The number of real solutions of (cot)- 1 √x (x + 4)+(cos)- 1 â¡ √x2 + 4 x + 1=(π /2) is equal to

Question

The number of real solutions of (cot)- 1 √x (x + 4)+(cos)- 1 â¡ √x2 + 4 x + 1=(π /2) is equal to (A) 0 (B) 1 (C) 2 (D) Infinite

Tardigrade Admin · Accepted Answer

Let, f(x)=(cot)- 1 √x (x + 4)+(cos)- 1 ⁡ √x2 + 4 x + 1 For f(x) to be defined, we have x2+4x≥ 0 and 0≤ x2+4x+1≤ 1 Which is only possible if x2+4x=0⇒ x=0,-4 Also these values satisfies the equation ∴ Number of solutions =2

Q. The number of real solutions of $(co t)^{- 1} x (x + 4) + (cos)^{- 1} x^{2} + 4 x + 1 = \frac{π}{2}$ is equal to

$0$

$1$

$2$

Infinite

Let, $f (x) = (co t)^{- 1} x (x + 4) + (cos)^{- 1} x^{2} + 4 x + 1$
For $f (x)$ to be defined, we have
$x^{2} + 4 x \geq 0$ and $0 \leq x^{2} + 4 x + 1 \leq 1$
Which is only possible if
$x^{2} + 4 x = 0 \Rightarrow x = 0, - 4$
Also these values satisfies the equation
$∴$ Number of solutions $= 2$

Q. The number of real solutions of (cot)−1x(x+4)​+(cos)−1⁡x2+4x+1​=2π​ is equal to

0

1

2

Infinite

Let, f(x)=(cot)−1x(x+4)​+(cos)−1⁡x2+4x+1​ For f(x) to be defined, we have x2+4x≥0 and 0≤x2+4x+1≤1 Which is only possible if x2+4x=0⇒x=0,−4 Also these values satisfies the equation ∴ Number of solutions =2

Q. The number of real solutions of $(co t)^{- 1} x (x + 4) + (cos)^{- 1} x^{2} + 4 x + 1 = \frac{π}{2}$ is equal to

$0$

$1$

$2$

Let, $f (x) = (co t)^{- 1} x (x + 4) + (cos)^{- 1} x^{2} + 4 x + 1$
For $f (x)$ to be defined, we have
$x^{2} + 4 x \geq 0$ and $0 \leq x^{2} + 4 x + 1 \leq 1$
Which is only possible if
$x^{2} + 4 x = 0 \Rightarrow x = 0, - 4$
Also these values satisfies the equation
$∴$ Number of solutions $= 2$