Ex 5.2 – Continuity and Differentiability | ML Aggarwal Class 12 Solutions

Solution: The greatest integer function is defined as: \[ f(x) = \lfloor x \rfloor = \text{greatest integer less than or equal to } x \] Checking continuity at \( x = 0 \): Left-hand limit: \[ \lim\limits_{x \to 0^-} f(x) = \lfloor 0^- \rfloor = -1 \] Right-hand limit: \[ \lim\limits_{x \to 0^+} f(x) = \lfloor 0^+ \rfloor = 0 \] Function value: \[ f(0) = \lfloor 0 \rfloor = 0 \] Since \( \lim\limits_{x \to 0^-} f(x) \neq \lim\limits_{x \to 0^+} f(x) \), \( f(x) \) is not continuous at \( x = 0 \).

Solution: \[ \text{Left-hand limit: } \lim_{x \to 1^-} f(x) = \lfloor 1^- \rfloor = 0 \] \[ \text{Right-hand limit: } \lim_{x \to 1^+} f(x) = \lfloor 1^+ \rfloor = 1 \] \[ f(1) = \lfloor 1 \rfloor = 1 \] Since the left-hand and right-hand limits are not equal, \( f(x) \) is discontinuous at \( x=1 \).

Solution: For \( x=\frac{1}{2} \): \[ f\left(\frac{1}{2}\right) = \lfloor 0.5 \rfloor = 0 \] \[ \lim_{x \to \left(\frac{1}{2}\right)^-} f(x) = 0,\quad \lim_{x \to \left(\frac{1}{2}\right)^+} f(x) = 0 \] Since both limits equal \( f\left(\frac{1}{2}\right) \), \( f(x) \) is continuous at \( x=\frac{1}{2} \).

Solution: For \( x<0 \) and \( x>0 \), each piece is continuous. At \( x=0 \): \[ \lim_{x\to 0^-} f(x)=\lim_{x\to 0^-} x^2=0,\quad \lim_{x\to 0^+} f(x)=\lim_{x\to 0^+} x=0, \] \[ f(0)=0. \] Since the limits and \( f(0) \) coincide, \( f(x) \) is continuous at \( x=0 \).

Solution: The function is clearly continuous for \( x<1 \) and \( x>1 \). Checking at \( x=1 \): \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x+5) = 1+5 = 6 \] \[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x-5) = 1-5 = -4 \] \[ f(1) = 1+5 = 6 \] Since \( \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x) \), the function is discontinuous at \( x=1 \).

Solution: For \( x<2 \) and \( x>2 \), the function is continuous. At \( x=2 \): \[ \lim_{x\to 2^-} f(x)=2(2)-1=3,\quad f(2)=\frac{3}{2}\cdot2=3 \] \[ \lim_{x\to 2^+} f(x)=\frac{3}{2}\cdot2=3. \] Since the limits and \( f(2) \) agree, \( f(x) \) is continuous at \( x=2 \).

Solution: For \( x\neq 0 \), \( f(x)=\sin x-\cos x \) is continuous. At \( x=0 \): \[ \lim_{x\to 0} (\sin x-\cos x)=0-1=-1, \quad f(0)=-1. \] The limit equals \( f(0) \), so the function is continuous at \( x=0 \).

Solution: The absolute value function can be written as: \[ |x|=\begin{cases} x, & x\geq 0, \\ -x, & x<0. \end{cases} \] Each branch (\(x\) and \(-x\)) is continuous on its domain. At \( x=0 \), evaluate the limits: \[ \lim_{x\to 0^-}|x|=\lim_{x\to 0^-}(-x)=0,\quad \lim_{x\to 0^+}|x|=\lim_{x\to 0^+}x=0, \] and note that \( f(0)=0 \). Since both one-sided limits equal \( f(0) \), the function is continuous at \( x=0 \) and hence continuous for all \( x \).

Solution: The function \( f(x)=|x-5| \) can be viewed as a translation of the absolute value function \( |x| \). Since \( |x| \) is continuous for all \( x \) and the transformation \( x-5 \) is continuous, their composition is continuous. Alternatively, writing it as: \[ |x-5|=\begin{cases} x-5, & x\geq 5, \\ 5-x, & x<5, \end{cases} \] both branches are continuous and agree at \( x=5 \): \[ \lim_{x\to 5^-} |x-5| = 0, \quad \lim_{x\to 5^+} |x-5| = 0, \quad f(5)=0. \]

Solution: Factorize the numerator: \[ x^2-25=(x-5)(x+5). \] For \( x\neq -5 \): \[ f(x)=\frac{(x-5)(x+5)}{x+5}=x-5, \] which is continuous everywhere. At \( x=-5 \), \( f(x) \) is undefined.

