The number of real roots of the equationtan -1 √x(x+1)+ sin -1 √x2+x+1=(π/4) is :

Question

The number of real roots of the equationtan -1 √x(x+1)+ sin -1 √x2+x+1=(π/4) is : (A) 1 (B) 2 (C) 4 (D) 0

Tardigrade Admin · Accepted Answer

tan -1 √x(x+1)+ sin -1 √x2+x+1=(π/4) For equation to be defined, x2+x ≥ 0 ⇒ x2+x+1 ≥ 1 ∴ only possibility that the equation is defined x2+x=0 ⇒ x=0 ; x=-1 None of these values satisfy ∴ No of roots =0

Q. The number of real roots of the equationtan $^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = \frac{π}{4}$ is :

1

2

4

0

$tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = \frac{π}{4}$
For equation to be defined,
$x^{2} + x \geq 0$
$\Rightarrow x^{2} + x + 1 \geq 1$
$∴$ only possibility that the equation is defined
$x^{2} + x = 0 \Rightarrow x = 0; x = - 1$
None of these values satisfy
$∴$ No of roots $= 0$

Q. The number of real roots of the equationtan −1x(x+1)​+sin−1x2+x+1​=4π​ is :

1

2

4

0

tan−1x(x+1)​+sin−1x2+x+1​=4π​ For equation to be defined, x2+x≥0 ⇒x2+x+1≥1 ∴ only possibility that the equation is defined x2+x=0⇒x=0;x=−1 None of these values satisfy ∴ No of roots =0

Q. The number of real roots of the equationtan $^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = \frac{π}{4}$ is :

$tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = \frac{π}{4}$
For equation to be defined,
$x^{2} + x \geq 0$
$\Rightarrow x^{2} + x + 1 \geq 1$
$∴$ only possibility that the equation is defined
$x^{2} + x = 0 \Rightarrow x = 0; x = - 1$
None of these values satisfy
$∴$ No of roots $= 0$