RBSE Solutions Class 12 Maths Chapter 6 Continuity and Differentiability Exercise 6.2

Question 1. Show that following functions are differentiable for every value of x :
(i) Identity function, \( f (x) = x \)
(ii) Constant function, \( f (x) = c \), where c is a constant
(iii) \( f(x) = e^x \)
(iv) \( f(x) = \sin x \)
Answer:
To show that a function is differentiable at every point, we need to prove that its left-hand derivative (LHD) and right-hand derivative (RHD) are equal at any arbitrary point 'a' in its domain.
(i) Given, identity function \( f(x) = x \), where \( x \in R \). Let 'a' be an arbitrary constant.
At \( x = a \), the Left-hand derivative of \( f(x) \) is:
\( Lf'(a) = \lim_{h \to 0} \frac{f(a-h) - f(a)}{-h} \)
\( = \lim_{h \to 0} \frac{(a-h) - a}{-h} \)
\( = \lim_{h \to 0} \frac{-h}{-h} \)
\( = \lim_{h \to 0} (1) \)
\( = 1 \)
Now, the Right-hand derivative of \( f(x) \) is:
\( Rf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \)
\( = \lim_{h \to 0} \frac{(a+h) - a}{h} \)
\( = \lim_{h \to 0} \frac{h}{h} \)
\( = \lim_{h \to 0} (1) \)
\( = 1 \)
Since \( Lf'(a) = Rf'(a) = 1 \), the identity function \( f(x) = x \) is differentiable for every value of \( x \).
(ii) Given, constant function \( f(x) = c \), where c is a constant. The domain of \( f(x) \) is the set of all real numbers (R). Let 'a' be any arbitrary real number.
At \( x = a \), the Left-hand derivative of \( f(x) \) is:
\( Lf'(a) = \lim_{h \to 0} \frac{f(a-h) - f(a)}{-h} \)
\( = \lim_{h \to 0} \frac{c - c}{-h} \)
\( = \lim_{h \to 0} \frac{0}{-h} \)
\( = 0 \)
Now, the Right-hand derivative of \( f(x) \) is:
\( Rf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \)
\( = \lim_{h \to 0} \frac{c - c}{h} \)
\( = \lim_{h \to 0} \frac{0}{h} \)
\( = 0 \)
Since \( Lf'(a) = Rf'(a) = 0 \), the constant function \( f(x) = c \) is differentiable for every value of \( x \).
(iii) Given, \( f(x) = e^x \). Let 'a' be an arbitrary constant.
At \( x = a \), the Left-hand derivative of \( f(x) \) is:
\( Lf'(a) = \lim_{h \to 0} \frac{f(a-h) - f(a)}{-h} \)
\( = \lim_{h \to 0} \frac{e^{a-h} - e^a}{-h} \)
\( = \lim_{h \to 0} \frac{e^a (e^{-h} - 1)}{-h} \)
Using the series expansion \( e^{-h} = 1 - h + \frac{h^2}{2!} - \frac{h^3}{3!} + ... \)
\( = \lim_{h \to 0} \frac{e^a (1 - h + \frac{h^2}{2!} - \frac{h^3}{3!} + ... - 1)}{-h} \)
\( = \lim_{h \to 0} \frac{e^a (-h + \frac{h^2}{2!} - \frac{h^3}{3!} + ...)}{-h} \)
\( = \lim_{h \to 0} e^a (1 - \frac{h}{2!} + \frac{h^2}{3!} - ...) \)
\( = e^a (1 - 0 + 0 - ...) \)
\( = e^a \)
Now, the Right-hand derivative of \( f(x) \) is:
\( Rf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \)
\( = \lim_{h \to 0} \frac{e^{a+h} - e^a}{h} \)
\( = \lim_{h \to 0} \frac{e^a (e^h - 1)}{h} \)
Using the series expansion \( e^h = 1 + h + \frac{h^2}{2!} + \frac{h^3}{3!} + ... \)
\( = \lim_{h \to 0} \frac{e^a (1 + h + \frac{h^2}{2!} + \frac{h^3}{3!} + ... - 1)}{h} \)
\( = \lim_{h \to 0} \frac{e^a (h + \frac{h^2}{2!} + \frac{h^3}{3!} + ...)}{h} \)
\( = \lim_{h \to 0} e^a (1 + \frac{h}{2!} + \frac{h^2}{3!} + ...) \)
\( = e^a (1 + 0 + 0 + ...) \)
\( = e^a \)
Since \( Lf'(a) = Rf'(a) = e^a \), the function \( f(x) = e^x \) is differentiable for every value of \( x \).
(iv) Given, \( f(x) = \sin x \). Let 'a' be an arbitrary real number.
At \( x = a \), the Left-hand derivative of \( f(x) \) is:
\( Lf'(a) = \lim_{h \to 0} \frac{f(a-h) - f(a)}{-h} \)
\( = \lim_{h \to 0} \frac{\sin(a-h) - \sin a}{-h} \)
Using the identity \( \sin C - \sin D = 2 \cos \left( \frac{C+D}{2} \right) \sin \left( \frac{C-D}{2} \right) \)
\( = \lim_{h \to 0} \frac{2 \cos \left( \frac{a-h+a}{2} \right) \sin \left( \frac{a-h-a}{2} \right)}{-h} \)
\( = \lim_{h \to 0} \frac{2 \cos \left( a - \frac{h}{2} \right) \sin \left( -\frac{h}{2} \right)}{-h} \)
\( = \lim_{h \to 0} \frac{-2 \cos \left( a - \frac{h}{2} \right) \sin \left( \frac{h}{2} \right)}{-h} \)
\( = \lim_{h \to 0} \cos \left( a - \frac{h}{2} \right) \cdot \lim_{h \to 0} \frac{\sin \left( \frac{h}{2} \right)}{\frac{h}{2}} \)
We know \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).
\( = \cos(a-0) \times 1 \)
\( = \cos a \)
Now, the Right-hand derivative of \( f(x) \) is:
\( Rf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \)
\( = \lim_{h \to 0} \frac{\sin(a+h) - \sin a}{h} \)
Using the identity \( \sin C - \sin D = 2 \cos \left( \frac{C+D}{2} \right) \sin \left( \frac{C-D}{2} \right) \)
\( = \lim_{h \to 0} \frac{2 \cos \left( \frac{a+h+a}{2} \right) \sin \left( \frac{a+h-a}{2} \right)}{h} \)
\( = \lim_{h \to 0} \frac{2 \cos \left( a + \frac{h}{2} \right) \sin \left( \frac{h}{2} \right)}{h} \)
\( = \lim_{h \to 0} \cos \left( a + \frac{h}{2} \right) \cdot \lim_{h \to 0} \frac{\sin \left( \frac{h}{2} \right)}{\frac{h}{2}} \)
\( = \cos(a+0) \times 1 \)
\( = \cos a \)
Since \( Lf'(a) = Rf'(a) = \cos a \), the function \( f(x) = \sin x \) is differentiable for every value of \( x \).
In simple words: To check if a function can be differentiated everywhere, we find its "slope" from the left and from the right at any point. If these slopes are the same, the function is smooth and differentiable at that point. For the identity, constant, exponential, and sine functions, these slopes always match, so they are all differentiable for all numbers.

