 2, & x = 0 \\ \frac{4(1 - \sqrt{1 - x})}{x}, & x < 0 \end{cases} \] is continuous at \( x = 0 \).
Answer: To show that \( f(x) \) is continuous at \( x = 0 \), we need to show that \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 2 \). Evaluating RHL (\( x > 0 \)): \[ \lim_{x \to 0^+} f(x) = \lim_{h \to 0} \left( \frac{\sin h}{h} + \cos h \right) = \lim_{h \to 0} \frac{\sin h}{h} + \lim_{h \to 0} \cos h = 1 + 1 = 2 \] Evaluating LHL (\( x < 0 \)): \[ \lim_{x \to 0^-} f(x) = \lim_{h \to 0} \frac{4(1 - \sqrt{1 - (-h)})}{-h} = \lim_{h \to 0} \frac{4(1 - \sqrt{1 + h})}{-h} \] Multiplying numerator and denominator by \( (1 + \sqrt{1 + h}) \): \[ = \lim_{h \to 0} \frac{4(1 - (1 + h))}{-h(1 + \sqrt{1 + h})} = \lim_{h \to 0} \frac{-4h}{-h(1 + \sqrt{1 + h})} \] \[ = \lim_{h \to 0} \frac{4}{1 + \sqrt{1 + h}} = \frac{4}{1 + 1} = 2 \] Since \( \text{LHL} = \text{RHL} = f(0) = 2 \), the function \( f(x) \) is continuous at \( x = 0 \).

Question. If the function defined by \[ f(x) = \begin{cases} 2x - 1, & x < 2 \\ a, & x = 2 \\ x + 1, & x > 2 \end{cases} \] is continuous at \( x = 2 \), find the value of \( a \). Also, discuss the continuity of \( f(x) \) at \( x = 3 \).
Answer: For continuity at \( x = 2 \): \[ \lim_{x \to 2^-} f(x) = f(2) = \lim_{x \to 2^+} f(x) \] \[ \lim_{x \to 2^-} (2x - 1) = a = \lim_{x \to 2^+} (x + 1) \] \[ 2(2) - 1 = a = 2 + 1 \Rightarrow a = 3 \] Now, let's discuss continuity at \( x = 3 \): For \( x > 2 \), \( f(x) = x + 1 \). At \( x = 3 \): \[ \lim_{x \to 3} f(x) = \lim_{x \to 3} (x + 1) = 4 \] And \( f(3) = 3 + 1 = 4 \). Since \( \lim_{x \to 3} f(x) = f(3) \), the function \( f(x) \) is continuous at \( x = 3 \).

Question. For what value of \( k \) is the following function continuous at \( x = 2 \)? \[ f(x) = \begin{cases} 2x + 1, & x < 2 \\ k, & x = 2 \\ 3x - 1, & x > 2 \end{cases} \]
Answer: For \( f(x) \) to be continuous at \( x = 2 \), we must have: \[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) \] Evaluating LHL: \[ \lim_{x \to 2^-} f(x) = 2(2) + 1 = 5 \] Evaluating RHL: \[ \lim_{x \to 2^+} f(x) = 3(2) - 1 = 5 \] And \( f(2) = k \). Thus, equating the limits to \( f(2) \): \[ k = 5 \]

Question. If \( f(x) \) defined by the following is continuous at \( x = 0 \), find the value of \( a \), \( b \) and \( c \). \[ f(x) = \begin{cases} \frac{\sin(a + 1)x + \sin x}{x}, & \text{if } x < 0 \\ c, & \text{if } x = 0 \\ \frac{\sqrt{x + bx^2} - \sqrt{x}}{bx^{3/2}}, & \text{if } x > 0 \end{cases} \]
Answer: Since \( f(x) \) is continuous at \( x = 0 \), we must have: \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \] Evaluating LHL: \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \left[ \frac{\sin(a + 1)x}{x} + \frac{\sin x}{x} \right] \] \[ = (a+1) \lim_{x \to 0^-} \frac{\sin(a+1)x}{(a+1)x} + \lim_{x \to 0^-} \frac{\sin x}{x} \] \[ = (a+1)(1) + 1 = a + 2 \quad \dots(1) \] Evaluating RHL: \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x(1 + bx)} - \sqrt{x}}{bx\sqrt{x}} \] \[ = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{1 + bx} - 1)}{bx\sqrt{x}} = \lim_{x \to 0^+} \frac{\sqrt{1 + bx} - 1}{bx} \] Rationalizing the numerator: \[ = \lim_{x \to 0^+} \frac{(\sqrt{1 + bx} - 1)(\sqrt{1 + bx} + 1)}{bx(\sqrt{1 + bx} + 1)} \] \[ = \lim_{x \to 0^+} \frac{(1 + bx) - 1}{bx(\sqrt{1 + bx} + 1)} \] \[ = \lim_{x \to 0^+} \frac{bx}{bx(\sqrt{1 + bx} + 1)} = \lim_{x \to 0^+} \frac{1}{\sqrt{1 + bx} + 1} = \frac{1}{1 + 1} = \frac{1}{2} \quad \dots(2) \] And \( f(0) = c \quad \dots(3) \). From equations (1), (2), and (3): \[ c = \frac{1}{2} \] \[ a + 2 = \frac{1}{2} \Rightarrow a = -\frac{3}{2} \] Since \( b \) cancels out in the limit computation for RHL, \( b \) can be any non-zero real number (i.e., \( b \in \mathbb{R} \setminus \{0\} \)).

