CBSE Class 12 Mathematics Continuity and Differentiability VBQs Set A

Continuity and Differentiability

BASIC CONCEPTS

• Continuity and Discontinuity of Function: A function \( y = f(x) \) is said to be continuous in an interval if for every value of \( x \) in that interval \( y \) exist. If we plot the points, the graph is drawn without lifting the pencil. If we have to lift the pencil on drawing the curve, then the function is said to be a discontinuous function.
• Continuity and Discontinuity of a Function at a Point: A function \( f(x) \) is said to be continuous at a point \( a \) of its domain if \( \lim_{x \to a^-} f(x), \lim_{x \to a^+} f(x), f(a) \) exist and \( \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) \).

A function \( f(x) \) is said to be discontinuous at \( x = a \) if it is not continuous at \( x = a \).

Properties of Continuous Function:

If \( f \) and \( g \) are two continuous functions at a point \( a \), then

(i) \( f + g \) is continuous at \( a \).

(ii) \( f - g \) is continuous at \( a \).

(iii) \( f \cdot g \) is continuous at \( a \).

(iv) \( \frac{f}{g} \) is continuous at \( a \), provided \( g(a) \neq 0 \).

(v) \( c \cdot f \) is continuous at \( a \), where \( c \) is a constant.

(vi) \( |f| \) is continuous function at \( a \).

• Every constant function is continuous function.
• Every polynomial function is continuous function.
• Identity function is continuous function.
• Every logarithmic and exponential function is a continuous function.

Important Series which are Frequently Used in Limits:

(i) \( (1 + x)^n = 1 + nx + \frac{n(n-1)x^2}{2!} + \frac{n(n-1)(n-2)x^3}{3!} + ... \)

(ii) \( e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + ... \infty \) and \( e^{-x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + ... \)

(iii) \( a^x = 1 + x \log_e a + \frac{x^2}{2!} (\log_e a)^2 + ... \) and \( \log |1 + x| = x - \frac{x^2}{2} + \frac{x^3}{3} - ... \)

(iv) \( \log |1 - x| = -x - \frac{x^2}{2} - \frac{x^3}{3} - ... \) and \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ... \)

(v) \( \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ... \) and \( \tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + ... \)

Differentiation from First Principle or Ab-initio Method or by Delta Method:

Given a function \( f(x) \) and if there is a small increment \( h \) in \( x \), let their corresponding increment is \( f(x + h) \) in \( f(x) \) i.e., \( f(x) \to f(x + h) \), then, \( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \) is called the differential coefficient of \( f(x) \) with respect to \( x \).

List of Useful Formulae:

(i) \( \frac{d}{dx}(x^n) = nx^{n-1} \)

(ii) \( \frac{d}{dx}(ax + b)^n = n(ax + b)^{n-1} \cdot a \)

(iii) \( \frac{d}{dx}(e^x) = e^x \)

(iv) \( \frac{d}{dx}(e^{ax}) = a \cdot e^{ax} \)

(v) \( \frac{d}{dx}(a^x) = a^x \cdot \log_e a \)

(vi) \( \frac{d}{dx}(a^{bx}) = b a^{bx} \log_e a \)

(vii) \( \frac{d}{dx} \log_e x = \frac{1}{x} \) and \( \frac{d}{dx} \log_e ax = \frac{1}{x} \)

(viii) \( \frac{d}{dx} \log_a x = \frac{1}{x \log_e a} \) and \( \frac{d}{dx} \log_a bx = \frac{1}{x \log_e a} \)

(i) \( \frac{d}{dx} \sin x = \cos x \) and \( \frac{d}{dx} \sin ax = a \cos ax \)

(ii) \( \frac{d}{dx} \cos x = -\sin x \) and \( \frac{d}{dx} \cos ax = -a \sin ax \)

(iii) \( \frac{d}{dx} \tan x = \sec^2 x \) and \( \frac{d}{dx} \tan ax = a \sec^2 ax \)

(iv) \( \frac{d}{dx} \cot x = -\csc^2 x \) and \( \frac{d}{dx} \cot ax = -a \csc^2 ax \)

(v) \( \frac{d}{dx} \sec x = \sec x \tan x \) and \( \frac{d}{dx} \sec ax = a \sec ax \tan ax \)

(vi) \( \frac{d}{dx} \csc x = -\csc x \cot x \) and \( \frac{d}{dx} \csc ax = -a \csc ax \cot ax \)

(i) \( \frac{d}{dx} \sin^{-1} x = \frac{1}{\sqrt{1 - x^2}} \) and \( \frac{d}{dx} \sin^{-1} ax = \frac{a}{\sqrt{1 - a^2x^2}} \)

(ii) \( \frac{d}{dx} \cos^{-1} x = \frac{-1}{\sqrt{1 - x^2}} \) and \( \frac{d}{dx} \cos^{-1} ax = \frac{-a}{\sqrt{1 - a^2x^2}} \)

(iii) \( \frac{d}{dx} \tan^{-1} x = \frac{1}{1 + x^2} \) and \( \frac{d}{dx} \tan^{-1} ax = \frac{a}{1 + a^2x^2} \)

