Let α ∈ (0 , (π /2)) and f(x)=√x2 + x+((tan)2 α /√x2 + x),x0 . If the least value of f(x) is 2√3, then α is equal to

Question

Let α ∈ (0 , (π /2)) and f(x)=√x2 + x+((tan)2 α /√x2 + x),x>0 . If the least value of f(x) is 2√3, then α is equal to (A) (π /3) (B) (π /8) (C) (π /6) (D) (π /4)

Tardigrade Admin · Accepted Answer

By AM-GM inequality √x2 + x+(tan2 α /√x2 + x)≥ 2⋅ √√x2 + x ⋅ (tan2 α /√x2 + x)=2tanα Since the least value of f(x) is 2tan α =2√3 (given) Hence, α =(π /3)

Q. Let $α \in (0, \frac{π}{2})$ and $f (x) = x^{2} + x + \frac{( t an ) ^{2} α}{x ^{2} + x}, x > 0$ . If the least value of $f (x)$ is $23,$ then $α$ is equal to

$\frac{π}{3}$

$\frac{π}{8}$

$\frac{π}{6}$

$\frac{π}{4}$

By AM-GM inequality
$x^{2} + x + \frac{t a n ^{2} α}{x ^{2} + x} \geq 2 \cdot x^{2} + x \cdot \frac{t a n ^{2} α}{x ^{2} + x} = 2 t an α$
Since the least value of $f (x)$ is $2 t an α = 23$ (given)
Hence, $α = \frac{π}{3}$

Q. Let α∈(0,2π​) and f(x)=x2+x​+x2+x​(tan)2α​,x>0 . If the least value of f(x) is 23​, then α is equal to

3π​

8π​

6π​

4π​