Question

The equation ${\sin }^{-1}x=|x-a|$ will have atleast one solution if

• A. $a\in [-1,1]$
• B. $a\in [-\frac{\pi }{2},\frac{\pi }{2}]$
• C. $a\in [1,1+\frac{\pi }{2}]$
• D. $a\in [1-\frac{\pi }{2},1+\frac{\pi }{2}]$

Answer 1

The correct option is D

$a\in [1-\frac{\pi }{2},1+\frac{\pi }{2}]$

Here, for atleast one solution to exist the curve $y=|x-a|$
should intersect the curve $y={\sin }^{-1}x$.
So, the best method to solve this question is by plotting the graph of both the curves in a single cartesian plane and then find the range of values of $a$.
So, visualizing the graph a little, we can see we can have three cases .
Case.(1)
here, we have assumed $a<0$.
Here, the modulus function intersects the x-axis at $a$ which is less than 0. If we assume a to be a very negative value , $|x-a|$ wouldn't intersect ${\sin }^{-1}x$. Now as we shift the value of $a$ to the right we will reach a value of $a$ such that $|x-a|$ intersects ${\sin }^{-1}x$. Notice from the graph that that in that case the point of intersection for both the curves will be $(1,\frac{\pi }{2})$
i.e. when a satisfies $\frac{\pi }{2}=|1-a|$ for $a<0$
If we shift the value of $a$ to the right of this value both these curves continue to intersect.

Case (2) $a=0$
${\sin }^{-1}x=|x|$ . It is clear that 0 is the solution.
Case(3). $a>0$
Here, in this case , the modulus function keeps on intersecting ${\sin }^{-1}x$ as we shift value of $a$ to the right until we reach a limiting value. How do we find it?
The same way we did for the case $a<0$. Again when this happens, the point of intersection for both the curves will be $(1,\frac{\pi }{2})$
i.e. when a satisfies $\frac{\pi }{2}=|1-a|$ for $a>0$
We thus obtain the two limiting values of a by solving $\frac{\pi }{2}=|1-a|$
So for $a<0$ we get $a=1-\frac{\pi }{2}$
for $a>0$ we get $a=1+\frac{\pi }{2}$
Thus, combining all cases we have
$\Rightarrow a\in [1-\frac{\pi }{2},1+\frac{\pi }{2}]$

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