(iv) \( \frac{d}{dx} \cot^{-1} x = \frac{-1}{1 + x^2} \) and \( \frac{d}{dx} \cot^{-1} ax = \frac{-a}{1 + a^2x^2} \)

(v) \( \frac{d}{dx} \sec^{-1} x = \frac{1}{x\sqrt{x^2 - 1}} \) and \( \frac{d}{dx} \sec^{-1} ax = \frac{1}{x\sqrt{a^2x^2 - 1}} \)

(vi) \( \frac{d}{dx} \csc^{-1} x = \frac{-1}{x\sqrt{x^2 - 1}} \) and \( \frac{d}{dx} \csc^{-1} ax = \frac{-1}{x\sqrt{a^2x^2 - 1}} \)

Product Rule:

Let \( u \) and \( v \) be two functions of \( x \), then \( \frac{d}{dx}(u \cdot v) = u \frac{dv}{dx} + v \frac{du}{dx} \).
i.e., \( \frac{d}{dx} \text{(Product of two functions) = First function} \cdot \frac{d}{dx}\text{(Second function) + Second function} \cdot \frac{d}{dx}\text{(First function)} \)

Quotient Rule:

If \( u \) and \( v \) are functions of \( x \) then, \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \)
i.e., \( \frac{d}{dx} \left( \frac{N'}{D'} \right) = \frac{D' \frac{d(N')}{dx} - N' \frac{d(D')}{dx}}{(D')^2} \)

Chain Rule:

Chain rule is applied when the given function is the function of function i.e.,
if \( y \) is a function of \( x \), then \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \) or \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx} \)

Logarithmic Differentiation:

Logarithmic differentiations are used for differentiation of functions which consists of the product or quotients of a number of functions and/or the given function is of type \( [f(x)]^{g(x)} \), where \( f(x) \) and \( g(x) \) both are differentiable functions of \( x \).
Therefore, in this method, we take the logarithm on both the sides of the function and then differentiate it with respect to ‘x’. So, this process is called logarithmic differentiation.
General method: If \( y = [f(x)]^{g(x)} \) then \( \frac{dy}{dx} = y \left[ \log f(x) \cdot g'(x) + g(x) \cdot \frac{1}{f(x)} \cdot f'(x) \right] \)

Parametric Form:

Sometimes we come across the function when both \( x \) and \( y \) are expressed in terms of another variable say \( t \) i.e., \( x = \phi(t) \) and \( y = \psi(t) \). This form of a function is called parametric form and \( t \) is called the parameter.
To obtain \( \frac{dy}{dx} \) in parametric type of functions we follow any one of the following two steps:
(i) Try to obtain a relationship between \( x \) and \( y \) by eliminating the parameter and then proceed to get \( \frac{dy}{dx} \) which is already discussed.
(ii) If it is not convenient to obtain such a relation between \( x \) and \( y \), then differentiate \( x \) and \( y \) both with respect to parameter \( t \) to get \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) (treating \( t \) as independent variable and \( x \) and \( y \) as dependent variables). Finally, divide \( \frac{dy}{dt} \) by \( \frac{dx}{dt} \) to get \( \frac{dy}{dx} \) i.e., \( \frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} \).
or sometimes \( \frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta} \), where \( \theta \) is an independent variable.

Rolle’s Theorem:

If \( f(x) \) be a real valued function, defined in a closed interval \( [a, b] \) such that:
(i) it is continuous in closed interval \( [a, b] \).
(ii) it is differentiable in open interval \( (a, b) \).
(iii) \( f(a) = f(b) \). Then there exists at least one value \( c \in (a, b) \) such that \( f'(c) = 0 \).
[Note: \( f'(c) = 0 \) means tangent at \( c \) is parallel to x-axis.]

Lagrange’s Mean Value Theorem:

If \( f(x) \) is a real valued function defined in the closed interval \( [a, b] \) such that:
(i) it is continuous in the closed interval \( [a, b] \).
(ii) it is differentiable in the open interval \( (a, b) \).
Then there exists at least one real value \( c \in (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).
[Note: \( f'(c) = \frac{f(b) - f(a)}{b - a} \) means tangent at \( c \) is parallel to chord \( AB \). i.e., Slope of tangent at \( c = \text{Slope of chord } AB \)]

Limits:

Let \( f(x) \) be a function of \( x \). Let \( a \) and \( l \) be two constants such that as \( x \to a \), we have \( f(x) \to l \), i.e., the numerical difference between \( f(x) \) and \( l \) can be made as small as we wish by taking \( x \) sufficiently close to \( a \). In such a case, we say that the limit of function \( f(x) \) as \( x \) approaches \( a \) is \( l \). We write this as \( \lim_{x \to a} f(x) = l \).

Procedure to Find \( \lim_{x \to a} f(x) \):

(i) Putting \( x = a \) in the given function. If \( f(a) \) is a finite value, then \( \lim_{x \to a} f(x) = f(a) \).

(ii) To find LHL of \( f(x) \) at \( x = a \) we put \( x = a - h, h \to 0 \) and find \( \lim_{h \to 0} f(a - h) \) after simplification.