Question. If the following function \( f(x) \) is continuous at \( x = 0 \), find the value of \( k \). \[ f(x) = \begin{cases} \frac{1 - \cos 2x}{2x^2}, & x \neq 0 \\ k, & x = 0 \end{cases} \]
Answer: Since \( f(x) \) is continuous at \( x = 0 \), we have: \[ f(0) = \lim_{x \to 0} f(x) \] Evaluating the limit: \[ \lim_{x \to 0} \frac{1 - \cos 2x}{2x^2} = \lim_{x \to 0} \frac{2\sin^2 x}{2x^2} \] \[ = \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^2 = 1^2 = 1 \] Equating this to \( f(0) = k \): \[ k = 1 \]

Question. Find the value of \( k \) if the function \[ f(x) = \begin{cases} kx^2, & x \ge 1 \\ 4, & x < 1 \end{cases} \] is continuous at \( x = 1 \).
Answer: Since \( f(x) \) is continuous at \( x = 1 \), we have: \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \] Evaluating LHL: \[ \lim_{x \to 1^-} f(x) = 4 \] Evaluating RHL: \[ \lim_{x \to 1^+} f(x) = k(1)^2 = k \] Equating LHL and RHL: \[ k = 4 \]

Question. If \( f(x) = \begin{cases} \frac{x^2 - 25}{x - 5}, & \text{if } x \neq 5 \\ k, & \text{if } x = 5 \end{cases} \) is continuous at \( x = 5 \), find the value of \( k \).
Answer: Since \( f(x) \) is continuous at \( x = 5 \), we must have: \[ \lim_{x \to 5} f(x) = f(5) \] Given \( f(5) = k \). Let's evaluate the limit: \[ \lim_{x \to 5} \frac{x^2 - 25}{x - 5} = \lim_{x \to 5} \frac{(x - 5)(x + 5)}{x - 5} \] \[ = \lim_{x \to 5} (x + 5) = 5 + 5 = 10 \] Equating this to \( f(5) \): \[ k = 10 \]

Question. Find the values of \( a \) and \( b \), if the function \( f \) defined by \[ f(x) = \begin{cases} x^2 + 3x + a, & x \le 1 \\ bx + 2, & x > 1 \end{cases} \] is differentiable at \( x = 1 \).
Answer: Since \( f(x) \) is differentiable at \( x = 1 \), it must also be continuous at \( x = 1 \). Continuity at \( x = 1 \): \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \] \[ 1^2 + 3(1) + a = b(1) + 2 \] \[ 4 + a = b + 2 \] \[ a - b + 2 = 0 \quad \dots(1) \] Differentiability at \( x = 1 \): \[ \text{LHD at } x = 1 = \text{RHD at } x = 1 \] Evaluating LHD: \[ \lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^-} \frac{(x^2 + 3x + a) - (4 + a)}{x - 1} \] \[ = \lim_{x \to 1^-} \frac{x^2 + 3x - 4}{x - 1} = \lim_{x \to 1^-} \frac{(x - 1)(x + 4)}{x - 1} \] \[ = \lim_{x \to 1^-} (x + 4) = 5 \] Evaluating RHD: \[ \lim_{x \to 1^+} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^+} \frac{(bx + 2) - (4 + a)}{x - 1} \] From (1), we have \( 4 + a = b + 2 \). Substituting this in the numerator: \[ = \lim_{x \to 1^+} \frac{(bx + 2) - (b + 2)}{x - 1} = \lim_{x \to 1^+} \frac{b(x - 1)}{x - 1} = b \] Equating LHD and RHD: \[ b = 5 \] Substituting \( b = 5 \) in equation (1): \[ a - 5 + 2 = 0 \Rightarrow a = 3 \] Hence, \( a = 3 \) and \( b = 5 \).

