Sample 2022-23 Class 12 Maths Q21 Solution: sin⁻¹[sin(13π/7)], or an onto function on N that is not one-one

The question

(a) Find the value of $\sin^{-1}\left[\sin\left(\frac{13\pi}{7}\right)\right]$

OR

(b) Prove that the function f is surjective, where $f: N \to N$ such that

f(n) = \begin{cases} \dfrac{n + 1}{2}, & \text{if } n \text{ is odd} \\ \dfrac{n}{2}, & \text{if } n \text{ is even} \end{cases}

Is the function injective? Justify your answer.

Exam answer

(a)

\begin{aligned} &\sin^{-1}\left[\sin\left(\frac{13\pi}{7}\right)\right] \\ &= \sin^{-1}\left[\sin\left(2\pi - \frac{\pi}{7}\right)\right] \\ &= \sin^{-1}\left[\sin\left(-\frac{\pi}{7}\right)\right] \end{aligned}

Why?

$\frac{13\pi}{7}$ is bigger than $\frac{\pi}{2}$ ⁠, so it is not in the principal value branch of $\sin^{-1}$ ⁠. First we replace it by an angle with the same sine that is in the branch.

$\frac{13\pi}{7} = \frac{14\pi}{7} - \frac{\pi}{7} = 2\pi - \frac{\pi}{7}$ ⁠. Sine repeats every $2\pi$ ⁠, and sine is odd, so

\begin{aligned} \sin(2\pi - \theta) &= \sin(-\theta) \\ &= -\sin\theta \end{aligned}

As $-\frac{\pi}{7} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ⁠, the principal value branch of $\sin^{-1}$ ⁠,

Why?

The rule used is

\sin^{-1}(\sin\theta) = \theta

and it holds only when $\theta$ lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ⁠. That is why the angle had to be changed first.

Watch out: $\frac{13\pi}{7}$ is not the answer. A value of $\sin^{-1}$ is never bigger than $\frac{\pi}{2}$ ⁠.

$\therefore\ \sin^{-1}\left[\sin\left(\frac{13\pi}{7}\right)\right] = -\frac{\pi}{7}$

(b) Let $y \in N$ (co-domain). Then $2y \in N$ (domain) and $2y$ is even, so

f(2y) = \frac{2y}{2} = y

Why?

Surjective (onto) means every element of the co-domain has at least one pre-image in the domain.

So we take any $y$ in the co-domain and look for an input that gives $y$ ⁠. The input $2y$ is an even natural number, so $f$ uses the rule $\frac{n}{2}$ for it, and that gives back $y$ ⁠.

So every element of the co-domain has a pre-image. Hence, $f$ is surjective.

$1, 2 \in N$ (domain) and

f(1) = \frac{1 + 1}{2} = 1,\quad f(2) = \frac22 = 1

Why?

Injective (one-one) means different inputs always give different outputs:

f(x_1) = f(x_2) \Rightarrow x_1 = x_2

To show a function is not injective, one pair of different inputs with the same output is enough. Here 1 is odd, so $f(1)$ uses $\frac{n + 1}{2}$ ⁠, and 2 is even, so $f(2)$ uses $\frac{n}{2}$ ⁠.

So $f(1) = f(2)$ but $1 \ne 2$ ⁠. Hence, $f$ is not injective.

$\therefore$ $f$ is surjective but not injective.

Where this comes from

Chapter 2, Inverse Trigonometric Functions: Principal value of sin⁻¹; onto, one-one functions.

Question type: (a) An inverse of its own trig function; (b) Prove one-one and onto + Not one-one or not onto (combined)

How to spot it: click to see

(a) Ch 2, type 5. An inverse of its own trig function: move the angle into the range first

Spot it: $\sin^{-1}(\sin\ldots)$ ⁠, $\cos^{-1}(\cos\ldots)$ or $\tan^{-1}(\tan\ldots)$ with a given angle.

Here: $\frac{13\pi}{7}$ is outside $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ ⁠. Write it as $2\pi - \frac{\pi}{7}$ ⁠, which has the same sine as $-\frac{\pi}{7}$ ⁠, an angle inside the range.

(b) Ch 1, type 5. Prove one-one and onto: the bijective function

Spot it: "show that $f$ is one-one and onto", or "is $f$ bijective?", with a formula like $f(x) = 3 - 4x$ ⁠.

Here: For onto, take any $y$ in $N$ ⁠: the even input $2y$ gives $f(2y) = y$ ⁠.

(b) Ch 1, 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: For one-one, one example is enough: $f(1) = f(2) = 1$ ⁠, so $f$ is not injective.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 1, Relations and Functions.

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Q21: sin⁡−1[sin⁡13π7]\sin^{-1}\left[\sin\frac{13\pi}{7}\right]sin−1[sin713π​]⁠, or an onto function on NNN that is not one-one

Q21: $\sin^{-1}\left[\sin\frac{13\pi}{7}\right]$ ⁠, or an onto function on $N$ that is not one-one