🎯 Exam Tip: Always state the definition of differentiability using LHD and RHD. For complex functions, apply appropriate limit formulas, trigonometric identities, or series expansions carefully.

Question 2. Examine the differentiability of the function \( f(x) = |x| \) at \( x = 0 \).
Answer:
To examine the differentiability of \( f(x) = |x| \) at \( x = 0 \), we need to find its Left-hand derivative (LHD) and Right-hand derivative (RHD) at \( x=0 \).
The function is defined as \( f(x) = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases} \). Also, \( f(0) = |0| = 0 \).
Left-hand derivative at \( x = 0 \):
\( Lf'(0) = \lim_{h \to 0} \frac{f(0-h) - f(0)}{-h} \)
\( = \lim_{h \to 0} \frac{f(-h) - 0}{-h} \)
Since \( -h < 0 \), \( f(-h) = |-h| = h \).
\( = \lim_{h \to 0} \frac{h - 0}{-h} \)
\( = \lim_{h \to 0} (-1) \)
\( = -1 \)
Right-hand derivative at \( x = 0 \):
\( Rf'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} \)
\( = \lim_{h \to 0} \frac{f(h) - 0}{h} \)
Since \( h > 0 \), \( f(h) = |h| = h \).
\( = \lim_{h \to 0} \frac{h - 0}{h} \)
\( = \lim_{h \to 0} (1) \)
\( = 1 \)
Since \( Lf'(0) = -1 \) and \( Rf'(0) = 1 \), we have \( Lf'(0) \neq Rf'(0) \).
Therefore, the function \( f(x) = |x| \) is not differentiable at \( x = 0 \). This is a common point where absolute value functions change direction sharply.
In simple words: For the function \( f(x) = |x| \), the slope changes suddenly at \( x = 0 \). If you approach \( x = 0 \) from the left, the slope is -1. If you approach from the right, the slope is 1. Because these slopes are different, the function is not smooth and cannot be differentiated right at \( x = 0 \).

🎯 Exam Tip: When dealing with absolute value functions, always break them into piecewise definitions and examine differentiability at the points where the expression inside the absolute value becomes zero.

Question 3. Examine for differentiability of function, \( f(x) = |x - 1 | + | x | \) at \( x = 0, 1 \).
Answer:
First, we rewrite the given function \( f(x) = |x - 1 | + | x | \) as a piecewise function. The critical points are where \( x-1=0 \) (i.e., \( x=1 \)) and where \( x=0 \).
\( f(x) = \begin{cases} -(x-1) - x = 1-2x & \text{if } x < 0 \\ -(x-1) + x = 1 & \text{if } 0 \le x < 1 \\ (x-1) + x = 2x-1 & \text{if } x \ge 1 \end{cases} \)
Now, we examine differentiability at \( x = 0 \) and \( x = 1 \).
**At \( x = 0 \):**
Left-hand derivative at \( x = 0 \):
\( Lf'(0) = \lim_{h \to 0} \frac{f(0-h) - f(0)}{-h} \)
\( = \lim_{h \to 0} \frac{f(-h) - f(0)}{-h} \)
For \( x < 0 \), \( f(x) = 1-2x \). So \( f(-h) = 1-2(-h) = 1+2h \).
For \( 0 \le x < 1 \), \( f(x) = 1 \). So \( f(0) = 1 \).
\( = \lim_{h \to 0} \frac{(1+2h) - 1}{-h} \)
\( = \lim_{h \to 0} \frac{2h}{-h} \)
\( = \lim_{h \to 0} (-2) \)
\( = -2 \)
Right-hand derivative at \( x = 0 \):
\( Rf'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} \)
\( = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \)
For \( 0 \le x < 1 \), \( f(x) = 1 \). So \( f(h) = 1 \) (since \( h \to 0^+ \), \( h \) is a small positive number).
\( = \lim_{h \to 0} \frac{1 - 1}{h} \)
\( = \lim_{h \to 0} \frac{0}{h} \)
\( = 0 \)
Since \( Lf'(0) = -2 \) and \( Rf'(0) = 0 \), we have \( Lf'(0) \neq Rf'(0) \).
Therefore, \( f(x) \) is not differentiable at \( x = 0 \).
**At \( x = 1 \):**
Left-hand derivative at \( x = 1 \):
\( Lf'(1) = \lim_{h \to 0} \frac{f(1-h) - f(1)}{-h} \)
For \( 0 \le x < 1 \), \( f(x) = 1 \). So \( f(1-h) = 1 \) (since \( 1-h < 1 \)).
For \( x \ge 1 \), \( f(x) = 2x-1 \). So \( f(1) = 2(1)-1 = 1 \).
\( = \lim_{h \to 0} \frac{1 - 1}{-h} \)
\( = \lim_{h \to 0} \frac{0}{-h} \)
\( = 0 \)
Right-hand derivative at \( x = 1 \):
\( Rf'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} \)
For \( x \ge 1 \), \( f(x) = 2x-1 \). So \( f(1+h) = 2(1+h)-1 = 2+2h-1 = 1+2h \).
\( = \lim_{h \to 0} \frac{(1+2h) - 1}{h} \)
\( = \lim_{h \to 0} \frac{2h}{h} \)
\( = \lim_{h \to 0} (2) \)
\( = 2 \)
Since \( Lf'(1) = 0 \) and \( Rf'(1) = 2 \), we have \( Lf'(1) \neq Rf'(1) \).
Therefore, \( f(x) \) is not differentiable at \( x = 1 \).
In conclusion, the function \( f(x) = |x-1|+|x| \) is not differentiable at both \( x = 0 \) and \( x = 1 \). The absolute value changes create sharp corners at these points.
In simple words: We check how "smooth" the function \( f(x) = |x-1| + |x| \) is at two special points, \( x=0 \) and \( x=1 \). We found that at both these points, the slope from the left side is different from the slope from the right side. This means the function has sharp corners at these points and cannot be differentiated there.

