Ex 5.4 – Continuity and Differentiability | ML Aggarwal Class 12 Solutions

Solution: If a function \( f \) is differentiable at a point \( c \), then by definition the derivative \[ f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c} \] exists. For this limit to exist, the limit \[ \lim_{x\to c} f(x) \] must equal \( f(c) \). This equality is exactly the definition of continuity at \( c \). Hence, differentiability at a point implies continuity at that point.

Solution: Continuity at a point means that the limit of the function as it approaches the point equals the function’s value at that point. However, differentiability requires a function to have a well-defined and unique tangent (slope) at that point. A classic counterexample is the function \( f(x)=|x| \), which is continuous everywhere but not differentiable at \( x=0 \) because it has a sharp corner there. Therefore, continuity does not guarantee differentiability.

Solution: The function \( f(x)=|x| \) can be written as \[ f(x)= \begin{cases} x, & \text{if } x \geq 0, \\ -x, & \text{if } x < 0. \end{cases} \] To check differentiability at \( x=0 \), we need to compute the left-hand and right-hand derivatives. For \( x>0 \), the derivative is: \[ f'(x)=1. \] Hence, the right-hand derivative at \( x=0 \) is: \[ f’_+(0)=1. \] For \( x<0 \), the derivative is: \[ f'(x)=-1. \] Hence, the left-hand derivative at \( x=0 \) is: \[ f'_-(0)=-1. \] Since \( f'_+(0) \neq f'_-(0) \), the derivative at \( x=0 \) does not exist.

Solution: The function \( f(x)=\cot x \) is defined as \[ f(x)=\frac{\cos x}{\sin x}. \] At \( x=0 \), the sine function \( \sin x \) equals 0, making \( \cot x \) undefined at that point. Since the function is not even defined at \( x=0 \), it cannot be differentiable there.

Solution: The function \( f(x)=\sec x \) is defined as \[ \sec x=\frac{1}{\cos x}. \] At \( x=\frac{\pi}{2} \), the cosine function \( \cos \frac{\pi}{2}=0 \), making \( \sec \frac{\pi}{2} \) undefined. Since the function is not defined at \( x=\frac{\pi}{2} \), it cannot be differentiable there.

Solution: A function \( f(x) \) is called derivable (or differentiable) at a point \( c \) if the following limit exists and is finite: \[ f'(c) = \lim_{x \to c} \frac{f(x) – f(c)}{x – c}. \] This limit represents the slope of the tangent to the curve at \( c \). If this limit exists, the function has a unique tangent at that point and is therefore said to be derivable there.

Solution: The function \( \sin x \) is a smooth trigonometric function with a well-defined derivative. Its derivative is given by: \[ (\sin x)’ = \cos x. \] Since \( \cos x \) is defined for all \( x \), \( \sin x \) is differentiable for every real number.

Solution: The function \( \cos x \) is a smooth trigonometric function, and its derivative is given by: \[ (\cos x)’ = -\sin x. \] Since \( -\sin x \) is defined for all real numbers, \( \cos x \) is differentiable everywhere.

Solution: The function \( \tan x \) is defined as: \[ \tan x = \frac{\sin x}{\cos x}. \] Differentiating both sides, \[ (\tan x)’ = \sec^2 x. \] The function \( \sec^2 x \) is not defined where \( \cos x = 0 \), i.e., at \( x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots \). Hence, \( \tan x \) is not differentiable at these points.

Solution: The function \( \cot x \) is defined as: \[ \cot x = \frac{\cos x}{\sin x}. \] Differentiating both sides, \[ (\cot x)’ = -\csc^2 x. \] The function \( \csc^2 x \) is not defined where \( \sin x = 0 \), i.e., at \( x = k\pi, \, k \in \mathbb{Z} \). Hence, \( \cot x \) is not differentiable at these points.