(iii) To find RHL of \( f(x) \) at \( x = a \) we put \( x = a + h, h \to 0 \) and find \( \lim_{h \to 0} f(a + h) \) after simplification.

(iv) If \( \text{LHL = RHL} = k \) (say), then \( \lim_{x \to a} f(x) = k \).

Fundamental Theorems on Limits:

Some important theorems are given below which are frequently used in limits:

(i) \( \lim_{x \to a} c = c \), i.e., the limit of a constant quantity is constant itself.

(ii) \( \lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) \). i.e., the limit of sum of two functions is equal to the sum of their limits.

(iii) \( \lim_{x \to a} [f(x) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x) \). i.e., the limit of difference of two functions is equal to the difference of their limits.

(iv) \( \lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \). i.e., the limit of the product of two functions is equal to the product of their limits.

(v) \( \lim_{x \to a} \left[ \frac{f(x)}{g(x)} \right] = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \). i.e., the limit of quotient of two functions is equal to quotient of their limits, provided \( \lim_{x \to a} g(x) \) is a finite value not equal to zero.

(vi) \( \lim_{x \to a} [c f(x)] = c \lim_{x \to a} f(x) \), where \( c \) is a constant. i.e., the limit of the product of a constant and the function is equal to the product of the constant and the limit of the function.

(vii) \( \lim_{x \to a} \sqrt{f(x)} = \sqrt{\lim_{x \to a} f(x)} \)

(viii) \( \lim_{x \to 0} f(-x) = \lim_{x \to 0^-} f(x) \)

Evaluation of Limits:

(i) Direct substituting method: We substitute the value of the point in the given expression and if we get a finite number, then this number is the limit of the given function.

(ii) Factorisation method: On substituting \( x = a \) in the given expression, if we get \( \frac{0}{0}, \frac{\infty}{\infty} \), etc. form, then we factorize the numerator and denominator and take \( (x - a) \) as a common factor from numerator and denominator. After cancelling out \( (x - a) \), we put \( x = a \). If we get a finite number, then it is the required value otherwise repeat the step till we get a finite number.

(iii) Rationalisation method: Rationalisation method is applicable when (a) numerator, denominator or both in square root or (b) after substituting the value of limit if we get the negative number in square root. Hence, after simplifying in both the cases, we get the required value.

(iv) L’ HOSPITAL Rule: With the help of this rule, if we have to evaluate \( \lim_{x \to c} \frac{f(x)}{g(x)} \) such that it takes indeterminate form, i.e., \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then we differentiate numerator and denominator to get \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \), if it is determinate form then it is required value, otherwise repeat the step till we get a determinate form and thus required value. [Note: According to L’ HOSPITAL rule \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \). where \( g'(x) \neq 0 \forall x \in Z \) with \( x \neq c \)]

Some Standard Results:

(i) (a) \( \lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}, a > 0, n \in Q \)

(b) \( \lim_{x \to a} \frac{x^m - a^m}{x^n - a^n} = \frac{m}{n} a^{m-n}, m, n \in Q \)

(c) \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)

(d) \( \lim_{x \to 0} \cos x = 1 \)

(e) \( \lim_{x \to 0} \frac{\tan x}{x} = 1 \)

(ii) (a) \( \lim_{x \to 0} \frac{\sin^{-1} x}{x} = 1 \)

(b) \( \lim_{x \to 0} \frac{\tan^{-1} x}{x} = 1 \)

(iii) (a) \( \lim_{x \to 0} e^x = 1 \)

(b) \( \lim_{x \to 0} \frac{e^x - 1}{x} = 1 \)

(c) \( \lim_{x \to 0} \frac{\log |1 + x|}{x} = 1 \)

(d) \( \lim_{x \to 0} \frac{(1 + x)^n - 1}{x} = n \)

(e) \( \lim_{x \to 0} \frac{a^x - 1}{x} = \log_e a \)

(iv) Limits at infinity: This method is applied when \( x \to \infty \). Procedure to solve the infinite limits:

• (a) Write the given expression in the form of rational function.
• (b) Divide the numerator and denominator by highest power of \( x \).
• (c) Use the result \( \lim_{x \to \infty} \frac{1}{x^n} = 0 \), where \( n > 0 \).
• (d) Simplify and get the required result.

Question. Show that the function f defined by \( f(x) = |1 - x + |x|| \), where \( x \) is any real number, is a continuous function.
Answer: Define \( g \) by \( g(x) = 1 - x + |x| \) and \( h \) by \( h(x) = |x| \) for all real \( x \). Then \( (hog)(x) = h(g(x)) = h(1 - x + |x|) = |1 - x + |x|| = f(x) \). We know that \( h \) is a continuous function. Hence \( g \) being a sum of a polynomial function and the modulus function is continuous. But then \( f \) being a composite of two continuous functions is continuous.

