Board 2025 Class 12 Maths Q20 Solution: Value of k for continuity at x = 5

The question

Select the correct answer from the options (A), (B), (C) and (D) as given below.

(A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

(C) Assertion (A) is true, but Reason (R) is false.

(D) Assertion (A) is false, but Reason (R) is true.

Assertion (A) : $f(x) = \begin{cases} 3x - 8, & x \le 5 \\ 2k, & x > 5 \end{cases}$ is continuous at $x = 5$ for $k = \frac{5}{2}$ ⁠.

Reason (R) : For a function f to be continuous at $x = a$ ⁠, $\lim\limits_{x \to a^-} f(x) = \lim\limits_{x \to a^+} f(x) = f(a)$ ⁠.

Exam answer

For $x \le 5$ ⁠, $f(x) = 3x - 8$ ⁠; for $x > 5$ ⁠, $f(x) = 2k$ ⁠.

\begin{aligned} \lim_{x \to 5^-} f(x) &= f(5) = 15 - 8 = 7 \\ \lim_{x \to 5^+} f(x) &= 2k \end{aligned}

Why?

Approaching 5 from the left, and at 5 itself, the rule $3x - 8$ applies, because $x \le 5$ includes 5. It is a polynomial, so put $x = 5$ ⁠: $3(5) - 8 = 7$ ⁠.

From the right, $x > 5$ ⁠, the function is the constant 2k, so its limit is 2k.

For continuity at $x = 5$ ⁠: $2k = 7$ ⁠, so $k = \frac{7}{2} \ne \frac{5}{2}$ ⁠.

With $k = \frac{5}{2}$ ⁠, RHL $= 2 \cdot \frac{5}{2} = 5 \ne 7$ ⁠, so f is not continuous at $x = 5$ ⁠.

Why?

Continuity at $x = 5$ needs LHL = RHL = $f(5)$ ⁠. The left limit and $f(5)$ are both 7, so the right limit 2k must be 7 as well.

\begin{aligned} 2k &= 7 \\ k &= \frac{7}{2} \quad (\text{divide by 2}) \\[4pt] 2 \cdot \frac{5}{2} &= \frac{2 \times 5}{2} = \frac{10}{2} = 5 \end{aligned}

With the asserted $k = \frac52$ ⁠, the graph jumps from 7 down to 5 at $x = 5$ ⁠.

So Assertion (A) is false. Reason (R) is true: it is the definition of continuity at $x = a$ ⁠.

Why?

The Assertion is false and the Reason is true. That is code (D). Once the Assertion is false, codes (A), (B) and (C) are all ruled out, since each needs a true Assertion.

Answer: (D) Assertion (A) is false, but Reason (R) is true.

Where this comes from

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

Question type: Find $k$ so that $f$ is continuous

How to spot it: click to see

Type 2. Find $k$ so that $f$ is continuous

Spot it: a function in pieces with an unknown $k$ (or $a$ and $b$ ⁠), and the words "find the value of $k$ ".

Here: $3x - 8$ gives LHL and $f(5)$ equal to 7, and $2k$ gives RHL. Continuity needs $2k = 7$ ⁠, so $k = \frac72$ ⁠, not $\frac52$ ⁠: A is false, R is true, option (D).

Quick lesson · Worked NCERT solutions · Interactive lesson

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →

Q20: Value of k for continuity at x=5x = 5x=5

Q20: Value of k for continuity at $x = 5$