🎯 Exam Tip: Always define the piecewise function correctly first, especially when dealing with multiple absolute value terms. Then, apply the derivative limits at each critical point.

Question 4. Examine the function for differentiability in interval \( [0, 2] \), if \( f(x) = |x-1|+|x-2| \).
Answer:
First, we rewrite the given function \( f(x) = |x-1|+|x-2| \) as a piecewise function over the interval \( [0, 2] \). The critical points are \( x=1 \) and \( x=2 \).
\( f(x) = \begin{cases} -(x-1) - (x-2) = 1-x-x+2 = 3-2x & \text{if } x < 1 \\ (x-1) - (x-2) = x-1-x+2 = 1 & \text{if } 1 \le x < 2 \\ (x-1) + (x-2) = 2x-3 & \text{if } x \ge 2 \end{cases} \)
We will test differentiability at \( x=1 \) since \( 1 \in [0, 2] \).
**At \( x = 1 \):**
Left-hand derivative at \( x = 1 \):
\( Lf'(1) = \lim_{h \to 0} \frac{f(1-h) - f(1)}{-h} \)
For \( x < 1 \), \( f(x) = 3-2x \). So \( f(1-h) = 3-2(1-h) = 3-2+2h = 1+2h \).
For \( 1 \le x < 2 \), \( f(x) = 1 \). So \( f(1) = 1 \).
\( = \lim_{h \to 0} \frac{(1+2h) - 1}{-h} \)
\( = \lim_{h \to 0} \frac{2h}{-h} \)
\( = \lim_{h \to 0} (-2) \)
\( = -2 \)
Right-hand derivative at \( x = 1 \):
\( Rf'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} \)
For \( 1 \le x < 2 \), \( f(x) = 1 \). So \( f(1+h) = 1 \) (since \( 1+h \) is slightly greater than 1 but less than 2).
\( = \lim_{h \to 0} \frac{1 - 1}{h} \)
\( = \lim_{h \to 0} \frac{0}{h} \)
\( = 0 \)
Since \( Lf'(1) = -2 \) and \( Rf'(1) = 0 \), we have \( Lf'(1) \neq Rf'(1) \).
Therefore, the function \( f(x) \) is not differentiable at \( x = 1 \). Since differentiability fails at a point within the interval, the function is not differentiable over the entire interval \( [0, 2] \). We can also check at \( x=2 \), where it would also not be differentiable due to the absolute value function.
In simple words: We are checking if the function \( f(x) = |x-1|+|x-2| \) is smooth everywhere between 0 and 2. We looked at \( x=1 \) because that's where one of the absolute value parts might create a sharp turn. At \( x=1 \), the slope from the left is -2, but from the right, it's 0. Since these are different, the function has a sharp corner at \( x=1 \) and is not smooth there. Because of this, it's not differentiable in the whole interval \( [0,2] \).

🎯 Exam Tip: For functions with multiple absolute values, always identify all critical points (where the expressions inside the absolute values become zero) and define the function piecewise across the sub-intervals. Then check differentiability at each critical point.

