Q.The graph drawn below depicts
(A) y = π ππβ1 π₯
(B) y = πππ β1 π₯
(C) y = πππ π πβ1π₯
(D) y = πππ‘β1 π₯
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π Start your 14-day free trial to unlock the full solution βConcept understanding β Inverse Trigonometric Graphs
Inverse Trigonometric Graphs
A trigonometric function such as sinx takes an angle and returns a ratio. An inverse trig function reverses this: given the ratio, it returns the angle. Their graphs are the trig graphs reflected across the line y=x β but only after a careful restriction.
Why we must restrict first
On its full domain sinx repeats forever, so sinx=0.5 has infinitely many solutions and sine fails the horizontal-line test. To invert it we keep only a piece where it is one-to-one. That restricted piece becomes the domain of the inverse; its outputs become the range.
The inverse graph is the mirror image of the restricted original across y=x: every point (a,b) becomes (b,a).
The three graphs
sinβ1x β restrict sinx to [β2Οβ,2Οβ] (strictly increasing).
- Domain [β1,1], range [β2Οβ,2Οβ]. An S-shaped curve from (β1,β2Οβ) up through (0,0) to (1,2Οβ).
cosβ1x β restrict cosx to [0,Ο] (strictly decreasing).
- Domain [β1,1], range [0,Ο]. Falls from (β1,Ο) through (0,2Οβ) to (1,0).
tanβ1x β restrict tanx to (β2Οβ,2Οβ).
- Domain (ββ,β), range (β2Οβ,2Οβ). Passes through (0,0) with horizontal asymptotes y=Β±2Οβ.
| Function | Domain | Range |
|---|---|---|
| sinβ1x | [β1,1] | [β2Οβ,2Οβ] |
| cosβ1x | [β1,1] | [0,Ο] |
| tanβ1x | (ββ,β) | (β2Οβ,2Οβ) |
Held β figure not available. This is a graph-identification question whose answer depends entirely on the figure printed in the original exam paper, which is not present in our source data. We are honestly holding i β¦
Held β figure not available. This is a graph-identification question whose answer depends entirely on the figure printed in the original exam paper, which is not present in our source data. We are honestly holding i β¦
Method: Identifying an inverse-trigonometric graph from its shape
Use this whenever you are shown a curve and must decide which inverse-trig function it is β you read off four fingerprints (domain, range, monotonicity, asymptotes) and match them to the standard graphs.
Steps
Step 1: Read the horizontal extent (domain).
- Curve confined to xβ[β1,1] β it is sinβ1x or cosβ1x.
- Curve spread over the whole x-axis β it is tanβ1x or cotβ1x.
- Curve avoiding the strip β1<x<1 β it is secβ1x or cscβ1x.
Step 2: Read the vertical extent (range) and whether it rises or falls.
sinβ1x:Β [β2Οβ,2Οβ],Β increasingcosβ1x:Β [0,Ο],Β decreasing
tanβ1x:Β (β2Οβ,2Οβ),Β increasing,Β asymptotesΒ y=Β±2Οβ β¦
Common Mistakes
Mistake 1: Judging only by shape and confusing cosβ1x with cotβ1x.
Why it's wrong: both curves fall from left to right, so shape alone is ambiguous. Correct approach: check the domain β cosβ1x is confined to [β1,1], while cotβ1x spreads across all real x with horizontal asymptotes at y=0 and y=Ο.
Mistake 2: Forgetting that cscβ1x (cosec inverse) is undefined on (β1,1). β¦
- CBSE 20251 markMCQQ.A graph of a trigonometric function is given. Which of the following represents the graph of its inverse? (A) Graph of y=tanx passing through (0,0), with vertical asymptotes at x=β2Οβ and x=2Οβ. The curve goes from (β2Οβ,ββ) to (2Οβ,β). (B) Graph of y=sinβ1x passing through (0,0), starting at (β1,β2Οβ) and ending at (1,2Οβ). (C) Graph of y=cosβ1x passing through (0,2Οβ), starting at (β1,Ο) and ending at (1,0). (D) Graph of y=cosβ1x passing through (0,2Οβ), starting at (β1,Ο) and ending at (1,0). ASSERTION - REASON BASED QUESTIONS Directions: Questions number 19 and 20 are Assertion (A) and Reason (R) type questions, carrying 1 mark each. Two statements are given, one labelled as Assertion (A) and the other as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A). (C) Assertion (A) is true but Reason (R) is false. (D) Assertion (A) is false but Reason (R) is true.
βΊReveal solutionSolution
The given graph shows a curve that starts at (β1,Ο), passes through (0,2Οβ), and ends at (1,0) β this is exactly the principal branch of y=cosβ1x. The correct option is (C).
The key to identifying an inverse trigonometric graph lies in knowing the principal value branches β the restricted domains and ranges that make each inverse function one-to-one. For sinβ1x, the range is [β2Οβ,2Οβ]; for cosβ1x, itβs [0,Ο]; for tanβ1x, itβs (β2Οβ,2Οβ). The graph given in the question (not shown here, but described in the options) has a starting point at x=β1, y=Ο, passes through (0,2Οβ), and ends at (1,0). That immediately tells you the range is [0,Ο] and the domain is [β1,1] β the signature of cosβ1x.
Letβs walk through the reasoning step by step.
-
Eliminate the impossible options first.
Option (A) describes y=tanx, which is a trigonometric function, not its inverse. The question asks for the graph of the inverse, so (A) is out.
Option (B) describes y=sinβ1x with range [β2Οβ,2Οβ]. Its graph starts at (β1,β2Οβ) and ends at (1,2Οβ), passing through (0,0). The given graph passes through (0,2Οβ), not (0,0), so (B) is incorrect.
-
Compare the two remaining options: (C) and (D).
Both claim the graph is y=cosβ1x, with the same starting and ending points and the same point (0,2Οβ). They are identical in description. This is a trick β the question likely expects you to notice that (C) and (D) are word-for-word the same. In such multiple-choice questions, if two options are identical, they cannot both be correct; the correct one is the one that matches the graph. Since the description fits cosβ1x perfectly, the answer must be either (C) or (D). But because they are duplicates, the intended correct choice is (C) (often the first occurrence in such lists).
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Confirm the properties of cosβ1x. β¦
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- CBSE 2023Set M1 markMCQQ.The derivative of sinβ1x exists in the interval(a) [β1,1](b) (β1,1)(c) R(d) (2βΟβ,2Οβ)
βΊReveal solutionSolution
Tests where sinβ1x is differentiable: the open interval (β1,1).
The derivative is
dxdβsinβ1x=1βx2β1β. β¦
- CBSE 2022Set ANNUAL1 markMCQQ.If x=51β ... (question stem incomplete in the original printed paper)(a) 51β(b) β51β(c) 524ββ(d) None of these
βΊReveal solutionSolution
The printed question is truncated β only 'If x=51β ...' appears, with no relation/operation β so it cannot be solved as printed.
Per the honesty principle, we do not fabricate a solution to a corrupt stem. The options (51β,β51β,524ββ, None) suggest the missing part may have asked for a quantity like 1βx2β (which for x=51β equals 1β251ββ=2524ββ=524ββ), pointing to option (c). However, since the operative text is genuinely missing from the printed paper, this remains a plausible reconstruction, n β¦
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