If the area enclosed between f(x)= min .( cos -1( cos x)., . cot -1( cot x)) and x-axis in x ∈(π, 2 π) is (π2/k) where k ∈ N, then k is equal to

Question

If the area enclosed between f(x)= min .( cos -1( cos x)., . cot -1( cot x)) and x-axis in x ∈(π, 2 π) is (π2/k) where k ∈ N, then k is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Given, f(x) begincasesx-π ; π < x ≤ (3 π/2) 2 π-x ; (3 π/2) < x < 2 π endcases <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/2dafbb9f10376981a0a61aa574d77143-.png" /> Clearly, required area = area of shaded portion of triangle A B C=(1/2) × (π2/2) =(π2/4)=(π2/k) ∴ On comparing, we get k=4.

Q. If the area enclosed between $f (x) = min . (cos^{- 1} (cos x)$ , $cot^{- 1} (cot x))$ and $x$ -axis in $x \in (π, 2 π)$ is $\frac{π ^{2}}{k}$ where $k \in N$ , then $k$ is equal to

Given, $f (x) {x - π 2 π - x; π < x \leq \frac{3 π}{2}; \frac{3 π}{2} < x < 2 π$

Clearly, required area
$=$ area of shaded portion of $△ A BC = \frac{1}{2} \times \frac{π ^{2}}{2}$
$= \frac{π ^{2}}{4} = \frac{π ^{2}}{k}$
$∴$ On comparing, we get $k = 4$ .

Q. If the area enclosed between f(x)=min.(cos−1(cosx), cot−1(cotx)) and x-axis in x∈(π,2π) is kπ2​ where k∈N, then k is equal to

Given, f(x){x−π2π−x​;π<x≤23π​;23π​<x<2π​ Clearly, required area = area of shaded portion of △ABC=21​×2π2​ =4π2​=kπ2​ ∴ On comparing, we get k=4.

Q. If the area enclosed between $f (x) = min . (cos^{- 1} (cos x)$ , $cot^{- 1} (cot x))$ and $x$ -axis in $x \in (π, 2 π)$ is $\frac{π ^{2}}{k}$ where $k \in N$ , then $k$ is equal to

Given, $f (x) {x - π 2 π - x; π < x \leq \frac{3 π}{2}; \frac{3 π}{2} < x < 2 π$

Clearly, required area
$=$ area of shaded portion of $△ A BC = \frac{1}{2} \times \frac{π ^{2}}{2}$
$= \frac{π ^{2}}{4} = \frac{π ^{2}}{k}$
$∴$ On comparing, we get $k = 4$ .