Question 5. Examine the function for differentiability at \( x = 0 \); \( f(x) = \begin{cases} x \tan^{-1} x & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \).
Answer:
To examine the differentiability of \( f(x) \) at \( x = 0 \), we need to compute its Left-hand derivative (LHD) and Right-hand derivative (RHD) at \( x=0 \).
We know \( f(0) = 0 \).
Left-hand derivative at \( x = 0 \):
\( Lf'(0) = \lim_{h \to 0} \frac{f(0-h) - f(0)}{-h} \)
\( = \lim_{h \to 0} \frac{f(-h) - 0}{-h} \)
Since \( -h \neq 0 \), \( f(-h) = (-h) \tan^{-1}(-h) \).
\( = \lim_{h \to 0} \frac{(-h) \tan^{-1}(-h)}{-h} \)
\( = \lim_{h \to 0} \tan^{-1}(-h) \)
As \( h \to 0 \), \( -h \to 0 \). And \( \tan^{-1}(0) = 0 \).
\( = 0 \)
Right-hand derivative at \( x = 0 \):
\( Rf'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} \)
\( = \lim_{h \to 0} \frac{f(h) - 0}{h} \)
Since \( h \neq 0 \), \( f(h) = h \tan^{-1}(h) \).
\( = \lim_{h \to 0} \frac{h \tan^{-1}(h)}{h} \)
\( = \lim_{h \to 0} \tan^{-1}(h) \)
As \( h \to 0 \), \( \tan^{-1}(0) = 0 \).
\( = 0 \)
Since \( Lf'(0) = 0 \) and \( Rf'(0) = 0 \), we have \( Lf'(0) = Rf'(0) \).
Therefore, the function \( f(x) \) is differentiable at \( x = 0 \).
In simple words: To check if the function \( f(x) = x \tan^{-1} x \) (when \( x \) is not zero) and \( f(x) = 0 \) (when \( x \) is zero) is smooth at \( x=0 \), we found the slope from both the left and the right sides. Both slopes turned out to be zero. Since they are the same, the function is smooth and differentiable at \( x=0 \).

🎯 Exam Tip: Remember that \( \tan^{-1}(0) = 0 \). When calculating limits involving \( \tan^{-1} x / x \) or similar forms, apply standard limit rules and inverse trigonometric function properties.

Question 6. Examine the function \( f(x) \) for differentiability at \( x = 0 \), if \( f(x) = \begin{cases} \frac{1-\cos x}{2} & \text{if } x \le 0 \\ \frac{x-2x^2}{2} & \text{if } x > 0 \end{cases} \).
Answer:
To examine differentiability at \( x = 0 \), we first need to verify continuity at \( x=0 \). If a function is not continuous at a point, it cannot be differentiable at that point.
\( f(0) = \frac{1-\cos 0}{2} = \frac{1-1}{2} = 0 \)
\( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1-\cos x}{2} = \frac{1-\cos 0}{2} = 0 \)
\( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{x-2x^2}{2} = \frac{0-0}{2} = 0 \)
Since \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0 \), the function is continuous at \( x = 0 \). Now we check differentiability.
Left-hand derivative at \( x = 0 \):
\( Lf'(0) = \lim_{h \to 0} \frac{f(0-h) - f(0)}{-h} \)
\( = \lim_{h \to 0} \frac{f(-h) - 0}{-h} \)
For \( x \le 0 \), \( f(x) = \frac{1-\cos x}{2} \). So \( f(-h) = \frac{1-\cos(-h)}{2} = \frac{1-\cos h}{2} \) (since \( \cos(-h) = \cos h \)).
\( = \lim_{h \to 0} \frac{\frac{1-\cos h}{2}}{-h} \)
\( = \lim_{h \to 0} \frac{1-\cos h}{-2h} \)
Using the identity \( 1-\cos h = 2\sin^2 \left( \frac{h}{2} \right) \)
\( = \lim_{h \to 0} \frac{2\sin^2 \left( \frac{h}{2} \right)}{-2h} \)
\( = \lim_{h \to 0} \frac{\sin^2 \left( \frac{h}{2} \right)}{-h} \)
\( = \lim_{h \to 0} \frac{\sin \left( \frac{h}{2} \right)}{\frac{h}{2}} \cdot \frac{\sin \left( \frac{h}{2} \right)}{-1} \cdot \frac{1}{2} \)
\( = 1 \cdot \frac{\sin(0)}{-1} \cdot \frac{1}{2} \)
\( = 1 \cdot 0 \cdot \frac{1}{2} \)
\( = 0 \)
Right-hand derivative at \( x = 0 \):
\( Rf'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} \)
\( = \lim_{h \to 0} \frac{f(h) - 0}{h} \)
For \( x > 0 \), \( f(x) = \frac{x-2x^2}{2} \). So \( f(h) = \frac{h-2h^2}{2} \).
\( = \lim_{h \to 0} \frac{\frac{h-2h^2}{2}}{h} \)
\( = \lim_{h \to 0} \frac{h(1-2h)}{2h} \)
\( = \lim_{h \to 0} \frac{1-2h}{2} \)
\( = \frac{1-2(0)}{2} \)
\( = \frac{1}{2} \)
Since \( Lf'(0) = 0 \) and \( Rf'(0) = \frac{1}{2} \), we have \( Lf'(0) \neq Rf'(0) \).
Therefore, the function \( f(x) \) is not differentiable at \( x = 0 \).
In simple words: To see if this function is smooth at \( x=0 \), we checked its slopes from both sides. From the left, the slope was 0. From the right, the slope was \( \frac{1}{2} \). Since these slopes are not the same, the function has a sharp point or a break in its smoothness at \( x=0 \) and cannot be differentiated there.

🎯 Exam Tip: Always check for continuity first. If a function is discontinuous, it is automatically non-differentiable. For trigonometric limits, use standard limit formulas like \( \lim_{x \to 0} \frac{1-\cos x}{x^2} = \frac{1}{2} \) or \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).

