Match List I with List II:

	List - I	List - II
(I)	The number of solutions of the equation where  is	(P) 4
(II)	equals	(Q) 3
(III)	The number of solutions of (x, y), which satisfy |y| = sin x and y = cos-1(cos x), where -2π ≤ x ≤ 2π is	(R) 2
(IV)	For the equation , the number of real solutions is	(S) 1
		(T) 0

 

Calculation

(I)

⇒ ⇒

But for all values of x, is not defined.

Hence, no solution exist.

(I) → T

(II)

=

=

=

=

=

=

=

=

=

(II) → P

(III) In [0, π], |y| = sin x, y = cos^-1(cos x) = x

In [π, 2π], |y| = sin x, y = cos^-1(cos(2π - x)) = 2π - x

In [-π, 0], |y| = sin x, y = cos^-1(cos(-x)) = -x

In [-2π, -π], |y| = sin x, y = cos^-1(cos(2π + x)) = 2π + x

Plotting the graph, we have

Total 3 solutions

(III) → Q

(IV) Given equation is

⇒

⇒

⇒

⇒

On squaring both sides, we get

⇒

⇒

⇒

⇒

∴ But x = 0 does not satisfy the given equation.

So, the number of real solutions is zero.

(IV) → T 

Hence option 3 is correct