Solution: Domain: \( x-5\neq0 \) \[ \Rightarrow x\neq5. \] For \( x\neq5 \), \( f(x) \) is a quotient of continuous functions. At \( x=5 \), \( f(x) \) is undefined.

Solution: For \( x > 0 \), \( f(x) = \frac{\sin 2x}{x} \) is defined and continuous. For \( x \leq 0 \), \( f(x) = 2 \) is a constant function, hence continuous. Check continuity at \( x = 0 \): \[ \lim_{x \to 0^-} f(x) = 2. \] \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sin 2x}{x}. \] Using \( \lim_{x \to 0} \frac{\sin kx}{x} = k \), we get: \[ \lim_{x \to 0^+} f(x) = 2. \] Since both limits are equal to \( f(0) = 2 \), \( f(x) \) is continuous at \( x=0 \).

Solution: For \( x > 0 \), \( f(x) = \frac{x}{|x|} = 1 \), which is constant and continuous. For \( x < 0 \), \( f(x) = \frac{x}{|x|} = -1 \), which is also constant and continuous. At \( x = 0 \), check left-hand and right-hand limits: \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-1) = -1. \] \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (1) = 1. \] Since \( \lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x) \), the limit does not exist, so \( f(x) \) is discontinuous at \( x = 0 \).

Solution: For \( x \leq 1 \), \( f(x) = x^{10} – 1 \) is a polynomial, hence continuous. For \( x > 1 \), \( f(x) = x^2 \) is also a polynomial and continuous. At \( x = 1 \), check left-hand and right-hand limits: \[ \lim_{x \to 1^-} f(x) = (1)^{10} – 1 = 0. \] \[ \lim_{x \to 1^+} f(x) = (1)^2 = 1. \] Since \( \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x) \), the limit does not exist, so \( f(x) \) is discontinuous at \( x = 1 \).

Solution: The function \( f(x) = \tan x \) is defined and continuous for all \( x \) except at points where \( \tan x \) is undefined. In the given interval \( 0 \leq x \leq \frac{\pi}{4} \), \( \tan x \) is well-defined and continuous since there are no points where it becomes undefined. Hence, \( f(x) \) is continuous in the entire given domain.

Solution: For \( x \leq 2 \), \( f(x) = 2x + 3 \) is a linear function and continuous. For \( x > 2 \), \( f(x) = 2x – 3 \) is also a linear function and continuous. Check continuity at \( x = 2 \): \[ \lim_{x \to 2^-} f(x) = 2(2) + 3 = 7. \] \[ \lim_{x \to 2^+} f(x) = 2(2) – 3 = 1. \] Since \( \lim_{x \to 2^-} f(x) \neq \lim_{x \to 2^+} f(x) \), the limit does not exist, so \( f(x) \) is discontinuous at \( x = 2 \).

Solution: For \( x \leq 2 \), \( f(x) = x^3 – 3 \) is a polynomial, hence continuous. For \( x > 2 \), \( f(x) = x^2 + 1 \) is also a polynomial and continuous. Check continuity at \( x = 2 \): \[ \lim_{x \to 2^-} f(x) = (2)^3 – 3 = 5. \] \[ \lim_{x \to 2^+} f(x) = (2)^2 + 1 = 5. \] Since \( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) = 5 \), the function is continuous at \( x = 2 \). No discontinuity found.

Solution: For \( x \neq 2 \), \[ f(x) = \frac{x^4 – 16}{x – 2}. \] Factorizing the numerator: \[ x^4 – 16 = (x^2 – 4)(x^2 + 4) = (x – 2)(x + 2)(x^2 + 4). \] Canceling \( (x – 2) \), we get: \[ f(x) = (x + 2)(x^2 + 4), \quad x \neq 2. \] Check continuity at \( x = 2 \): \[ \lim_{x \to 2} f(x) = (2 + 2)(2^2 + 4) = 4 \times 8 = 32. \] Since \( \lim_{x \to 2} f(x) = 32 \neq f(2) = 16 \), \( f(x) \) is discontinuous at \( x = 2 \).

