The number of values of x for which tan-1-x2 - x - sin-1-x2-x-1 - -2 is

We have, tan^ 1√(x^2 + x) + sin^ 1√(x^2+x+1) = π2 This equation holds, if x^2 +x ≥ 0 ⋯(1) and 0 ≤ x^2 + x + 1 ≤ 1 ⋯(2) From (1) and (2), we get x^2 +x = ...

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Standard XII

Mathematics

Inverse Function

The number of...

Question

The number of value(s) of $x$ for which ${tan}^{- 1} \sqrt{x^{2} + x} + {sin}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}$ is

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2

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3

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Solution

The correct option is B $2$
We have,
${tan}^{- 1} \sqrt{x^{2} + x} + {sin}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}$
This equation holds, if
$x^{2} + x \geq 0 \dots (1)$
and $0 \leq x^{2} + x + 1 \leq 1 \dots (2)$
From $(1)$ and $(2),$ we get
$x^{2} + x = 0$
$\Rightarrow x = 0, - 1$
Clearly, both these values satisfy the given equation.
Hence, $x = 0, - 1$ are the solutions of the given equation.

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Similar questions

Q. If

{sin}^{- 1} \sqrt{1 + x + x^{2}} + {tan}^{- 1} \sqrt{x + x^{2}} = \frac{π}{2}

t h e n x =

Q. Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

.
(a)

3 {s i n}^{- 1} \frac{2 x}{1 + x^{2}} - 4 {c o s}^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 {t a n}^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

(b)

s i n [(1 / 5) {c o s}^{- 1} x] = 1

(c)

{t a n}^{- 1} \sqrt{x^{2} + x} + {s i n}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}

Q. Solve the equation

{tan}^{- 1} \sqrt{x^{2} + x} + {sin}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}

Q. Assertion :

s i n^{- 1} [x - \frac{x^{2}}{2} + \frac{x^{3}}{4} . . . .] = π / 2 - c o s^{- 1} [x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} . . . .]

for

0 < | x | < \sqrt{2}

has a unique solution. Reason:

t a n^{- 1} \sqrt{x (x + 1)} + s i n^{- 1} \sqrt{x^{2} + x + 1} = π / 2

has no solution for

- \sqrt{2} < x < 0

Q. The set of values of

x

for which

{tan}^{- 1} \frac{x}{\sqrt{1 - x^{2}}} = {sin}^{- 1} x

holds, is

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Inverse Function

Standard XII Mathematics

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The number of value(s) of x for which tan−1√x2+x+sin−1√x2+x+1=π2 is

The correct option is B 2We have, tan−1√x2+x+sin−1√x2+x+1=π2 This equation holds, if x2+x≥0 ⋯(1) and 0≤x2+x+1≤1 ⋯(2) From (1) and (2), we get x2+x=0 ⇒x=0,−1 Clearly, both these values satisfy the given equation. Hence, x=0,−1 are the solutions of the given equation.

The number of value(s) of $x$ for which ${tan}^{- 1} \sqrt{x^{2} + x} + {sin}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}$ is