Solution: The function \( \sec x \) is defined as: \[ \sec x = \frac{1}{\cos x}. \] Differentiating both sides, \[ (\sec x)’ = \sec x \tan x. \] The function \( \sec x \tan x \) is not defined where \( \cos x = 0 \), i.e., at \( x = \frac{\pi}{2} + k\pi, \, k \in \mathbb{Z} \). Hence, \( \sec x \) is not differentiable at these points.

Solution: The function \( \csc x \) is defined as: \[ \csc x = \frac{1}{\sin x}. \] Differentiating both sides, \[ (\csc x)’ = -\csc x \cot x. \] The function \( \csc x \cot x \) is not defined where \( \sin x = 0 \), i.e., at \( x = k\pi, \, k \in \mathbb{Z} \). Hence, \( \csc x \) is not differentiable at these points.

Solution: A function is differentiable at \( x = 2 \) if the left-hand derivative (LHD) and right-hand derivative (RHD) exist and are equal. **Step 1: Compute Left-Hand Derivative (LHD)** The left-hand derivative is given by: \[ \lim_{h \to 0^-} \frac{f(2+h) – f(2)}{h}. \] Since \( x \leq 2 \), we use \( f(x) = 1 + x \), \[ f(2) = 1 + 2 = 3, \quad f(2+h) = 1 + (2+h) = 3 + h. \] Therefore, \[ \text{LHD} = \lim_{h \to 0^-} \frac{(3 + h) – 3}{h} = \lim_{h \to 0^-} \frac{h}{h} = 1. \] **Step 2: Compute Right-Hand Derivative (RHD)** The right-hand derivative is given by: \[ \lim_{h \to 0^+} \frac{f(2+h) – f(2)}{h}. \] Since \( x > 2 \), we use \( f(x) = 5 – x \), \[ f(2) = 3, \quad f(2+h) = 5 – (2+h) = 3 – h. \] Therefore, \[ \text{RHD} = \lim_{h \to 0^+} \frac{(3 – h) – 3}{h} = \lim_{h \to 0^+} \frac{-h}{h} = -1. \] **Step 3: Compare LHD and RHD** Since \( \text{LHD} = 1 \) and \( \text{RHD} = -1 \), they are not equal. Hence, \( f(x) \) is not differentiable at \( x = 2 \).

Solution: Given that \( f(2) = 4 \) and \( f'(2) = 4 \), we need to evaluate: \[ \lim_{x \to 2} \frac{x f(2) – 2 f(x)}{x – 2}. \] **Step 1: Substitute the given value of \( f(2) \)** \[ \lim_{x \to 2} \frac{x(4) – 2 f(x)}{x – 2}. \] This simplifies to: \[ \lim_{x \to 2} \frac{4x – 2 f(x)}{x – 2}. \] **Step 2: Express in terms of \( f(x) \)** We rewrite the numerator as: \[ 4x – 2 f(x) = 2(2x – f(x)). \] So the expression becomes: \[ \lim_{x \to 2} \frac{2(2x – f(x))}{x – 2}. \] **Step 3: Use the definition of the derivative** We know that: \[ f'(2) = \lim_{x \to 2} \frac{f(x) – f(2)}{x – 2}. \] Substituting \( f(2) = 4 \) and \( f'(2) = 4 \), we get: \[ 4 = \lim_{x \to 2} \frac{f(x) – 4}{x – 2}. \] Rearranging, \[ \lim_{x \to 2} \frac{4x – f(x)}{x – 2} = 4. \] **Step 4: Compute the final limit** \[ \lim_{x \to 2} \frac{2(2x – f(x))}{x – 2} = 2 \times 4 = 8. \]

