NIMCET 2023 Question Paper with solutions

Question : 12 of 120

Marks: +1, -0

n_1

and

n_2

are the number of real valued solutions of equations

x={| sin \;^{-1} x |}

and

x= sin \;(x)

respectively, then the value of

n_2-n_1

is:

Solution:

f(x)\;=x- sin \;x

f^{'}(x)\;=1-\cos x ≥ 0

This is increasing function on

\; {\R}-\{2 n π: n ∈ z\}

\;⇒ f^{'}(x)>f(x) \;\text " for all "\; x>0

\;\;\text " Thus "\; x> sin \;x \;\text " for all "\; x>0

\;\;\text " and "\; x< sin \;x \;\text " for "\; x ≠-2 n π

\;\;\text " Thus, "\; x- sin \;x ≠ 0 \;\text " for all "\; x ∈ {\R}-2 n π

\;\;\text " Hence, "\; x- sin \;x=0 \;\text " for "\; x=0 \;\text " only "\;

\; ∴ \;\; n_2=1

We should know that graph of

f^{-1}(x)

is mirror image of

f(x)

with respect to

y=x

as a mirror.

Number of solutions

n_1=0

(only

(0,0)

is the solution of this equation) of equation

x={| sin \;^{-1} x |}

Hence

n_2-n_1=1

Go to Question:

Prev Question

Next Question