The number of real solutions of tan -1 √x(x+1)+ sin -1 √x2+x+1=(π/2) is :

Question

The number of real solutions of tan -1 √x(x+1)+ sin -1 √x2+x+1=(π/2) is : (A) zero (B) one (C) two (D) infinite

Tardigrade Admin · Accepted Answer

tan -1 √x(x+1)+ sin -1 √x2+x+1=(π/2) x ( x +1) ≥ 0.....(1) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/35adf2b585a213bbbbfa6634ce3d902a-.png" /> Also 0 ≤ x 2+ x +1 ≤ 1 but x2+x+1 is always ≥ 0 ∴ x2+x+1 ≤ 1 x(x-1) ≤ 0.....(2) operatornamesum(1) text and (2) x(x+1)=0 ⇒ x=0 text or x=-1

Q. The number of real solutions of $tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = \frac{π}{2}$ is :

zero

one

two

infinite

$tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = \frac{π}{2}$
$x (x + 1) \geq 0$ .....(1)

Also $0 \leq x^{2} + x + 1 \leq 1$
but $x^{2} + x + 1$ is always $\geq 0$
$∴ x^{2} + x + 1 \leq 1$
$x (x - 1) \leq 0$ .....(2)
$sum (1) and (2) x (x + 1) = 0$
$\Rightarrow x = 0 or x = - 1$

Q. The number of real solutions of tan−1x(x+1)​+sin−1x2+x+1​=2π​ is :

zero

one

two

infinite

tan−1x(x+1)​+sin−1x2+x+1​=2π​ x(x+1)≥0.....(1) Also 0≤x2+x+1≤1 but x2+x+1 is always ≥0 ∴x2+x+1≤1 x(x−1)≤0.....(2) sum(1) and (2)x(x+1)=0 ⇒x=0 or x=−1

Q. The number of real solutions of $tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = \frac{π}{2}$ is :

$tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = \frac{π}{2}$
$x (x + 1) \geq 0$ .....(1)

Also $0 \leq x^{2} + x + 1 \leq 1$
but $x^{2} + x + 1$ is always $\geq 0$
$∴ x^{2} + x + 1 \leq 1$
$x (x - 1) \leq 0$ .....(2)
$sum (1) and (2) x (x + 1) = 0$
$\Rightarrow x = 0 or x = - 1$