Solution: **Step 1: Checking Continuity at \( x = 1 \)** Since the function \( f(x) \) is not defined at \( x = 1 \), it is automatically **discontinuous** at \( x = 1 \). **Step 2: Checking Differentiability at \( x = 2 \)** A function is differentiable at \( x = 2 \) if its left-hand derivative (LHD) and right-hand derivative (RHD) exist and are equal. **Left-Hand Derivative (LHD) at \( x = 2 \):** Since \( 1 < x < 2 \), we use the function \( f(x) = 4x^2 - 3x \). First, calculate \( f(2) \): \[ f(2) = 4(2)^2 - 3(2) = 16 - 6 = 10. \] The left-hand derivative is: \[ \lim_{h \to 0^-} \frac{f(2 + h) - f(2)}{h}. \] Expanding \( f(2 + h) \): \[ f(2 + h) = 4(2+h)^2 - 3(2+h). \] Expanding: \[ = 4(4 + 4h + h^2) - 6 - 3h = 16 + 16h + 4h^2 - 6 - 3h. \] \[ = 10 + 16h + 4h^2 - 3h. \] Substituting in the LHD formula: \[ \lim_{h \to 0^-} \frac{(10 + 16h + 4h^2 - 3h) - 10}{h}. \] \[ = \lim_{h \to 0^-} \frac{16h + 4h^2 - 3h}{h}. \] \[ = \lim_{h \to 0^-} (13 + 4h) = 13. \] So, \( \text{LHD} = 13 \). **Right-Hand Derivative (RHD) at \( x = 2 \):** Since \( x \geq 2 \), we use \( f(x) = 3x + 4 \). \[ f(2) = 3(2) + 4 = 6 + 4 = 10. \] The right-hand derivative is: \[ \lim_{h \to 0^+} \frac{f(2+h) - f(2)}{h}. \] Expanding \( f(2+h) \): \[ f(2+h) = 3(2+h) + 4 = 6 + 3h + 4 = 10 + 3h. \] Substituting in the RHD formula: \[ \lim_{h \to 0^+} \frac{(10 + 3h) - 10}{h}. \] \[ = \lim_{h \to 0^+} \frac{3h}{h} = 3. \] So, \( \text{RHD} = 3 \). **Conclusion:** Since \( \text{LHD} \neq \text{RHD} \) (i.e., \( 13 \neq 3 \)), the function is **not differentiable** at \( x = 2 \).

Solution: **Step 1: Checking Continuity at \( x = 0 \)** A function is continuous at \( x = 0 \) if: \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0). \] The function is defined as: \[ f(x) = 2x – |x|. \] **For \( x > 0 \)**: Since \( |x| = x \), we get: \[ f(x) = 2x – x = x. \] Thus, \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0. \] **For \( x < 0 \)**: Since \( |x| = -x \), we get: \[ f(x) = 2x - (-x) = 3x. \] Thus, \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 3x = 0. \] Evaluating \( f(0) \): \[ f(0) = 2(0) - |0| = 0. \] Since, \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0), \] the function is **continuous at \( x = 0 \)**. **Step 2: Checking Differentiability at \( x = 0 \)** A function is differentiable at \( x = 0 \) if the left-hand derivative (LHD) and right-hand derivative (RHD) exist and are equal. **Left-Hand Derivative (LHD) at \( x = 0 \):** \[ \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h}. \] Since \( f(x) = 3x \) for \( x < 0 \), \[ f(h) = 3h, \quad f(0) = 0. \] \[ \text{LHD} = \lim_{h \to 0^-} \frac{3h - 0}{h} = \lim_{h \to 0^-} 3 = 3. \] **Right-Hand Derivative (RHD) at \( x = 0 \):** \[ \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h}. \] Since \( f(x) = x \) for \( x > 0 \), \[ f(h) = h, \quad f(0) = 0. \] \[ \text{RHD} = \lim_{h \to 0^+} \frac{h – 0}{h} = \lim_{h \to 0^+} 1 = 1. \] Since \( \text{LHD} \neq \text{RHD} \) (i.e., \( 3 \neq 1 \)), the function is **not differentiable at \( x = 0 \)**.

