CBSE Sample Paper 2025-26 · Section A · 1 mark

Q7: $k$ for continuity of $kx + 1$ and $\cos x$ at $x = \pi$

Sample 2025-26Ch 5 · Continuity and Differentiability

The question

If a function defined by $f(x) = \begin{cases} kx + 1, & x \le \pi \\ \cos x, & x > \pi \end{cases}$ is continuous at $x = \pi$ ⁠, then the value of $k$ is

(A) $\pi$

(B) $\frac{-1}{\pi}$

(D) $\frac{-2}{\pi}$

Exam answer

$f$ is continuous at $x = \pi$ ⁠, so

\begin{aligned} &\lim_{x \to \pi^-}(kx + 1) = \lim_{x \to \pi^+}\cos x \end{aligned}

Why?

Continuity at $x = \pi$ means the left-hand limit, the right-hand limit and $f(\pi)$ are all equal.

Just left of $\pi$ (and at $\pi$ itself, since $x \le \pi$ ⁠) the rule is $kx + 1$ ⁠. Just right of $\pi$ it is $\cos x$ ⁠. So each side uses its own formula.

\begin{aligned} \Rightarrow\ &k\pi + 1 = \cos\pi = -1 \end{aligned}

Why?

$kx + 1$ and $\cos x$ are both continuous, so each limit is found by putting $x = \pi$ ⁠. The left side, $k\pi + 1$ ⁠, is also $f(\pi)$ ⁠.

Watch out: $\cos\pi = -1$ ⁠, not 0 or 1.

\begin{aligned} \Rightarrow\ &k\pi = -1 - 1 = -2 \\ \Rightarrow\ &k = \frac{-2}{\pi} \end{aligned}

Why?

Subtract 1 from both sides, then divide by $\pi$ ⁠.

Answer: (D) $\dfrac{-2}{\pi}$

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: $kx + 1$ gives LHL $k\pi + 1$ ⁠, and $\cos x$ gives RHL $\cos\pi = -1$ ⁠. Set $k\pi + 1 = -1$ ⁠: $k = -\frac{2}{\pi}$ ⁠.

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Q7: kkk for continuity of kx+1kx + 1kx+1 and cos⁡x\cos xcosx at x=πx = \pix=π

Q7: $k$ for continuity of $kx + 1$ and $\cos x$ at $x = \pi$