Number of real value of x satisfying the equation- arctan-x x-1 - arcsin-x x-1 - 1 - -2 is

The correct option is C 2Let f(x)=tan^ 1(√(x(x+1)))+sin^ 1(√(x(x+1)+1)) π2 Hence f(x)=0 for x= 1,0. Where denotes singleton set.Hence we have two so ...

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Standard XII

Mathematics

Graphs of Basic Inverse Trigonometric Functions

Number of rea...

Question

Number of real value of $x$ satisfying the equation, $arctan \sqrt{x (x + 1)} + arcsin \sqrt{x (x + 1) + 1} = \frac{π}{2}$ is

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more than 2

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Solution

The correct option is C 2
Let
$f (x) = t a n^{- 1} (\sqrt{x (x + 1)}) + s i n^{- 1} (\sqrt{x (x + 1) + 1}) - \frac{π}{2}$
Hence $f (x) = 0$ for x={-1,0}.
Where {} denotes singleton set.
Hence we have two solutions for the above given equation.

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Number of real value of x satisfying the equation, arctan√x(x+1)+arcsin√x(x+1)+1=π2 is

The correct option is C 2Letf(x)=tan−1(√x(x+1))+sin−1(√x(x+1)+1)−π2Hence f(x)=0 for x={-1,0}. Where {} denotes singleton set.Hence we have two solutions for the above given equation.

Number of real value of $x$ satisfying the equation, $arctan \sqrt{x (x + 1)} + arcsin \sqrt{x (x + 1) + 1} = \frac{π}{2}$ is

The correct option is C 2
Let
$f (x) = t a n^{- 1} (\sqrt{x (x + 1)}) + s i n^{- 1} (\sqrt{x (x + 1) + 1}) - \frac{π}{2}$
Hence $f (x) = 0$ for x={-1,0}.
Where {} denotes singleton set.
Hence we have two solutions for the above given equation.