Question. Find all points of discontinuity of \( f \) where \( f \) is defined by the following function:
\[ f(x) = \begin{cases} \frac{x}{|x|}, & \text{if } x < 0 \\ -1, & \text{if } x \ge 0 \end{cases} \]
Answer: For \( x < 0, f(x) = \frac{x}{|x|} \) is continuous and for \( x > 0, f(x) \) is a constant function so it is continuous.
At \( x = 0 \),
LHL \( = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \left( \frac{x}{|x|} \right) = \lim_{x \to 0^-} \frac{x}{-x} = -1 \), \( (\because |x| = -x, \text{ when } x < 0) \)
RHL \( = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (-1) = -1 \)
and \( f(0) = -1 \)
\( \therefore \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \Rightarrow f(x) \) is continuous at \( x = 0 \).
Hence, \( f(x) \) has no points of discontinuity.

Question. For what value of \( \lambda \) is the function defined by
\[ f(x) = \begin{cases} \lambda(x^2 - 2x), & \text{if } x \le 0 \\ 4x + 1, & \text{if } x > 0 \end{cases} \]
continuous at \( x = 0 \)? What about continuity at \( x = 1 \)? 
Answer: \( \because f(x) \) to be continuous at \( x = 0 \)
\( \therefore \text{LHL = RHL} = f(0) \)
i.e., \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \) ...(i)
Now LHL \( = \lim_{x \to 0^-} \lambda(x^2 - 2x) = \lambda(0 - 0) = 0 \) ...(ii)
RHL \( = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (4x + 1) = 4 \times 0 + 1 = 1 \) ...(iii)
and \( f(0) = 0 \) ...(iv)
from (ii) and (iii), \( \text{LHL} \neq \text{RHL} \)
So, \( f(x) \) is not continuous for any value of \( \lambda \).
For continuity at \( x = 1 \),
\( f(1) = 4 \times 1 + 1 = 4 + 1 = 5 \)
and \( \lim_{x \to 1} f(x) = \lim_{x \to 1} (4x + 1) = 4 \times 1 + 1 = 5 \)
As \( \lim_{x \to 1} f(x) = f(1) \)
So, it is continuous at \( x = 1 \).

Question. Show that the function defined by \( g(x) = x - [x] \) is discontinuous at all integral points. Here \( [x] \) denotes the greatest integer less than or equal to \( x \).
Answer: Let \( g(x) = x - [x] \) be the greatest integer function. Let \( n \) be any integer. Then
\( \lim_{x \to n^-} g(x) = \lim_{h \to 0} g(n - h) = \lim_{h \to 0} (n - h) - [n - h] = \lim_{h \to 0} (n - h) - (n - 1) = n - (n - 1) = 1 \)
\( \lim_{x \to n^+} g(x) = \lim_{h \to 0} g(n + h) = \lim_{h \to 0} (n + h) - [n + h] = \lim_{h \to 0} (n + h) - n = n - n = 0 \)
\( \therefore \lim_{x \to n^-} g(x) \neq \lim_{x \to n^+} g(x) \)
LHL \( \neq \) RHL so, \( f \) is discontinuous at all integral points.

Question. Find the value of \( a \) and \( b \) such that the function \( f(x) \) defined by:
\[ f(x) = \begin{cases} 5, & \text{if } x \le 2 \\ ax + b, & \text{if } 2 < x < 10 \\ 21, & \text{if } x \ge 10 \end{cases} \]
is a continuous function 
Answer: Since, \( f(x) \) is continuous.
\( \Rightarrow f(x) \) is continuous at \( x = 2 \) and \( x = 10 \)
\( \Rightarrow (\text{LHL of } f(x) \text{ at } x = 2) = (\text{RHL of } f(x) \text{ at } x = 2) = f(2) \)
\( \Rightarrow \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) \) ...(i)
Similarly, \( \lim_{x \to 10^-} f(x) = \lim_{x \to 10^+} f(x) = f(10) \) ...(ii)
\( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} 5 = 5 \), \( [\because f(x) = 5 \text{ if } x \le 2] \)
\( \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (ax + b) = 2a + b \), \( [\because f(x) = ax + b \text{ if } x > 2] \)
\( f(2) = 5 \)
Putting these values in (i), we get \( 2a + b = 5 \) ...(iii)
Again \( \lim_{x \to 10^-} f(x) = \lim_{x \to 10^-} (ax + b) = 10a + b \), \( [\because f(x) = ax + b \text{ if } x < 10] \)
\( \lim_{x \to 10^+} f(x) = \lim_{x \to 10^+} 21 = 21 \), \( [\because f(x) = 21 \text{ if } x > 10] \)
\( f(10) = 21 \)
Putting these values in (ii), we get \( 10a + b = 21 \) ...(iv)
Subtracting (iii) from (iv), we get
\( 10a + b = 21 \)
\( -2a - b = -5 \)
\( 8a = 16 \Rightarrow a = 2 \)
\( \therefore b = 5 - 2 \times 2 = 1 \)
Hence, the value of \( a = 2 \) and \( b = 1 \).

