Question 1

Which of the following is NOT a necessary condition for a function $f(x)$ to be continuous at $x = a$?

Accepted Answer

$f'(a)$ exists — Continuity at $x = a$ requires that $f(a)$ is defined, the limit $\lim_{x 	o a} f(x)$ exists, and that the limit equals $f(a)$. Differentiability (existence of $f'(a)$) is not required for continuity.

Question 2

If a function $f(x)$ has a jump discontinuity at $x = c$, which of the following statements is true?

Accepted Answer

$\lim_{x 	o c^-} f(x) 
eq \lim_{x 	o c^+} f(x)$ — A jump discontinuity occurs when the left-hand limit and right-hand limit at $x = c$ exist but are not equal, causing a 'jump' in the graph.

Question 3

State the three conditions for continuity of a function $f(x)$ at a point $x = a$.

Accepted Answer

The three conditions for continuity at $x = a$ are:
1) $f(a)$ is defined.
2) $\lim_{x 	o a} f(x)$ exists.
3) $\lim_{x 	o a} f(x) = f(a)$. — These conditions ensure the function has no breaks, jumps, or holes at $x = a$, making it continuous there.

Question 4

Explain with an example what is meant by removable discontinuity.

Accepted Answer

A removable discontinuity occurs at $x = a$ if the limit $\lim_{x 	o a} f(x)$ exists but is not equal to $f(a)$ or $f(a)$ is not defined. For example, the function $f(x) = \frac{x^2 - 1}{x - 1}$ has a removable discontinuity at $x = 1$ because $f(1)$ is undefined but $\lim_{x 	o 1} f(x) = 2$. — Removable discontinuities can be 'fixed' by redefining the function value at the point to equal the limit.

Question 5

If $f(x) = \sin x$ and $g(x) = x^2$, what is the value of $(g \circ f)(\pi/2)$ and is the composite function continuous at $x = \pi/2$?

Accepted Answer

First, compute $f(\pi/2) = \sin(\pi/2) = 1$. Then, $g(f(\pi/2)) = g(1) = 1^2 = 1$. Both $f$ and $g$ are continuous at these points, so the composite function $(g \circ f)(x)$ is continuous at $x = \pi/2$. — Since $f$ is continuous at $\pi/2$ and $g$ is continuous at $f(\pi/2) = 1$, by the property of continuity of composite functions, $(g \circ f)$ is continuous at $\pi/2$.

Question 6

Prove that if $f$ is continuous at $x = a$ and $g$ is continuous at $f(a)$, then the composite function $g(f(x))$ is continuous at $x = a$.

Accepted Answer

Proof:
Given $f$ is continuous at $x = a$, so $\lim_{x 	o a} f(x) = f(a)$.
Given $g$ is continuous at $f(a)$, so $\lim_{y 	o f(a)} g(y) = g(f(a))$.
Consider $\lim_{x 	o a} g(f(x))$.
Using limit property, $\lim_{x 	o a} g(f(x)) = g(\lim_{x 	o a} f(x)) = g(f(a))$.
Since $g(f(a)) = (g \circ f)(a)$, the composite function is continuous at $x = a$.
Hence proved. — The proof uses the definition of continuity and limit properties to show that continuity of inner and outer functions ensures continuity of the composite.

Question 7

Define differentiability of a function $f$ at a point $x = a$.

Accepted Answer

A function $f$ is said to be differentiable at $x = a$ if the derivative $f'(a)$ exists, i.e., the limit $\lim_{h 	o 0} \frac{f(a + h) - f(a)}{h}$ exists and is finite. — Differentiability means the function has a well-defined instantaneous rate of change or slope of the tangent at $x = a$.

Question 8

Is the function $f(x) = |x|$ differentiable at $x = 0$? Justify your answer.

Accepted Answer

No, $f(x) = |x|$ is not differentiable at $x = 0$ because the left-hand derivative at 0 is $-1$ and the right-hand derivative at 0 is $1$. Since these two one-sided derivatives are not equal, the derivative at 0 does not exist. — The function has a sharp corner at $x = 0$, so although it is continuous there, it is not differentiable.

Question 9

Explain why differentiability implies continuity but continuity does not necessarily imply differentiability.

Accepted Answer

If $f$ is differentiable at $x = a$, then the derivative $f'(a)$ exists, which means the limit $\lim_{h 	o 0} \frac{f(a+h) - f(a)}{h}$ exists. This implies that $\lim_{h 	o 0} f(a+h) = f(a)$, so $f$ is continuous at $a$.

However, a function can be continuous at $a$ but not differentiable there. For example, $f(x) = |x|$ is continuous at $x = 0$ but not differentiable due to a corner.

Thus, differentiability is a stronger condition than continuity. — The existence of a derivative requires the function to be continuous, but continuity alone does not guarantee the existence of a derivative.

Question 10

Assertion (A): If a function is differentiable at $x = a$, then it is continuous at $x = a$.
Reason (R): Differentiability requires the limit of the difference quotient to exist, which implies the function must be continuous at that point.

Accepted Answer

A) Both A and R are true and R is the correct explanation of A — Differentiability at a point implies the difference quotient limit exists, which requires the function to be continuous at that point. Hence, both statements are true and R correctly explains A.

Question 11

Using the definition of derivative, find $f'(2)$ for the function $f(x) = x^2$.

Accepted Answer

Using the definition:
$f'(2) = \lim_{h 	o 0} \frac{(2 + h)^2 - 2^2}{h} = \lim_{h 	o 0} \frac{4 + 4h + h^2 - 4}{h} = \lim_{h 	o 0} \frac{4h + h^2}{h} = \lim_{h 	o 0} (4 + h) = 4$. — Apply the limit definition stepwise to simplify the difference quotient and evaluate the limit.

Question 12

Find the derivative of the composite function $y = \sin(x^2)$ using the chain rule.

Accepted Answer

Let $u = x^2$, then $y = \sin u$.
Using chain rule, $\frac{dy}{dx} = \frac{dy}{du} 	imes \frac{du}{dx} = \cos u 	imes 2x = 2x \cos(x^2)$. — Differentiate outer function $\sin u$ with respect to $u$, then multiply by derivative of inner function $x^2$.

Question 13

State the chain rule for differentiation of composite functions.

Accepted Answer

If $y = g(f(x))$, where both $f$ and $g$ are differentiable, then $\frac{dy}{dx} = g'(f(x)) 	imes f'(x)$. — The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

Question 14

Find the derivative of $y = \sin^{-1} x$ using implicit differentiation.

Accepted Answer

Let $y = \sin^{-1} x$, then $\sin y = x$.
Differentiating both sides w.r.t $x$:
$\cos y \cdot \frac{dy}{dx} = 1$.
So, $\frac{dy}{dx} = \frac{1}{\cos y}$.
Using $\sin^2 y + \cos^2 y = 1$, we get $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$.
Therefore, $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$, valid for $|x| < 1$. — Implicit differentiation and trigonometric identities are used to find the derivative of inverse sine.

Question 15

Match the following functions with their derivatives:

Accepted Answer

— Matching derivatives with functions tests understanding of basic derivative formulas.

Continuity and Differentiability

Continuity and Differentiability — Study Notes

Continuity of a Function

Continuity of Composite Functions

Differentiability

Practice Questions — Continuity and Differentiability

All 6 Chapters in Mathematics Part-I