Solution: **Step 1: Checking Continuity at \( x = 5 \)** A function is continuous at \( x = 5 \) if: \[ \lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = f(5). \] The function is given as: \[ f(x) = |x – 5|. \] **For \( x > 5 \)**: Since \( x – 5 \) is positive, we have: \[ f(x) = x – 5. \] Thus, \[ \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (x – 5) = 0. \] **For \( x < 5 \)**: Since \( x - 5 \) is negative, we take the absolute value: \[ f(x) = -(x - 5) = 5 - x. \] Thus, \[ \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (5 - x) = 0. \] Evaluating \( f(5) \): \[ f(5) = |5 - 5| = 0. \] Since, \[ \lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = f(5), \] the function is **continuous at \( x = 5 \)**. **Step 2: Checking Differentiability at \( x = 5 \)** A function is differentiable at \( x = 5 \) if the left-hand derivative (LHD) and right-hand derivative (RHD) exist and are equal. **Left-Hand Derivative (LHD) at \( x = 5 \):** \[ \lim_{h \to 0^-} \frac{f(5+h) - f(5)}{h}. \] Since \( f(x) = 5 - x \) for \( x < 5 \), \[ f(5 + h) = 5 - (5 + h) = -h, \quad f(5) = 0. \] \[ \text{LHD} = \lim_{h \to 0^-} \frac{-h - 0}{h} = \lim_{h \to 0^-} -1 = -1. \] **Right-Hand Derivative (RHD) at \( x = 5 \):** \[ \lim_{h \to 0^+} \frac{f(5+h) - f(5)}{h}. \] Since \( f(x) = x - 5 \) for \( x > 5 \), \[ f(5 + h) = (5 + h) – 5 = h, \quad f(5) = 0. \] \[ \text{RHD} = \lim_{h \to 0^+} \frac{h – 0}{h} = \lim_{h \to 0^+} 1 = 1. \] Since \( \text{LHD} \neq \text{RHD} \) (i.e., \( -1 \neq 1 \)), the function is **not differentiable at \( x = 5 \)**.

Solution: **Step 1: Checking Continuity at \( x = 3 \) and \( x = 4 \)** The function \( f(x) = |x – 3| + |x – 4| \) consists of absolute values, which are continuous everywhere. Hence, \( f(x) \) is continuous at \( x = 3 \) and \( x = 4 \). **Step 2: Checking Differentiability at \( x = 3 \) and \( x = 4 \)** A function is differentiable at a point if the left-hand derivative (LHD) and right-hand derivative (RHD) exist and are equal. **Case 1: Checking Differentiability at \( x = 3 \)** The function is split into different expressions based on the value of \( x \): – For \( x < 3 \): \[ f(x) = (3 - x) + (4 - x) = 7 - 2x \] Differentiating: \[ f'(x) = -2. \] - For \( x > 3 \): \[ f(x) = (x – 3) + (4 – x) = 1. \] Differentiating: \[ f'(x) = 0. \] Now, calculating the left-hand and right-hand derivatives: \[ \text{LHD} = \lim_{h \to 0^-} \frac{f(3 + h) – f(3)}{h} = -2. \] \[ \text{RHD} = \lim_{h \to 0^+} \frac{f(3 + h) – f(3)}{h} = 0. \] Since \( \text{LHD} \neq \text{RHD} \), \( f(x) \) is **not differentiable at \( x = 3 \)**. **Case 2: Checking Differentiability at \( x = 4 \)** – For \( x < 4 \): \[ f(x) = (x - 3) + (4 - x) = 1. \] Differentiating: \[ f'(x) = 0. \] - For \( x > 4 \): \[ f(x) = (x – 3) + (x – 4) = 2x – 7. \] Differentiating: \[ f'(x) = 2. \] Now, calculating the left-hand and right-hand derivatives: \[ \text{LHD} = \lim_{h \to 0^-} \frac{f(4 + h) – f(4)}{h} = 0. \] \[ \text{RHD} = \lim_{h \to 0^+} \frac{f(4 + h) – f(4)}{h} = 2. \] Since \( \text{LHD} \neq \text{RHD} \), \( f(x) \) is **not differentiable at \( x = 4 \)**.