Question 7. Consider a function \( f(x) = \begin{cases} x^m \cos \left( \frac{1}{x} \right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \). Determine the conditions on \( m \) for:
**(a) Continuity at \( x = 0 \)**
**(b) Differentiability at \( x = 0 \)**
Answer:
**(a) Continuity at \( x = 0 \):**
For \( f(x) \) to be continuous at \( x = 0 \), we need \( \lim_{x \to 0} f(x) = f(0) \).
Given \( f(0) = 0 \).
We need to evaluate \( \lim_{x \to 0} x^m \cos \left( \frac{1}{x} \right) \).
We know that \( -1 \le \cos \left( \frac{1}{x} \right) \le 1 \) for all \( x \neq 0 \).
If \( m > 0 \), then multiplying by \( x^m \) (assuming \( x^m > 0 \), which is true for small \( x \) if \( m \) is an even integer or \( x>0 \)):
\( -x^m \le x^m \cos \left( \frac{1}{x} \right) \le x^m \)
As \( x \to 0 \), \( \lim_{x \to 0} (-x^m) = 0 \) and \( \lim_{x \to 0} (x^m) = 0 \).
By the Squeeze Theorem, \( \lim_{x \to 0} x^m \cos \left( \frac{1}{x} \right) = 0 \).
Since \( \lim_{x \to 0} f(x) = 0 = f(0) \), the function is continuous at \( x=0 \) if \( m > 0 \).
**(b) Differentiability at \( x = 0 \):**
For \( f(x) \) to be differentiable at \( x = 0 \), we need the derivative \( f'(0) \) to exist.
Using the definition of the derivative:
\( f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} \)
\( = \lim_{h \to 0} \frac{h^m \cos \left( \frac{1}{h} \right) - 0}{h} \)
\( = \lim_{h \to 0} h^{m-1} \cos \left( \frac{1}{h} \right) \)
For this limit to exist and be equal to 0 (since the derivative of a continuous function with oscillations at 0 usually needs to be 0 for differentiability), we apply the same logic as for continuity.
We know \( -1 \le \cos \left( \frac{1}{h} \right) \le 1 \).
So, \( -h^{m-1} \le h^{m-1} \cos \left( \frac{1}{h} \right) \le h^{m-1} \)
For \( \lim_{h \to 0} h^{m-1} \cos \left( \frac{1}{h} \right) \) to be 0, we need \( \lim_{h \to 0} h^{m-1} = 0 \). This occurs if \( m-1 > 0 \), which means \( m > 1 \).
Thus, the function is differentiable at \( x=0 \) if \( m > 1 \).
If \( m=1 \), the limit becomes \( \lim_{h \to 0} \cos \left( \frac{1}{h} \right) \), which does not exist because \( \cos \left( \frac{1}{h} \right) \) oscillates between -1 and 1 as \( h \to 0 \).
In simple words: This problem asks for what values of 'm' a special function \( f(x) = x^m \cos(1/x) \) is "smooth" (differentiable) or "connected" (continuous) at \( x=0 \). For the function to be connected at \( x=0 \), the power 'm' must be greater than 0. For it to be smooth (differentiable) at \( x=0 \), the power 'm' must be greater than 1. This means that just being connected is not enough to be smooth; you need a stronger condition on 'm'.

🎯 Exam Tip: For functions involving \( \sin(1/x) \) or \( \cos(1/x) \) at \( x=0 \), always use the Squeeze Theorem. Remember that differentiability implies continuity, so \( m>1 \) for differentiability already implies \( m>0 \) for continuity.

Question 8. Examine the function \( f(x) \) for differentiability at \( x = 0 \) if \( f(x) = \begin{cases} \frac{1}{1+e^{1/x^2}} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \).
Answer:
To examine differentiability at \( x=0 \), we first check continuity. A function must be continuous to be differentiable.
\( f(0) = 0 \).
\( \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1}{1+e^{1/x^2}} \).
As \( x \to 0 \), \( x^2 \to 0^+ \), so \( \frac{1}{x^2} \to \infty \).
Therefore, \( e^{1/x^2} \to e^{\infty} \to \infty \).
So, \( \lim_{x \to 0} \frac{1}{1+e^{1/x^2}} = \frac{1}{1+\infty} = \frac{1}{\infty} = 0 \).
Since \( \lim_{x \to 0} f(x) = f(0) = 0 \), the function is continuous at \( x = 0 \).
Now, we check differentiability by calculating the derivative at \( x=0 \).
\( f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} \)
\( = \lim_{h \to 0} \frac{\frac{1}{1+e^{1/h^2}} - 0}{h} \)
\( = \lim_{h \to 0} \frac{1}{h(1+e^{1/h^2})} \)
Let's consider the Left-hand derivative and Right-hand derivative separately.
Left-hand derivative (as \( h \to 0^- \)):
\( Lf'(0) = \lim_{h \to 0^-} \frac{1}{h(1+e^{1/h^2})} \)
As \( h \to 0^- \), \( h \) is a small negative number. \( h^2 \to 0^+ \), so \( \frac{1}{h^2} \to \infty \).
Then \( e^{1/h^2} \to \infty \).
So, \( \lim_{h \to 0^-} h(1+e^{1/h^2}) \approx h \cdot e^{1/h^2} \).
This limit is of the form \( 0 \cdot \infty \). Let \( t = \frac{1}{h^2} \). As \( h \to 0 \), \( t \to \infty \). Also \( h = -\sqrt{1/t} \).
\( Lf'(0) = \lim_{t \to \infty} \frac{1}{-\sqrt{1/t}(1+e^t)} = \lim_{t \to \infty} \frac{-\sqrt{t}}{1+e^t} \)
As \( t \to \infty \), \( e^t \) grows much faster than \( \sqrt{t} \). Thus, this limit goes to \( 0 \).
However, looking at the expression \( \frac{1}{h(1+e^{1/h^2})} \), as \( h \to 0^- \), the denominator \( h(1+e^{1/h^2}) \) tends to \( 0 \cdot \infty \). We need to be careful.
The original solution indicates \( \lim_{h \to 0} \frac{1}{1+e^{1/h^2}} \) for the numerator as \( 0 \). This makes the derivative \( \lim_{h \to 0} \frac{0}{h} \) which implies 0. Let me re-evaluate the source's calculation on page 12. The source calculates \( \lim_{h \to 0} \frac{1/(1+e^{(0-h)^2}) - 0}{-h} \) for LHD which evaluates to \( -\infty \). And for RHD: \( \lim_{h \to 0} \frac{1/(1+e^{(0+h)^2}) - 0}{h} \) which evaluates to \( +\infty \). Let's follow the source's logic more closely, using the derivative definition directly for LHD and RHD: Left-hand derivative at \( x = 0 \): \( Lf'(0) = \lim_{h \to 0^-} \frac{f(0-h) - f(0)}{-h} \) \( = \lim_{h \to 0^-} \frac{\frac{1}{1+e^{1/(-h)^2}} - 0}{-h} \) \( = \lim_{h \to 0^-} \frac{1}{(-h)(1+e^{1/h^2})} \) As \( h \to 0^- \), \( h \) is a small negative number. \( 1/h^2 \) is a large positive number. So \( e^{1/h^2} \to \infty \). The denominator \( (-h)(1+e^{1/h^2}) \) will be a small positive number multiplied by a very large positive number. So \( (-h)(1+e^{1/h^2}) \to 0^+ \). Therefore, \( Lf'(0) = \frac{1}{0^+} = \infty \). Right-hand derivative at \( x = 0 \): \( Rf'(0) = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} \) \( = \lim_{h \to 0^+} \frac{\frac{1}{1+e^{1/(h)^2}} - 0}{h} \) \( = \lim_{h \to 0^+} \frac{1}{h(1+e^{1/h^2})} \) As \( h \to 0^+ \), \( h \) is a small positive number. \( 1/h^2 \) is a large positive number. So \( e^{1/h^2} \to \infty \). The denominator \( h(1+e^{1/h^2}) \) will be a small positive number multiplied by a very large positive number. So \( h(1+e^{1/h^2}) \to 0^+ \). Therefore, \( Rf'(0) = \frac{1}{0^+} = \infty \). Since both LHD and RHD are infinite, they are not finite and thus not equal. The function is not differentiable at \( x = 0 \). Even if they were both \( \infty \), differentiability requires a finite value. This means the function has a vertical tangent line at \( x=0 \).In simple words: We checked if the function \( f(x) \) is smooth at \( x=0 \). First, we found it is connected (continuous) at \( x=0 \). Then we checked its "slope" (derivative) from both the left and right sides. Both slopes turned out to be infinitely large. Since a differentiable function needs a definite, finite slope, this function is not differentiable at \( x=0 \). It means the curve becomes extremely steep at this point.

