Board 2024 Class 12 Maths Q3 Solution: Continuity of f(x) = x² + 3 (x ≠ 0), f(0) = 1

The question

Which of the following statements is true for the function

f(x) = \begin{cases} x^2 + 3, & x \ne 0 \\ 1, & x = 0 \end{cases}\ ?

(A) $f(x)$ is continuous and differentiable $\forall\ x \in \mathbb{R}$

(B) $f(x)$ is continuous $\forall\ x \in \mathbb{R}$

(C) $f(x)$ is continuous and differentiable $\forall\ x \in \mathbb{R} - \{0\}$

(D) $f(x)$ is discontinuous at infinitely many points

Exam answer

At $x = 0$ ⁠: $f(0) = 1$ ⁠, and

\begin{aligned} \lim_{x \to 0} f(x) &= \lim_{x \to 0} (x^2 + 3) \\ &= 0 + 3 \\ &= 3 \ne 1 = f(0) \end{aligned}

Why?

$f$ is continuous at $x = a$ only when the limit there equals the value there:

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

While $x \to 0$ ⁠, $x$ is never equal to $0$ ⁠, so the limit uses the rule $x^2 + 3$ ⁠. The value $f(0) = 1$ comes from the separate rule for $x = 0$ ⁠.

$x^2 + 3$ is a polynomial, so its limit is found by putting $x = 0$ ⁠:

\begin{aligned} (0)^2 + 3 &= 0 + 3 \\ &= 3 \end{aligned}

So $f$ is discontinuous at $x = 0$ ⁠, and hence not differentiable there.

Why?

A function that is differentiable at a point is always continuous there. So a function that is not continuous at a point cannot be differentiable there.

For $x \ne 0$ ⁠, $f(x) = x^2 + 3$ is a polynomial, so $f$ is continuous and differentiable at every $x \ne 0$ ⁠.

Why?

Any point $a \ne 0$ has a small interval around it that does not contain $0$ ⁠. On that interval $f(x) = x^2 + 3$ ⁠, a polynomial, and polynomials are continuous and differentiable everywhere. Here $f'(a) = 2a$ ⁠.

So (A) and (B) are false ( $f$ fails at $0$ ⁠), and (D) is false ( $0$ is the only point of discontinuity).

Answer: (C) $f(x)$ is continuous and differentiable $\forall\ x \in \mathbb{R} - \{0\}$

Where this comes from

Chapter 5, Continuity and Differentiability: Continuity of a piecewise function.

Question type: Is $f$ continuous at the joint? Compare three numbers

How to spot it: click to see

Type 1. Is $f$ continuous at the joint? Compare three numbers

Spot it: $f$ is given in pieces, and you are asked where it is continuous, or where it breaks.

Here: The pieces are $x^2 + 3$ away from 0 and the value 1 at $x = 0$ ⁠. Near 0 the limit is 3 but $f(0) = 1$ ⁠, so $f$ breaks only at 0, and so is not differentiable there either: option (C).

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Q3: Continuity of f(x)=x2+3f(x) = x^2 + 3f(x)=x2+3 (x≠0x \ne 0x=0⁠), f(0)=1f(0) = 1f(0)=1

Q3: Continuity of $f(x) = x^2 + 3$ ( $x \ne 0$ ⁠), $f(0) = 1$