Solution: **Step 1: Checking Continuity at \( x=1 \)** A function is continuous at \( x = 1 \) if: \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1). \] – Left-hand limit (\( LHL \)): \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x + a) = 1 + a. \] – Right-hand limit (\( RHL \)): \[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (a x^2 + 1) = a(1)^2 + 1 = a + 1. \] – Function value at \( x=1 \): \[ f(1) = a(1)^2 + 1 = a + 1. \] Since \( LHL = RHL = f(1) \), the function is continuous for all values of \( a \). **Step 2: Finding the Left-Hand and Right-Hand Derivatives at \( x=1 \)** – Left-hand derivative (\( LHD \)): \[ LHD = \lim_{h \to 0^-} \frac{f(1+h) – f(1)}{h} \] \[ = \lim_{h \to 0^-} \frac{(1+h) + a – (a+1)}{h} \] \[ = \lim_{h \to 0^-} \frac{h}{h} = 1. \] – Right-hand derivative (\( RHD \)): \[ RHD = \lim_{h \to 0^+} \frac{f(1+h) – f(1)}{h} \] \[ = \lim_{h \to 0^+} \frac{a(1+h)^2 + 1 – (a+1)}{h} \] Expanding \( (1+h)^2 = 1 + 2h + h^2 \), we get: \[ = \lim_{h \to 0^+} \frac{a(1 + 2h + h^2) + 1 – (a+1)}{h} \] \[ = \lim_{h \to 0^+} \frac{a + 2ah + a h^2 + 1 – a – 1}{h} \] \[ = \lim_{h \to 0^+} \frac{2ah + a h^2}{h} \] \[ = \lim_{h \to 0^+} (2a + a h). \] As \( h \to 0 \), we get \( RHD = 2a \). **Step 3: Condition for Differentiability at \( x=1 \)** For differentiability at \( x=1 \), \( LHD \) must be equal to \( RHD \): \[ 1 = 2a. \] Solving for \( a \): \[ a = \frac{1}{2}. \]

Solution: **Step 1: Checking Continuity at \( x=0 \)** A function is continuous at \( x=0 \) if: \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0). \] – Left-hand limit (\( LHL \)): \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{|x|}. \] Since \( x < 0 \Rightarrow |x| = -x \), we get: \[ \frac{x}{-x} = -1. \] - Right-hand limit (\( RHL \)): \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{x}{|x|}. \] Since \( x > 0 \Rightarrow |x| = x \), we get: \[ \frac{x}{x} = 1. \] – Function value at \( x=0 \): \[ f(0) = 1. \] Since \( LHL = -1 \neq RHL = 1 \), the left-hand and right-hand limits are not equal, so \( \lim_{x \to 0} f(x) \) does not exist. Hence, \( f(x) \) is **discontinuous** at \( x=0 \). **Step 2: Checking Differentiability at \( x=0 \)** A function is differentiable at \( x=0 \) if: \[ LHD = RHD. \] – Left-hand derivative (\( LHD \)): \[ LHD = \lim_{h \to 0^-} \frac{f(0+h) – f(0)}{h} \] \[ = \lim_{h \to 0^-} \frac{-1 – 1}{h} = \lim_{h \to 0^-} \frac{-2}{h}. \] Since \( h \to 0^- \), \( LHD \to -\infty \). – Right-hand derivative (\( RHD \)): \[ RHD = \lim_{h \to 0^+} \frac{f(0+h) – f(0)}{h} \] \[ = \lim_{h \to 0^+} \frac{1 – 1}{h} = \lim_{h \to 0^+} \frac{0}{h} = 0. \] Since \( LHD \neq RHD \), the function is **not differentiable at \( x=0 \)**.