Question. If \( y = \tan^{-1} \left[ \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right] \), \( x^2 \le 1 \), then find \( \frac{dy}{dx} \).
Answer: Given expression: \[ y = \tan^{-1} \left[ \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right] \] Let \( x^2 = \cos \theta \Rightarrow \theta = \cos^{-1}(x^2) \). Substituting this into the expression: \[ y = \tan^{-1} \left[ \frac{\sqrt{1 + \cos \theta} + \sqrt{1 - \cos \theta}}{\sqrt{1 + \cos \theta} - \sqrt{1 - \cos \theta}} \right] \] We know that \( \sqrt{1 + \cos \theta} = \sqrt{2}\cos\left(\frac{\theta}{2}\right) \) and \( \sqrt{1 - \cos \theta} = \sqrt{2}\sin\left(\frac{\theta}{2}\right) \). \[ y = \tan^{-1} \left[ \frac{\sqrt{2}\cos\left(\frac{\theta}{2}\right) + \sqrt{2}\sin\left(\frac{\theta}{2}\right)}{\sqrt{2}\cos\left(\frac{\theta}{2}\right) - \sqrt{2}\sin\left(\frac{\theta}{2}\right)} \right] \] \[ = \tan^{-1} \left[ \frac{\cos\left(\frac{\theta}{2}\right) + \sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right) - \sin\left(\frac{\theta}{2}\right)} \right] \] Dividing the numerator and denominator by \( \cos\left(\frac{\theta}{2}\right) \): \[ = \tan^{-1} \left[ \frac{1 + \tan\left(\frac{\theta}{2}\right)}{1 - \tan\left(\frac{\theta}{2}\right)} \right] \] \[ = \tan^{-1} \left[ \tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right) \right] \] \[ y = \frac{\pi}{4} + \frac{\theta}{2} \] Substituting back \( \theta = \cos^{-1}(x^2) \): \[ y = \frac{\pi}{4} + \frac{1}{2}\cos^{-1}(x^2) \] Differentiating with respect to \( x \) on both sides: \[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4}\right) + \frac{1}{2} \frac{d}{dx}(\cos^{-1}(x^2)) \] \[ = 0 + \frac{1}{2} \cdot \left[ -\frac{1}{\sqrt{1 - (x^2)^2}} \cdot \frac{d}{dx}(x^2) \right] \] \[ = \frac{1}{2} \cdot \left[ -\frac{1}{\sqrt{1 - x^4}} \cdot (2x) \right] \] \[ = -\frac{x}{\sqrt{1 - x^4}} \]

Question. If \( f(x) = \sqrt{x^2+1} \); \( g(x) = \frac{x+1}{x^2+1} \) and \( h(x) = 2x-3 \), then find \( f'[h'(g'(x))] \).
Answer: Here, \( f(x) = \sqrt{x^2+1} = (x^2+1)^{\frac{1}{2}} \)
\( \Rightarrow f'(x) = \frac{1}{2}(x^2+1)^{-\frac{1}{2}} \cdot 2x = \frac{x}{\sqrt{x^2+1}} \) ...(1)
\( g(x) = \frac{x+1}{x^2+1} \)
\( \Rightarrow g'(x) = \frac{(x^2+1)\cdot 1 - (x+1)\cdot 2x}{(x^2+1)^2} = \frac{-x^2-2x+1}{(x^2+1)^2} \) ...(2)
and \( h(x) = 2x - 3 \)
\( \Rightarrow h'(x) = 2 \) ...(3)
\( \therefore f'[h'(g'(x))] = f'\left[ h'\left( \frac{-x^2-2x+1}{(x^2+1)^2} \right) \right] \) [Using (2)]
\( = f'(2) \) [Using (3)]
\( = \frac{2}{\sqrt{2^2+1}} = \frac{2}{\sqrt{5}} \) [Using (1)]

Question. Show that the function \( f(x) = |x - 1| + |x + 1| \), for all \( x \in \mathbb{R} \), is not differentiable at the points \( x = -1 \) and \( x = 1 \).
Answer: The given function is \( f(x) = |x - 1| + |x + 1| \)
\( = \begin{cases} -(x-1) - (x+1), & x < -1 \\ -(x-1) + x + 1, & -1 \le x \le 1 \\ x - 1 + x + 1, & x > 1 \end{cases} \)
\( = \begin{cases} -2x, & x < -1 \\ 2, & -1 \le x \le 1 \\ 2x, & x > 1 \end{cases} \)
At \( x = 1 \),
\( f'(1^-) = \lim_{h \to 0} \frac{f(1-h) - f(1)}{-h} = \lim_{h \to 0} \frac{2 - 2}{-h} = 0 \)
\( f'(1^+) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{2(1+h) - 2}{h} = \lim_{h \to 0} \frac{2h}{h} = 2 \)
\( \therefore f'(1^-) \ne f'(1^+) \)
\( \Rightarrow f \) is not differentiable at \( x = 1 \).
At \( x = -1 \),
\( f'(-1^-) = \lim_{h \to 0} \frac{f(-1-h) - f(-1)}{-h} = \lim_{h \to 0} \frac{-2(-1-h) - 2}{-h} = \lim_{h \to 0} \frac{2h}{-h} = -2 \)
\( f'(-1^+) = \lim_{h \to 0} \frac{f(-1+h) - f(-1)}{h} = \lim_{h \to 0} \frac{2 - 2}{h} = 0 \)
\( \dots f'(-1^-) \ne f'(-1^+) \)
\( \Rightarrow f \) is not differentiable at \( x = -1 \).

