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CBSE Board Examination 2023 · Section B · 2 marks

Q21: The set B for a one-one onto f(x)=2xf(x) = 2x⁠, or sin⁡−1(sin⁡3π4)+cos⁡−1(cos⁡3π4)+tan⁡−11\sin^{-1}\left(\sin\frac{3\pi}{4}\right) + \cos^{-1}\left(\cos\frac{3\pi}{4}\right) + \tan^{-1}1

Board 2023Ch 1 · Relations and FunctionsCh 2 · Inverse Trigonometric Functions
The question

(a) A function f:A→Bf : A \to B defined as f(x)=2xf(x) = 2x is both one-one and onto. If A={1,2,3,4}A = \{1, 2, 3, 4\}⁠, then find the set B.

OR

(b) Evaluate :

sin⁡−1(sin⁡3π4)+cos⁡−1(cos⁡3π4)+tan⁡−1(1)\begin{aligned} &\sin^{-1}\left(\sin\frac{3\pi}{4}\right) \\ &\quad + \cos^{-1}\left(\cos\frac{3\pi}{4}\right) \\ &\quad + \tan^{-1}(1) \end{aligned}
Exam answer

(a)

f(1)=2, f(2)=4,f(3)=6, f(4)=8\begin{aligned} f(1) &= 2,\ f(2) = 4, \\ f(3) &= 6,\ f(4) = 8 \end{aligned}
Why?

Put each element of AA into the rule f(x)=2xf(x) = 2x⁠. For example, f(3)=2×3=6f(3) = 2 \times 3 = 6⁠.

f(1)=2×1=2f(2)=2×2=4f(3)=2×3=6f(4)=2×4=8\begin{aligned} f(1) &= 2 \times 1 = 2 \\ f(2) &= 2 \times 2 = 4 \\ f(3) &= 2 \times 3 = 6 \\ f(4) &= 2 \times 4 = 8 \end{aligned}

The four images are all different, which agrees with f being one-one.

Since f is onto, B is the range of f.

Why?

Onto means every element of B is the image of some element of A. So B holds the images f(1)f(1)⁠, f(2)f(2)⁠, f(3)f(3)⁠, f(4)f(4) and nothing else.

Watch out: without onto, B could be any bigger set containing 2, 4, 6 and 8, and the question would have no single answer.

∴ B={2,4,6,8}\therefore\ B = \{2, 4, 6, 8\}⁠.

(b)

sin⁡3π4=sin⁡π4\sin\frac{3\pi}{4} = \sin\frac{\pi}{4} and π4∈[−π2,π2]\frac{\pi}{4} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠, so

sin⁡−1(sin⁡3π4)=π4\sin^{-1}\left(\sin\frac{3\pi}{4}\right) = \frac{\pi}{4}
Why?

The principal value branch of sin⁡−1\sin^{-1} is [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. So

sin⁡−1(sin⁡θ)=θ\sin^{-1}(\sin\theta) = \theta

only when θ\theta lies in that interval. 3π4\frac{3\pi}{4} does not, so first find an angle in the interval with the same sine:

3π4=4π4−π4=π−π4sin⁡3π4=sin⁡(π−π4)=sin⁡π4\begin{aligned} \frac{3\pi}{4} &= \frac{4\pi}{4} - \frac{\pi}{4} = \pi - \frac{\pi}{4} \\[4pt] \sin\frac{3\pi}{4} &= \sin\left(\pi - \frac{\pi}{4}\right) \\ &= \sin\frac{\pi}{4} \end{aligned}

(The identity used is sin⁡(π−θ)=sin⁡θ\sin(\pi - \theta) = \sin\theta⁠.) Since π4\frac{\pi}{4} is in the interval, sin⁡−1(sin⁡π4)=π4\sin^{-1}\left(\sin\frac{\pi}{4}\right) = \frac{\pi}{4}⁠.

Watch out: the answer is not 3π4\frac{3\pi}{4}⁠.

3π4∈[0,π]\frac{3\pi}{4} \in [0, \pi]⁠, so

cos⁡−1(cos⁡3π4)=3π4\cos^{-1}\left(\cos\frac{3\pi}{4}\right) = \frac{3\pi}{4}
Why?

The principal value branch of cos⁡−1\cos^{-1} is [0,π][0, \pi]⁠, and

cos⁡−1(cos⁡θ)=θ\cos^{-1}(\cos\theta) = \theta

for every θ\theta in [0,π][0, \pi]⁠. 3π4\frac{3\pi}{4} is between 0 and π\pi⁠, so it comes straight back.

The two branches are different intervals, which is why sin⁡−1\sin^{-1} and cos⁡−1\cos^{-1} behave differently here.

Also tan⁡−1(1)=π4\tan^{-1}(1) = \frac{\pi}{4}⁠.

Why?

tan⁡π4=1\tan\frac{\pi}{4} = 1⁠, and π4\frac{\pi}{4} lies in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠, the principal value branch of tan⁡−1\tan^{-1}⁠.

Required value=π4+3π4+π4\begin{aligned} \text{Required value} &= \frac{\pi}{4} + \frac{3\pi}{4} + \frac{\pi}{4} \end{aligned}
Why?

The three fractions have the same denominator, so add the numerators:

π+3π+π4=5π4\frac{\pi + 3\pi + \pi}{4} = \frac{5\pi}{4}

∴\therefore Required value =5π4= \dfrac{5\pi}{4}⁠.

Where this comes from

Chapter 1, Relations and Functions: One-one onto function; inverse trig values.

Question type: (b) Adding principal values + An inverse of its own trig function (combined)

How to spot it: click to see

(b) Type 4. Adding principal values: one at a time, then add

Spot it: two or three inverses joined by plus or minus signs, and the question asks for the value.

Here: First, three inverses are joined by plus signs: find each principal value on its own line, then add them to get 5π4\frac{5\pi}{4}⁠.

(b) Type 5. An inverse of its own trig function: move the angle into the range first

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

Here: In sin⁡−1(sin⁡3π4)\sin^{-1}\left(\sin\frac{3\pi}{4}\right) the angle is outside the range, so change it to π4\frac{\pi}{4}⁠. In cos⁡−1(cos⁡3π4)\cos^{-1}\left(\cos\frac{3\pi}{4}\right) it is inside [0,π][0, \pi]⁠, so it stays.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 2, Inverse Trigonometric Functions.

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