Inverse Trigonometric FunctionsmediumFree

Inverse Trigonometric Functions: Let Defined Sgn Find Number Solutions Lying

JEE Maths question with a full step-by-step solution.

Question

Let

g:\mathbb{R}\to\mathbb{R}

be defined as

g(x)=\text{sgn}(x^{2}-5x+6)

. Find the number of solutions of

\sin x=\cos^{-1}(g(\sin^{-1}x))

lying in

[0,314]

Solution

Answer: 1

Step 1:

g(\sin^{-1}x)=\text{sgn}((\sin^{-1}x)^{2}-5\sin^{-1}x+6)=\text{sgn}((\sin^{-1}x-2)(\sin^{-1}x-3))

. Step 2: Domain of

\sin^{-1}x

[-1,1]

, with range

\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\approx[-1.57,1.57]

. Step 3: For

\sin^{-1}x\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]

: both

\sin^{-1}x-2<0

and

\sin^{-1}x-3<0

, so their product is positive. Hence

g(\sin^{-1}x)=1

always (for

x\in[-1,1]

). Step 4: The equation becomes:

\sin x=\cos^{-1}(1)=0\;\Rightarrow\;\sin x=0\;\Rightarrow\;x=n\pi,\;n\in\mathbb{Z}

Step 5: But the equation is only meaningful for

x\in[-1,1]

(the domain of

\sin^{-1}

). The only multiple of

\pi

[-1,1]

x=0

. Step 6: Only

1

solution. Answer:

1

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