Sample 2024-25 Class 12 Maths Q20 Solution: Is f(x) = sec x one-one on its domain?

The question

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

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

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

(C) (A) is true but (R) is false.

(D) (A) is false but (R) is true.

Assertion (A): The function

\begin{aligned} f &: R - \left\{(2n + 1)\frac{\pi}{2} : n \in Z\right\} \\ &\to (-\infty, -1] \cup [1, \infty) \end{aligned}

defined by $f(x) = \sec x$ is not one-one function in its domain.

Reason (R): The line $y = 2$ meets the graph of the function at more than one point.

Exam answer

\cos\frac{\pi}{3} = \cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}

\sec\frac{\pi}{3} = \sec\left(-\frac{\pi}{3}\right) = 2

Why?

$\sec x = \frac{1}{\cos x}$ ⁠, and cosine is an even function: $\cos(-x) = \cos x$ ⁠. So $\sec$ is even too, and $\frac{1}{1/2} = 2$ ⁠.

Both $\frac{\pi}{3}$ and $-\frac{\pi}{3}$ are in the domain, since neither is an odd multiple of $\frac{\pi}{2}$ ⁠.

$f\left(\frac{\pi}{3}\right) = f\left(-\frac{\pi}{3}\right)$ but $\frac{\pi}{3} \ne -\frac{\pi}{3}$ ⁠, so $f$ is not one-one. Assertion (A) is true.

Why?

$f$ is one-one when $f(x_1) = f(x_2)$ forces $x_1 = x_2$ ⁠. One pair of different inputs with the same output is enough to show that $f$ is not one-one.

The line $y = 2$ meets the graph of $f$ at $x = \frac{\pi}{3}$ and $x = -\frac{\pi}{3}$ ⁠, both in the domain. Reason (R) is true.

A horizontal line meeting the graph at more than one point shows that $f$ is not one-one, so the Reason explains the Assertion.

Why?

This is the horizontal line test. If the line $y = k$ meets the graph at two points, those two $x$ -values give the same output $k$ ⁠, which is exactly a pair like the one above. In fact $y = 2$ meets the graph at every $x = 2n\pi \pm \frac{\pi}{3}$ ⁠.

So R is true and it is the reason A holds: option (A).

Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A).

Where this comes from

Chapter 1, Relations and Functions: One-one functions: sec x and the line test.

Question type: Not one-one or not onto

How to spot it: click to see

Type 6. Not one-one or not onto: one example is enough

Spot it: functions such as $x^2$ ⁠, $|x|$ or $[x]$ on $\R$ ⁠, where two inputs share an output, or some numbers are never reached.

Here: $\sec\frac{\pi}{3} = \sec\left(-\frac{\pi}{3}\right) = 2$ ⁠: two different inputs share the output 2, which is the line $y = 2$ meeting the graph twice.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 2, Inverse Trigonometric Functions.

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Q20: Is f(x)=sec⁡xf(x) = \sec xf(x)=secx one-one on its domain?

Q20: Is $f(x) = \sec x$ one-one on its domain?