Question. Find whether the following function is differentiable at \( x = 1 \) and \( x = 2 \) or not.
\( f(x) = \begin{cases} x, & x < 1 \\ 2 - x, & 1 \le x \le 2 \\ -2 + 3x - x^2, & x > 2 \end{cases} \)
Answer: At \( x = 1 \):
\( f'(1^-) = \lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^-} \frac{x - 1}{x - 1} = 1 \)
\( f'(1^+) = \lim_{x \to 1^+} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^+} \frac{2 - x - 1}{x - 1} = \lim_{x \to 1^+} \frac{1 - x}{x - 1} = -1 \)
Since \( f'(1^-) \ne f'(1^+) \),
\( \therefore f(x) \) is not differentiable at \( x = 1 \).
At \( x = 2 \):
\( f'(2^-) = \lim_{x \to 2^-} \frac{f(x) - f(2)}{x - 2} = \lim_{x \to 2^-} \frac{2 - x - 0}{x - 2} = -1 \)
\( f'(2^+) = \lim_{x \to 2^+} \frac{f(x) - f(2)}{x - 2} = \lim_{x \to 2^+} \frac{-2 + 3x - x^2 - 0}{x - 2} = \lim_{x \to 2^+} \frac{(1-x)(x-2)}{x-2} = -1 \)
Since \( f'(2^-) = f'(2^+) \),
\( \therefore f(x) \) is differentiable at \( x = 2 \).

Question. For what value \( \lambda \) of the function defined by
\( f(x) = \begin{cases} \lambda(x^2 + 2), & \text{if } x \le 0 \\ 4x + 6, & \text{if } x > 0 \end{cases} \)
is continuous at \( x = 0 \)? Hence check the differentiability of \( f(x) \) at \( x = 0 \).
Answer: At \( x = 0 \), \( f(0) = \lambda(0^2 + 2) = 2\lambda \)
L.H.L. \( = \lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0-h) = \lim_{h \to 0} [\lambda((-h)^2 + 2)] = 2\lambda \)
R.H.L. \( = \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0+h) = \lim_{h \to 0} [4(h) + 6] = 6 \)
\( \therefore \) For \( f \) to be continuous at \( x = 0 \),
\( 2\lambda = 6 \Rightarrow \lambda = 3 \).
Hence the function becomes:
\( f(x) = \begin{cases} 3(x^2 + 2), & \text{if } x \le 0 \\ 4x + 6, & \text{if } x > 0 \end{cases} \)
\( f'(0^-) = \lim_{h \to 0} \frac{f(0-h) - f(0)}{-h} = \lim_{h \to 0} \frac{3(h^2 + 2) - 6}{-h} = \lim_{h \to 0} \frac{3h^2}{-h} = \lim_{h \to 0} (-3h) = 0 \)
and \( f'(0^+) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{4h + 6 - 6}{h} = 4 \)
\( \Rightarrow f'(0^-) \ne f'(0^+) \)
\( \therefore f(x) \) is not differentiable at \( x = 0 \).

Question. If \( \cos y = x \cos(a + y) \), where \( \cos a \ne \pm 1 \), prove that \( \frac{dy}{dx} = \frac{\cos^2(a + y)}{\sin a} \).
Answer: We have, \( \cos y = x \cos(a + y) \)
\( \Rightarrow x = \frac{\cos y}{\cos(a + y)} \)
Differentiating w.r.t. \( y \) on both sides, we get
\( \frac{dx}{dy} = \frac{\cos(a + y) \frac{d}{dy}(\cos y) - \cos y \frac{d}{dy}(\cos(a + y))}{\cos^2(a + y)} \)
\( \Rightarrow \frac{dx}{dy} = \frac{\cos(a + y)(-\sin y) - \cos y(-\sin(a + y))}{\cos^2(a + y)} \)
\( \Rightarrow \frac{dx}{dy} = \frac{\cos y \sin(a + y) - \cos(a + y) \sin y}{\cos^2(a + y)} \)
\( = \frac{\sin[(a + y) - y]}{\cos^2(a + y)} = \frac{\sin a}{\cos^2(a + y)} \)
\( \therefore \frac{dy}{dx} = \frac{\cos^2(a + y)}{\sin a} \)