Sin-13x-2-x2-cos-1x2-4x-3-4 can have a solution for x -

The correct option is D ( 2 √( 2 ) , 3 √( 5 ) #160; 2 #160;) ∪( 3 √( 5 ) #160; 2 ,2+√( 2 ) #160;) ∪{ 2 } For sin^ 1 ( 3x 2 x^ 2 ) ⇒ 1 ...

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Byju's Answer

Standard XII

Mathematics

Basic Inverse Trigonometric Functions

sin-13x-2-x2+...

Question

$s i n^{- 1} (3 x - 2 - x^{2}) + c o s^{- 1} (x^{2} - 4 x + 3) = \frac{π}{4}$ can have a solution for $x ϵ$

[1, 2]

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(\frac{3 + \sqrt{5}}{2}, 2 + \sqrt{2})

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(\frac{3 - \sqrt{5}}{2}, 2 - \sqrt{2})

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(2 - \sqrt{2}, \frac{3 - \sqrt{5}}{2}) \cup (\frac{3 - \sqrt{5}}{2}, 2 + \sqrt{2}) \cup {2}

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Solution

The correct option is D $(2 - \sqrt{2}, \frac{3 - \sqrt{5}}{2}) \cup (\frac{3 - \sqrt{5}}{2}, 2 + \sqrt{2}) \cup {2}$
For $s i n^{- 1} (3 x - 2 - x^{2}) \Rightarrow - 1 < 3 x - 2 - x^{2} < 1$
$3 x - 2 - x^{2} < 1$ imaginary solution
$3 x - 2 - x^{2} > - 1 \Rightarrow x ϵ {\frac{3 - \sqrt{5}}{2}, \frac{3 + \sqrt{5}}{2}}$ ...(1)
And for $c o s^{- 1} (x^{2} - 4 x + 3) \Rightarrow - 1 < (x^{2} - 4 x + 3) < 1$
$(x^{2} - 4 x + 3) < 1 \Rightarrow x = 2$ ...(2)
$x^{2} - 4 x + 3 > - 1 \Rightarrow x ϵ {2 - \sqrt{2}, 2 + \sqrt{2}}$ ...(3)
From (1),(2) and (3)
$x ϵ (2 - \sqrt{2}, \frac{3 - \sqrt{5}}{2}) \cup (\frac{3 - \sqrt{5}}{2}, 2 + \sqrt{2}) \cup {2}$

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Similar questions

Q. Choose the correct option, if List I denotes polar co-ordinates and List II contains cartesian co-ordinates:

LIST-I	LIST-II
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B) $(5, \frac{π}{4})$	2. $(- \sqrt{2,} \sqrt{2})$
C) $(3, \frac{π}{2})$	3. $(2, 2 \sqrt{3})$
D) $(4, \frac{π}{3})$	4. $(\frac{5}{\sqrt{2}}, \frac{5}{\sqrt{2}})$

Q. Assertion :Zeroes of

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are 5,-1 Reason: The polynomial whose zeroes are

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sin−1(3x−2−x2)+cos−1(x2−4x+3)=π4 can have a solution for xϵ

$s i n^{- 1} (3 x - 2 - x^{2}) + c o s^{- 1} (x^{2} - 4 x + 3) = \frac{π}{4}$ can have a solution for $x ϵ$