JEE Advanced 2020 Concept Recapitulation Test 1 Paper 2

We have f(x)={

3x	‌	x≥0
x	‌	x<0.

and g(x)={

‌ x 3	‌	x≥0
x	‌	x<0.

Clearly f and g are inverse of each other
Now, h(x)=f(g(x))={

3(‌ x 3 )=x	‌	x≥0
x	‌	x<0.

As, h(x)=x‌∀x∈R
⇒k(x)=1+‌

1

π

(cos−1x+cot−1x)
Domain of k(x)=[−1,1] and k(x) is decreasing function on [−1,1]
As k(x) is continuous function on [−1,1]
Now, kmin(x=1)=1+‌

1

π

(cos−11+cot−11)
=1+‌

1

π

(0+‌

π

4

)=1+‌

1

4

=‌

5

4

kmax.(x=−1)=1+‌

1

π

(cos−1(−1)+cot−1(−1))
=1+‌

1

π

[π+‌

3π

4

]=1+‌

7

4

=‌

11

4

⇒ Range of k(x)=[‌

5

4

,‌

11

4

]