Question:
The following graph represents :
The following graph represents :
Show Hint
Read the points (-1, pi), (0, pi/2) and (1, 0) from the graph. They are the values of cos^{-1} x at x = -1, 0, 1.
Updated On: Oct 22, 2026
- \( y = \cos^{-1} x \)
- \( y = \sec^{-1} x \)
- \( y = \sin^{-1} x \)
- \( y = \tan^{-1} x \)
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The Correct Option is A
Solution and Explanation
Step 1: Read the key points from the graph:
The curve starts at \((-1, \pi)\), crosses the y-axis at \((0, \pi/2)\) and ends at \((1, 0)\).
So the curve lives only for \(x\) between \(-1\) and \(1\), and its height falls from \(\pi\) to \(0\).
Step 2: Match with the standard graph:
For \(y = \cos^{-1} x\), the domain is \([-1, 1]\) and the range is \([0, \pi]\).
At \(x = -1\): \(\cos^{-1}(-1) = \pi\). At \(x = 0\): \(\cos^{-1}(0) = \pi/2\). At \(x = 1\): \(\cos^{-1}(1) = 0\).
These three values are exactly the three points on the graph.
Step 3: Check option (A).
\(y = \cos^{-1} x\) passes through all three marked points and is a decreasing curve. So (A) is TRUE.
Step 4: Check option (B).
\(\sec^{-1} x\) is defined only for \(|x| \geq 1\), so it has no point at \(x = 0\). The graph has one. So (B) is FALSE.
Step 5: Check option (C).
\(\sin^{-1} x\) has range \([-\pi/2, \pi/2]\) and gives \(\sin^{-1}(0) = 0\), not \(\pi/2\). It is also increasing. So (C) is FALSE.
Step 6: Check option (D).
\(\tan^{-1} x\) is defined for all real \(x\) with range \((-\pi/2, \pi/2)\), and it is increasing. The graph is limited to \([-1, 1]\) and decreasing. So (D) is FALSE.
Final Answer:
The graph is that of the inverse cosine function, option (A). \[ \boxed{y = \cos^{-1} x} \]
The curve starts at \((-1, \pi)\), crosses the y-axis at \((0, \pi/2)\) and ends at \((1, 0)\).
So the curve lives only for \(x\) between \(-1\) and \(1\), and its height falls from \(\pi\) to \(0\).
Step 2: Match with the standard graph:
For \(y = \cos^{-1} x\), the domain is \([-1, 1]\) and the range is \([0, \pi]\).
At \(x = -1\): \(\cos^{-1}(-1) = \pi\). At \(x = 0\): \(\cos^{-1}(0) = \pi/2\). At \(x = 1\): \(\cos^{-1}(1) = 0\).
These three values are exactly the three points on the graph.
Step 3: Check option (A).
\(y = \cos^{-1} x\) passes through all three marked points and is a decreasing curve. So (A) is TRUE.
Step 4: Check option (B).
\(\sec^{-1} x\) is defined only for \(|x| \geq 1\), so it has no point at \(x = 0\). The graph has one. So (B) is FALSE.
Step 5: Check option (C).
\(\sin^{-1} x\) has range \([-\pi/2, \pi/2]\) and gives \(\sin^{-1}(0) = 0\), not \(\pi/2\). It is also increasing. So (C) is FALSE.
Step 6: Check option (D).
\(\tan^{-1} x\) is defined for all real \(x\) with range \((-\pi/2, \pi/2)\), and it is increasing. The graph is limited to \([-1, 1]\) and decreasing. So (D) is FALSE.
Final Answer:
The graph is that of the inverse cosine function, option (A). \[ \boxed{y = \cos^{-1} x} \]
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