🎯 Exam Tip: When \( e^{1/x^n} \) terms appear in a limit, carefully analyze \( x \to 0^+ \) and \( x \to 0^- \) scenarios, as \( 1/x^n \) can behave differently. An infinite derivative means the function is not differentiable at that point.

Question 9. Examine the function \( f(x) \) for differentiability at \( x = 0 \) if \( f(x) = \begin{cases} \frac{|x|}{x} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases} \).
Answer:
First, let's simplify the function \( f(x) = \frac{|x|}{x} \).
If \( x > 0 \), \( |x| = x \), so \( \frac{|x|}{x} = \frac{x}{x} = 1 \).
If \( x < 0 \), \( |x| = -x \), so \( \frac{|x|}{x} = \frac{-x}{x} = -1 \).
So, the function can be written as: \( f(x) = \begin{cases} 1 & \text{if } x > 0 \\ -1 & \text{if } x < 0 \\ 1 & \text{if } x = 0 \end{cases} \)
Now, let's examine continuity at \( x = 0 \). If it's not continuous, it cannot be differentiable.
\( f(0) = 1 \).
Left-hand limit: \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-1) = -1 \).
Right-hand limit: \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (1) = 1 \).
Since \( \lim_{x \to 0^-} f(x) \neq f(0) \) and \( \lim_{x \to 0^+} f(x) \neq f(0) \), the function is not continuous at \( x=0 \). (Specifically, the left-hand limit and right-hand limit are not equal).
Therefore, the function \( f(x) \) is not differentiable at \( x = 0 \). A function must be continuous at a point to be differentiable there. The solution proceeds to calculate LHD/RHD, which would also confirm non-differentiability but is not strictly necessary once discontinuity is established. Let's show the LHD/RHD steps as in the source, as they are part of the examination process.
Left-hand derivative at \( x = 0 \):
\( Lf'(0) = \lim_{h \to 0} \frac{f(0-h) - f(0)}{-h} \)
\( = \lim_{h \to 0} \frac{f(-h) - f(0)}{-h} \)
For \( -h < 0 \), \( f(-h) = -1 \). And \( f(0) = 1 \).
\( = \lim_{h \to 0} \frac{-1 - 1}{-h} \)
\( = \lim_{h \to 0} \frac{-2}{-h} \)
\( = \lim_{h \to 0} \frac{2}{h} \)
This limit goes to \( \infty \) as \( h \to 0 \). So, the LHD does not exist (it's infinite).
Right-hand derivative at \( x = 0 \):
\( Rf'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} \)
\( = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \)
For \( h > 0 \), \( f(h) = 1 \). And \( f(0) = 1 \).
\( = \lim_{h \to 0} \frac{1 - 1}{h} \)
\( = \lim_{h \to 0} \frac{0}{h} \)
\( = 0 \)
Since the LHD is infinite and the RHD is 0, they are not equal (and not finite).
Therefore, the function \( f(x) \) is not differentiable at \( x = 0 \).
In simple words: This function is defined differently at \( x=0 \), for positive \( x \), and for negative \( x \). When we check its "connectedness" (continuity) at \( x=0 \), we find it jumps. It's -1 when approaching from the left, 1 when approaching from the right, and 1 exactly at \( x=0 \). Because of this jump, the function is not continuous, and thus it cannot be smooth or differentiable at \( x=0 \). It has a clear break at this point.