Question. Find the derivative of the function \( f(x) = \sin(x^2) \).
Answer: \( \because f(x) = \sin(x^2) \)
\( \therefore f'(x) = \cos x^2 \cdot 2x = 2x \cos x^2 \)

Question. Prove that the function \( f \) given by \( f(x) = |x - 1|, x \in \mathbb{R} \) is not differentiable at \( x = 1 \).
Answer: Here \( f(x) = |x - 1| = \begin{cases} x - 1 & \text{if } x \ge 1 \\ 1 - x & \text{if } x < 1 \end{cases} \)
LHD at \( x = 1 \)
\( \lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1} = \lim_{h \to 0} \frac{f(1 - h) - f(1)}{(1 - h) - 1} = \lim_{h \to 0} \frac{1 - (1 - h) - (1 - 1)}{-h} = \lim_{h \to 0} \frac{h}{-h} = -1 \)
RHD at \( x = 1 \)
\( \lim_{x \to 1^+} \frac{f(x) - f(1)}{x - 1} = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{(1 + h) - 1} = \lim_{h \to 0} \frac{(1 + h) - 1 - (1 - 1)}{h} = \lim_{h \to 0} \frac{h}{h} = 1 \)
\( \therefore \text{LHD} \neq \text{RHD} \)
Thus, \( f \) is not differentiable at \( x = 1 \).

Question. Find \( \frac{dy}{dx} \), if \( x - y = \pi \).
Answer: \( \because x - y = \pi \Rightarrow y = x - \pi \Rightarrow \frac{dy}{dx} = 1 \)

Question. Find \( \frac{dy}{dx} \) if \( y = \tan^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right), -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}} \).
Answer: \( y = \tan^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right) \), putting \( x = \tan \theta \Rightarrow \theta = \tan^{-1} x \)
\( y = \tan^{-1} \left( \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} \right) = \tan^{-1} [\tan 3\theta] = 3\theta = 3 \tan^{-1} x \)
Differentiating both sides w.r.t \( x \), we get
\( \frac{dy}{dx} = 3 \times \frac{1}{1 + x^2} = \frac{3}{1 + x^2} \)

Question. Find \( \frac{dy}{dx} \) if \( y = \sin^{-1}(2x\sqrt{1-x^2}); -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \).
Answer: Here \( y = \sin^{-1}(2x\sqrt{1-x^2}) \), putting \( x = \sin \theta \Rightarrow \theta = \sin^{-1} x \).
\( y = \sin^{-1} [2 \sin \theta \sqrt{1 - \sin^2 \theta}] = \sin^{-1} [2 \sin \theta \cos \theta] \)
\( \Rightarrow y = \sin^{-1} (\sin 2\theta) = 2\theta = 2 \sin^{-1} x \)
Differentiating both sides w.r.t. \( x \), we get
\( \frac{dy}{dx} = 2 \cdot \frac{1}{\sqrt{1 - x^2}} = \frac{2}{\sqrt{1 - x^2}} \)

Question. Differentiate the following with respect to \( x \): \( \sqrt{e^{\sqrt{x}}}, x > 0 \)
Answer: Let \( y = \sqrt{e^{\sqrt{x}}} \)
\( \Rightarrow \frac{dy}{dx} = \frac{d}{d(e^{\sqrt{x}})} (\sqrt{e^{\sqrt{x}}}) \times \frac{d}{d(\sqrt{x})} (e^{\sqrt{x}}) \times \frac{d}{dx} \sqrt{x} = \frac{1}{2\sqrt{e^{\sqrt{x}}}} \times e^{\sqrt{x}} \times \frac{1}{2\sqrt{x}} \)
\( \frac{dy}{dx} = \frac{e^{\sqrt{x}}}{4 \sqrt{x} e^{\sqrt{x}}} \)

Question. Differentiate with respect to \( x \): \( y = (x + \frac{1}{x})^x + x^{(1 + \frac{1}{x})} \).
Answer: Let \( u = (x + \frac{1}{x})^x \) and \( v = x^{(1 + \frac{1}{x})} \)
then \( y = u + v \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \) ...(i)
Now, \( u = (x + \frac{1}{x})^x \)
\( \log u = x \log(x + \frac{1}{x}) \) [Taking logarithm on both sides]
Differentiating both sides, w.r.t \( x \), we get
\( \frac{1}{u} \frac{du}{dx} = 1 \cdot \log(x + \frac{1}{x}) + x \cdot \frac{1}{x + \frac{1}{x}} \cdot (1 - \frac{1}{x^2}) \)
\( \frac{du}{dx} = u \left[ \log(x + \frac{1}{x}) + x \cdot \frac{x}{x^2 + 1} \cdot \frac{x^2 - 1}{x^2} \right] = (x + \frac{1}{x})^x \left[ \log(x + \frac{1}{x}) + \frac{x^2 - 1}{x^2 + 1} \right] \) ...(ii)
Now, \( v = (x)^{1 + \frac{1}{x}} \)
\( \log v = (1 + \frac{1}{x}) \log x \) [Taking logarithm on both sides]
Differentiating both sides w.r.t. \( x \), we get
\( \frac{1}{v} \frac{dv}{dx} = (1 - \frac{1}{x^2}) \log x + (1 + \frac{1}{x}) \cdot \frac{1}{x} \Rightarrow \frac{dv}{dx} = x^{(1 + \frac{1}{x})} \left[ \frac{x^2 - 1}{x^2} \log x + \frac{x + 1}{x^2} \right] \) ...(iii)
From (i), (ii) and (iii), we get
\( \frac{dy}{dx} = (x + \frac{1}{x})^x \left[ \log(x + \frac{1}{x}) + \frac{x^2 - 1}{x^2 + 1} \right] + x^{(1 + \frac{1}{x})} \left[ \frac{x^2 - 1}{x^2} \log x + \frac{x + 1}{x^2} \right] \)