Solution: For \( x > 0 \), \( f(x) = \frac{\sin 3x}{x} + \cos x \). The function is defined for all \( x > 0 \) and remains continuous as both terms are continuous. For \( x \leq 0 \), \( f(x) = 4 – 3x \), which is a polynomial and hence continuous. Check continuity at \( x = 0 \): \[ \lim_{x \to 0^-} f(x) = 4 – 3(0) = 4. \] \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0} \left(\frac{\sin 3x}{x} + \cos x \right). \] Using \( \lim_{x \to 0} \frac{\sin 3x}{x} = 3 \), \[ \lim_{x \to 0^+} f(x) = 3 + \cos 0 = 3 + 1 = 4. \] Since \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 4 \), \( f(x) \) is continuous at \( x = 0 \). No discontinuity found.

Solution: For \( x < 0 \), \( f(x) = 2x \) is a polynomial and continuous. For \( 0 \leq x \leq 1 \), \( f(x) = 0 \), which is a constant function and continuous. For \( x > 1 \), \( f(x) = 4x \) is a polynomial and continuous. Check continuity at \( x = 0 \): \[ \lim_{x \to 0^-} f(x) = 2(0) = 0, \quad \lim_{x \to 0^+} f(x) = 0. \] Since both limits are equal and \( f(0) = 0 \), the function is continuous at \( x = 0 \). Check continuity at \( x = 1 \): \[ \lim_{x \to 1^-} f(x) = 0, \quad \lim_{x \to 1^+} f(x) = 4(1) = 4. \] Since \( \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x) \), the function is discontinuous at \( x = 1 \).

Solution: For \( x \leq 1 \), \( f(x) = x + 2 \), which is a continuous function. For \( 1 < x < 2 \), \( f(x) = x - 2 \), which is also continuous. For \( x \geq 2 \), \( f(x) = 0 \), which is continuous. Now, check for continuity at \( x = 1 \): \[ \lim_{x \to 1^-} f(x) = 1 + 2 = 3. \] \[ \lim_{x \to 1^+} f(x) = 1 - 2 = -1. \] Since the left-hand and right-hand limits are not equal, the function is discontinuous at \( x = 1 \). Now, check for continuity at \( x = 2 \): \[ \lim_{x \to 2^-} f(x) = 2 - 2 = 0. \] \[ \lim_{x \to 2^+} f(x) = 0. \] Since the left-hand and right-hand limits are equal, the function is continuous at \( x = 2 \).

Solution: For continuity at \( x = 2 \), \[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2). \] Compute left-hand limit: \[ \lim_{x \to 2^-} f(x) = k(2)^2 = 4k. \] Compute right-hand limit: \[ \lim_{x \to 2^+} f(x) = 2 – 3 = -1. \] Equating limits for continuity: \[ 4k = -1. \] Solve for \( k \): \[ k = -\frac{1}{4}. \]

Solution: For continuity at \( x = -1 \): \[ \lim_{x \to -1^-} f(x) = -5. \] \[ \lim_{x \to -1^+} f(x) = a(-1) – b = -a – b. \] Equating limits: \[ -a – b = -5. \] For continuity at \( x = 3 \): \[ \lim_{x \to 3^-} f(x) = 3a – b. \] \[ \lim_{x \to 3^+} f(x) = 7. \] Equating limits: \[ 3a – b = 7. \] Solving the system: \[ -a – b = -5, \] \[ 3a – b = 7. \] Subtracting the equations: \[ (3a – b) – (-a – b) = 7 – (-5), \] \[ 4a = 12. \] \[ a = 3. \] Substituting \( a = 3 \) into \( -3 – b = -5 \): \[ b = 2. \]

Solution: For continuity at \( x = 0 \): \[ \lim_{x \to 0^-} f(x) = 0^2 = 0. \] \[ \lim_{x \to 0^+} f(x) = a(0) + b = b. \] Equating limits for continuity: \[ b = 0. \] Thus, \( b = 0 \). Now, check the derivative for smoothness at \( x = 0 \): For \( x \leq 0 \), the derivative of \( f(x) = x^2 \) is \( f'(x) = 2x \). For \( x > 0 \), the derivative of \( f(x) = ax + b \) is \( f'(x) = a \). At \( x = 0 \), the left-hand derivative is \( f'(0^-) = 0 \). The right-hand derivative is \( f'(0^+) = a \). For continuity and smoothness, \[ a = 0. \]