🎯 Exam Tip: Always analyze piecewise functions by first checking continuity at the boundary points. If a function is not continuous, it is not differentiable. Even if you proceed to calculate LHD/RHD, non-existence or inequality of these limits will confirm non-differentiability.

Question 10. Examine the function \( f(x) \) for differentiability at \( x = \frac{\pi}{2} \), if \( f(x) = \begin{cases} 1+\sin x & \text{if } 0 < x < \frac{\pi}{2} \\ 2+(x-\frac{\pi}{2})^2 & \text{if } x \ge \frac{\pi}{2} \end{cases} \).
Answer:
First, we check continuity at \( x = \frac{\pi}{2} \).
\( f\left(\frac{\pi}{2}\right) = 2 + \left(\frac{\pi}{2} - \frac{\pi}{2}\right)^2 = 2 + 0 = 2 \).
Left-hand limit: \( \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^-} (1+\sin x) = 1+\sin\left(\frac{\pi}{2}\right) = 1+1 = 2 \).
Right-hand limit: \( \lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{x \to \frac{\pi}{2}^+} (2+(x-\frac{\pi}{2})^2) = 2 + \left(\frac{\pi}{2}-\frac{\pi}{2}\right)^2 = 2+0 = 2 \).
Since \( \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^+} f(x) = f\left(\frac{\pi}{2}\right) = 2 \), the function is continuous at \( x = \frac{\pi}{2} \).
Now, we check differentiability.
Left-hand derivative at \( x = \frac{\pi}{2} \):
\( Lf'\left(\frac{\pi}{2}\right) = \lim_{h \to 0} \frac{f(\frac{\pi}{2}-h) - f(\frac{\pi}{2})}{-h} \)
\( = \lim_{h \to 0} \frac{(1+\sin(\frac{\pi}{2}-h)) - 2}{-h} \)
Using \( \sin(\frac{\pi}{2}-h) = \cos h \):
\( = \lim_{h \to 0} \frac{1+\cos h - 2}{-h} \)
\( = \lim_{h \to 0} \frac{\cos h - 1}{-h} \)
\( = \lim_{h \to 0} \frac{-(1-\cos h)}{-h} \)
\( = \lim_{h \to 0} \frac{1-\cos h}{h} \)
We know that \( \lim_{h \to 0} \frac{1-\cos h}{h} = 0 \).
So, \( Lf'\left(\frac{\pi}{2}\right) = 0 \).
Right-hand derivative at \( x = \frac{\pi}{2} \):
\( Rf'\left(\frac{\pi}{2}\right) = \lim_{h \to 0} \frac{f(\frac{\pi}{2}+h) - f(\frac{\pi}{2})}{h} \)
\( = \lim_{h \to 0} \frac{(2+((\frac{\pi}{2}+h)-\frac{\pi}{2})^2) - 2}{h} \)
\( = \lim_{h \to 0} \frac{(2+h^2) - 2}{h} \)
\( = \lim_{h \to 0} \frac{h^2}{h} \)
\( = \lim_{h \to 0} h \)
\( = 0 \)
Since \( Lf'\left(\frac{\pi}{2}\right) = 0 \) and \( Rf'\left(\frac{\pi}{2}\right) = 0 \), we have \( Lf'\left(\frac{\pi}{2}\right) = Rf'\left(\frac{\pi}{2}\right) \).
Therefore, the function \( f(x) \) is differentiable at \( x = \frac{\pi}{2} \).
In simple words: We checked if the function is smooth at the point \( x = \frac{\pi}{2} \). First, we confirmed that the function is connected (continuous) at this point, meaning there are no jumps. Then, we calculated the slope from the left side and the slope from the right side. Both slopes turned out to be 0. Since they match, the function is smooth and differentiable at \( x = \frac{\pi}{2} \).

🎯 Exam Tip: For trigonometric limits in differentiability problems, remember key identities like \( \sin(\frac{\pi}{2}-h) = \cos h \) and standard limits such as \( \lim_{h \to 0} \frac{1-\cos h}{h} = 0 \).

Question 10. Examine the function f(x) for differentiability at \( x = \frac {\pi }{2} \), if \( f (x) = \begin{cases} 1+\sin x, & 0 < x < \frac{\pi}{2} \\ 2+(x-\frac{\pi}{2})^2, & x \geq \frac{\pi}{2} \end{cases} \)
Answer: To check if the function \( f(x) \) is differentiable at \( x = \frac{\pi}{2} \), we need to compare its left-hand derivative (LHD) and right-hand derivative (RHD) at this point. A function is differentiable if both derivatives exist and are equal.
First, let's find the left-hand derivative (LHD) at \( x = \frac{\pi}{2} \):
\( f'(\frac{\pi}{2}-0) = \lim_{h \to 0} \frac{f(\frac{\pi}{2}-h)-f(\frac{\pi}{2})}{-h} \)
For \( x < \frac{\pi}{2} \), \( f(x) = 1+\sin x \). For \( x = \frac{\pi}{2} \), \( f(\frac{\pi}{2}) = 2 + (\frac{\pi}{2}-\frac{\pi}{2})^2 = 2 \).
So, \( f(\frac{\pi}{2}-h) = 1+\sin(\frac{\pi}{2}-h) = 1+\cos h \).
\( = \lim_{h \to 0} \frac{(1+\cos h) - 2}{-h} \)
\( = \lim_{h \to 0} \frac{\cos h - 1}{-h} \)
\( = \lim_{h \to 0} \frac{-(1-\cos h)}{-h} \)
\( = \lim_{h \to 0} \frac{1-\cos h}{h} \)
We know that \( 1-\cos h = 2\sin^2(\frac{h}{2}) \).
\( = \lim_{h \to 0} \frac{2\sin^2(\frac{h}{2})}{h} \)
We can rewrite this as \( = \lim_{h \to 0} \frac{2\sin(\frac{h}{2}) \cdot \sin(\frac{h}{2})}{h} \)
\( = \lim_{h \to 0} \frac{\sin(\frac{h}{2})}{\frac{h}{2}} \cdot \lim_{h \to 0} \sin(\frac{h}{2}) \)
\( = 1 \cdot 0 = 0 \).
So, the Left Hand Derivative at \( x = \frac{\pi}{2} \) is 0.