Question. If \( y = x^{\sin x} + (\sin x)^{\cos x} \), find \( \frac{dy}{dx} \).
OR
If \( y = x^{\cos x} + (\cos x)^{\sin x} \), find \( \frac{dy}{dx} \). 
Answer: In \( y = x^{\sin x} + (\sin x)^{\cos x} \), let \( u = x^{\sin x}, v = (\sin x)^{\cos x} \)
Now, \( y = u + v \)
and \( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \) ...(i)
We have, \( u = x^{\sin x} \)
Taking log on both sides, we get
\( \log u = \sin x \log x \)
Differentiating both sides with respect to \( x \), we get
\( \frac{1}{u} \frac{du}{dx} = \sin x \times \frac{1}{x} + \log x \cdot \cos x \)
\( \frac{du}{dx} = u \left( \frac{\sin x}{x} + \log x \cdot \cos x \right) = x^{\sin x} \left[ \frac{\sin x}{x} + \log x \cdot \cos x \right] \) ...(ii)
Again, \( v = (\sin x)^{\cos x} \)
Taking log on both sides with respect to \( x \), we get
\( \log v = \cos x \log \sin x \)
Differentiating both sides with respect to \( x \), we get
\( \frac{1}{v} \frac{dv}{dx} = \cos x \times \frac{1}{\sin x} \times \cos x + \log \sin x (-\sin x) \)
\( = \cos x \cdot \cot x + \log \sin x \cdot (-\sin x) = \cos x \cdot \cot x - \sin x \cdot \log \sin x \)
\( \therefore \frac{dv}{dx} = v [\cos x \cdot \cot x - \sin x \cdot \log \sin x] \)
\( \frac{dv}{dx} = (\sin x)^{\cos x} [\cos x \cdot \cot x - \sin x \cdot \log \sin x] \) ...(iii)
From (i), (ii) and (iii), we get
\( \frac{dy}{dx} = x^{\sin x} \left[ \frac{\sin x}{x} + \log x \cdot \cos x \right] + (\sin x)^{\cos x} [\cos x \cdot \cot x - \sin x \cdot \log \sin x] \)

Question. Differentiate w.r.t \( x : y = (x \cos x)^x + (x \sin x)^{1/x} \)
Answer: We have \( y = (x \cos x)^x + (x \sin x)^{1/x} \)
\( y = e^{\log (x \cos x)^x} + e^{\log (x \sin x)^{1/x}} \Rightarrow y = e^{x \log (x \cos x)} + e^{\frac{1}{x} \log (x \sin x)} \)
\( \Rightarrow \frac{dy}{dx} = e^{x \log (x \cos x)} \cdot \frac{d}{dx} [x \log (x \cos x)] + e^{\frac{1}{x} \log (x \sin x)} \cdot \frac{d}{dx} \left( \frac{1}{x} \log(x \sin x) \right) \)
\( \Rightarrow \frac{dy}{dx} = (x \cos x)^x [1 \cdot \log(x \cos x) + x \cdot \frac{1}{x \cos x} (1 \cdot \cos x - x \sin x)] + (x \sin x)^{1/x} \left[ \frac{-1}{x^2} \log (x \sin x) + \frac{1}{x} \cdot \frac{1}{x \sin x} \cdot (1 \cdot \sin x + x \cos x) \right] \)
\( \Rightarrow \frac{dy}{dx} = (x \cos x)^x [\log(x \cos x) + 1 - x \tan x] + (x \sin x)^{1/x} \left[ \frac{x \cot x + 1 - \log (x \sin x)}{x^2} \right] \)