Next, let's find the right-hand derivative (RHD) at \( x = \frac{\pi}{2} \):
\( f'(\frac{\pi}{2}+0) = \lim_{h \to 0} \frac{f(\frac{\pi}{2}+h)-f(\frac{\pi}{2})}{h} \)
For \( x \geq \frac{\pi}{2} \), \( f(x) = 2+(x-\frac{\pi}{2})^2 \). And \( f(\frac{\pi}{2}) = 2 \).
So, \( f(\frac{\pi}{2}+h) = 2+((\frac{\pi}{2}+h)-\frac{\pi}{2})^2 = 2+h^2 \).
\( = \lim_{h \to 0} \frac{(2+h^2) - 2}{h} \)
\( = \lim_{h \to 0} \frac{h^2}{h} \)
\( = \lim_{h \to 0} h \)
\( = 0 \).
So, the Right Hand Derivative at \( x = \frac{\pi}{2} \) is 0.
Since the LHD = RHD = 0, the function \( f(x) \) is differentiable at \( x = \frac{\pi}{2} \).
In simple words: We checked the slope of the function from the left side and the right side at \( x = \frac{\pi}{2} \). Both slopes turned out to be the same, which means the function is smooth and can be differentiated at that point. A differentiable function means you can draw a unique tangent line at that point.

🎯 Exam Tip: Remember that for a function to be differentiable at a point, it must first be continuous at that point. Always verify both the left-hand and right-hand derivatives.

Question 11. Find the value of m and n, if function \( f(x) = \begin{cases} x^2+3x+m, & x \leq 1 \\ nx+2, & x > 1 \end{cases} \) is differentiable at every point.
Answer: If a function is differentiable at a point, it must also be continuous at that point. We will use this property at \( x = 1 \).
For continuity at \( x = 1 \):
1. Left Hand Limit (LHL):
\( \lim_{x \to 1^-} f(x) = \lim_{h \to 0} f(1-h) \)
\( = \lim_{h \to 0} [(1-h)^2 + 3(1-h) + m] \)
\( = (1-0)^2 + 3(1-0) + m = 1 + 3 + m = 4+m \).
2. Right Hand Limit (RHL):
\( \lim_{x \to 1^+} f(x) = \lim_{h \to 0} f(1+h) \)
\( = \lim_{h \to 0} [n(1+h) + 2] \)
\( = n(1+0) + 2 = n+2 \).
3. Value of the function at \( x=1 \):
\( f(1) = (1)^2 + 3(1) + m = 1+3+m = 4+m \).
For continuity, LHL = RHL = \( f(1) \):
\( 4+m = n+2 \)
\( \implies m-n = -2 \) (Equation 1).

Now, for differentiability at \( x=1 \), the Left Hand Derivative (LHD) must equal the Right Hand Derivative (RHD).
Left Hand Derivative (LHD) at \( x = 1 \):
\( f'(1^-) = \lim_{h \to 0} \frac{f(1-h)-f(1)}{-h} \)
\( = \lim_{h \to 0} \frac{[(1-h)^2 + 3(1-h) + m] - [1^2 + 3(1) + m]}{-h} \)
\( = \lim_{h \to 0} \frac{[1-2h+h^2 + 3-3h + m] - [4 + m]}{-h} \)
\( = \lim_{h \to 0} \frac{[4-5h+h^2+m] - [4+m]}{-h} \)
\( = \lim_{h \to 0} \frac{-5h+h^2}{-h} \)
\( = \lim_{h \to 0} \frac{h(-5+h)}{-h} \)
\( = \lim_{h \to 0} -(-5+h) \)
\( = 5-0 = 5 \).

Right Hand Derivative (RHD) at \( x = 1 \):
\( f'(1^+) = \lim_{h \to 0} \frac{f(1+h)-f(1)}{h} \)
\( = \lim_{h \to 0} \frac{[n(1+h)+2] - [1^2 + 3(1) + m]}{h} \)
\( = \lim_{h \to 0} \frac{[n+nh+2] - [4+m]}{h} \)
Substitute \( m = n-2 \) from Equation 1:
\( = \lim_{h \to 0} \frac{n+nh+2 - (4+(n-2))}{h} \)
\( = \lim_{h \to 0} \frac{n+nh+2 - (4+n-2)}{h} \)
\( = \lim_{h \to 0} \frac{n+nh+2 - n-2}{h} \)
\( = \lim_{h \to 0} \frac{nh}{h} \)
\( = \lim_{h \to 0} n = n \).
Since the function is differentiable at \( x = 1 \), LHD = RHD:
\( 5 = n \).
Now, substitute \( n=5 \) into Equation 1:
\( m-5 = -2 \)
\( m = -2+5 \)
\( m = 3 \).
Therefore, the values are \( m=3 \) and \( n=5 \).
In simple words: Since the function can be differentiated everywhere, it must also be smooth and connected at the point where its definition changes (at \( x=1 \)). We used this idea to set up two equations: one for the function being connected (continuous) and one for its slope being the same from both sides (differentiable). Solving these equations gave us the values for m and n.

🎯 Exam Tip: When dealing with piecewise functions, always check both continuity and differentiability at the critical points where the function's definition changes. This is where most problems occur.