Question. Find \( \frac{dy}{dx} \) if \( x = \frac{\sin^3 t}{\sqrt{\cos 2t}}, y = \frac{\cos^3 t}{\sqrt{\cos 2t}} \).
Answer: Given \( x = \frac{\sin^3 t}{\sqrt{\cos 2t}} \)
Differentiating both sides w.r.t. \( t \), we have
\( \frac{dx}{dt} = \frac{\sqrt{\cos 2t} (3 \sin^2 t \cos t) - \sin^3 t \left( \frac{-2 \sin 2t}{2 \sqrt{\cos 2t}} \right)}{\cos 2t} \)
\( = \frac{3 (\cos 2t) \sin^2 t \cos t + \sin 2t \sin^3 t}{(\cos 2t)^{3/2}} \)
\( = \frac{3 (1 - 2 \sin^2 t) \sin^2 t \cos t + 2 \sin t \cos t \sin^3 t}{(\cos 2t)^{3/2}} = \frac{3 \sin^2 t \cos t - 4 \sin^4 t \cos t}{(\cos 2t)^{3/2}} \)
\( = \frac{\sin t \cos t (3 \sin t - 4 \sin^3 t)}{(\cos 2t)^{3/2}} = \frac{\sin t \cos t \sin 3t}{(\cos 2t)^{3/2}} \)
Now, \( \frac{dy}{dt} = \frac{d}{dt} \left( \frac{\cos^3 t}{\sqrt{\cos 2t}} \right) = \frac{\sqrt{\cos 2t} (-3 \cos^2 t \sin t) - \cos^3 t \left( \frac{-2 \sin 2t}{2 \sqrt{\cos 2t}} \right)}{\cos 2t} \)
\( = \frac{-3 (\cos 2t) \cos^2 t \sin t + \cos^3 t \sin 2t}{(\cos 2t)^{3/2}} \)
\( = \frac{-3 (2 \cos^2 t - 1) \cos^2 t \sin t + \cos^3 t (2 \sin t \cos t)}{(\cos 2t)^{3/2}} \)
\( = \frac{-6 \cos^4 t \sin t + 3 \cos^2 t \sin t + 2 \cos^4 t \sin t}{(\cos 2t)^{3/2}} = \frac{3 \cos^2 t \sin t - 4 \cos^4 t \sin t}{(\cos 2t)^{3/2}} \)
\( = \frac{\sin t \cos t (3 \cos t - 4 \cos^3 t)}{(\cos 2t)^{3/2}} = \frac{\sin t \cos t (-\cos 3t)}{(\cos 2t)^{3/2}} \)
\( \Rightarrow \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = \frac{-\sin t \cos t \cos 3t}{(\cos 2t)^{3/2}} \times \frac{(\cos 2t)^{3/2}}{\sin t \cos t \sin 3t} = - \cot 3t \)

Question. If \( x = \sqrt{a^{\sin^{-1} t}} \) and \( y = \sqrt{a^{\cos^{-1} t}} \), show that \( \frac{dy}{dx} = -\frac{y}{x} \).
Answer: \( x = \sqrt{a^{\sin^{-1} t}} \) ...(i)
\( y = \sqrt{a^{\cos^{-1} t}} \) ...(ii)
Multiplying, (i) and (ii) we get
\( x \cdot y = \sqrt{a^{\sin^{-1} t}} \times \sqrt{a^{\cos^{-1} t}} \Rightarrow x \cdot y = \sqrt{a^{\sin^{-1} t + \cos^{-1} t}} \)
\( \Rightarrow x \cdot y = \sqrt{a^{\pi/2}} \), \( [\because \sin^{-1} t + \cos^{-1} t = \pi/2] \)
On differentiating both sides, we get
\( x \frac{dy}{dx} + y \times 1 = 0 \Rightarrow \frac{dy}{dx} = \frac{-y}{x} \)

Question. If \( y = Ae^{mx} + Be^{nx} \), then show that \( \frac{d^2y}{dx^2} - (m + n) \frac{dy}{dx} + mny = 0 \). 
Answer: Given, \( y = Ae^{mx} + Be^{nx} \)
On differentiating with respect to \( x \), we have
\( \frac{dy}{dx} = Ame^{mx} + Bne^{nx} \)
Again, differentiating with respect to \( x \), we have
\( \frac{d^2y}{dx^2} = Am^2 e^{mx} + Bn^2 e^{nx} \)
Now, \( \text{LHS} = \frac{d^2y}{dx^2} - (m + n) \frac{dy}{dx} + mny \)
\( = Am^2 e^{mx} + Bn^2 e^{nx} - (m + n)(Ame^{mx} + Bne^{nx}) + mn(Ae^{mx} + Be^{nx}) \)
\( = Am^2 e^{mx} + Bn^2 e^{nx} - Am^2 e^{mx} - Amne^{mx} - Bmne^{nx} - Bn^2 e^{nx} + Amne^{mx} + Bmne^{nx} \)
\( = 0 = \text{RHS} \)

Question. If \( y = (\tan^{-1} x)^2 \), show that \( (x^2 + 1)^2 y_2 + 2x(x^2 + 1)y_1 = 2 \).
Answer: We have, \( y = (\tan^{-1} x)^2 \) ...(i)
Differentiating with respect to \( x \), we get
\( \frac{dy}{dx} = 2 \tan^{-1} x \cdot \frac{1}{1 + x^2} \) ...(ii)
or \( (1 + x^2) y_1 = 2 \tan^{-1} x \) \( \left[ \text{where } y_1 = \frac{dy}{dx} \right] \)
Again differentiating with respect to \( x \), we get
\( (1 + x^2) \cdot \frac{dy_1}{dx} + y_1 \cdot \frac{d}{dx}(1 + x^2) = 2 \cdot \frac{1}{1 + x^2} \)
\( \Rightarrow (1 + x^2) \cdot y_2 + y_1 \cdot 2x = \frac{2}{1 + x^2} \)
or \( (1 + x^2)^2 y_2 + 2x (1 + x^2) y_1 = 2 \) \( \left[ \text{where } y_2 = \frac{d^2 y}{dx^2} \text{ and } y_1 = \frac{dy}{dx} \right] \)

Question. Verify mean value theorem if \( f(x) = x^2 - 4x - 3 \) in the interval \([a, b]\), where \( a = 1 \) and \( b = 4 \).
Answer: Given, \( f(x) = x^2 - 4x - 3, x \in [a, b] \) i.e., \( [1, 4] \)
Since \( f(x) \) is a polynomial so it is differentiable everywhere and so it is also continuous.
So, all conditions of mean value theorem are satisfied.
Now, \( f(a) = f(1) = (1)^2 - 4 (1) - 3 = - 6 \)
\( f(b) = f(4) = (4)^2 - 4 \times 4 - 3 = - 3 \)
Now, for any \( c \in (a, b) \), we have
\( f'(c) = \frac{f(b) - f(a)}{b - a} \Rightarrow f'(c) = \frac{f(4) - f(1)}{4 - 1} \)
\( \Rightarrow 2c - 4 = \frac{-3 - (-6)}{3} \Rightarrow 2c - 4 = \frac{-3 + 6}{3} = 1 \)
\( \Rightarrow 2c = 5 \Rightarrow c = \frac{5}{2} \in (1, 4) \)

Question. If \( y = \cot^{-1} \left[ \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} \right] \), \( 0 < x < \frac{\pi}{2} \) then find the value of \( \frac{dy}{dx} \). 
Answer: Consider \( \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}, 0 < x < \frac{\pi}{2} \)
\( = \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} \times \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}} \)
\( = \frac{(\sqrt{1 + \sin x} + \sqrt{1 - \sin x})^2}{1 + \sin x - (1 - \sin x)} = \frac{1 + \sin x + 1 - \sin x + 2\sqrt{1 - \sin^2 x}}{1 + \sin x - 1 + \sin x} \)
\( = \frac{2 + 2 \cos x}{2 \sin x} = \frac{1 + \cos x}{\sin x} = \frac{2 \cos^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} = \cot \left( \frac{x}{2} \right) \)
\( \therefore y = \cot^{-1} \left[ \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} \right] = \cot^{-1} \left( \cot \frac{x}{2} \right) = \frac{x}{2} \)
\( \therefore y = \frac{x}{2} \Rightarrow \frac{dy}{dx} = \frac{1}{2} \)

Question. Determine if \( f \) defined by \( f(x) = \begin{cases} x^2 \sin \left( \frac{1}{x} \right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \) is a continuous function.
Answer: We note that domain of \( f \) is \( R \). Let \( c \) be any real number then two cases arise.
Case I: If \( c \neq 0 \) then
\( \therefore \lim_{x \to c} f(x) = \lim_{x \to c} x^2 \sin \frac{1}{x} = c^2 \sin \left( \frac{1}{c} \right) = f(c) \)
\( \Rightarrow f \) is continuous for \( c \in R \), where \( c \neq 0 \).
Case II : If \( c = 0 \), then \( f(c) = f(0) = 0 \)
and \( \lim_{x \to 0} x^2 \sin \left( \frac{1}{x} \right) = 0 \times \) (a number oscillating between \( -1 \) and \( 1 \))
\( = 0 = f(0) \)
\( \Rightarrow f \) is continuous at \( x = 0 \).
Thus, \( f \) is a continuous function at every point of its domain.

Question. Using mathematical induction, prove that \( \frac{d}{dx}(x^n) = nx^{n-1} \), for all positive integer \( n \). 
Answer: Let \( P(n) \) be statement such that
\( P(n): \frac{d}{dx}(x^n) = nx^{n-1} \)
For \( P(1) \): Putting \( n = 1 \), we get
\( \frac{d}{dx}(x^1) = 1.x^{1-1} = 1 \Rightarrow P(1) \) is true.
Let \( P(m) \) be true \( \Rightarrow \frac{d}{dx}(x^m) = mx^{m-1} \)
Now for \( P(m + 1) \):
\( \frac{d}{dx}(x^{m+1}) = \frac{d}{dx}(x^m \cdot x) = x^m \cdot 1 + x \cdot \frac{d}{dx}(x^m) \)
\( = x^m + x \cdot m \cdot x^{m-1} = x^m + m \cdot x^m = (m + 1) \cdot x^m = (m + 1) \cdot x^{(m+1)-1} \)
\( \Rightarrow P(m + 1) \) is true
Here, by PMI \( P(n) \) is true.
i.e., \( \frac{d}{dx}(x^n) = n.x^{n-1} \)

Question. If \( y = \begin{vmatrix} f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c \end{vmatrix} \), then prove that \( \frac{dy}{dx} = \begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix} \).
Answer: Given, \( y = \begin{vmatrix} f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c \end{vmatrix} \)
\( y = (mc - nb) f(x) - (lc - na) g(x) + (lb - ma) h(x) \)
\( \Rightarrow \frac{dy}{dx} = (mc - nb) f'(x) - (lc - na) g'(x) + (lb - ma) h'(x) \)
\( = \begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix} \)