Exams Joint Entrance Examination Mathematics Inverse Trigonometric Functions Question 152 Question 152 of 155 Hard MCQ
Let (a, b) ⊂ ( 0 , 2 π ) \subset(0,2 \pi) ⊂ ( 0 , 2 π ) be the largest interval for which sin − 1 ( sin θ ) − cos − 1 ( sin θ ) > 0 , θ ∈ ( 0 , 2 π ) \sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>0, \theta \in(0,2 \pi) sin − 1 ( sin θ ) − cos − 1 ( sin θ ) > 0 , θ ∈ ( 0 , 2 π ) ,
holds.
If α x 2 + β x + sin − 1 ( x 2 − 6 x + 10 ) + cos − 1 ( x 2 − 6 x + 10 ) = 0 \alpha x^{2}+\beta x+\sin ^{-1}\left(x^{2}-6 x+10\right)+\cos ^{-1}\left(x^{2}-6 x+10\right)=0 α x 2 + β x + sin − 1 ( x 2 − 6 x + 10 ) + cos − 1 ( x 2 − 6 x + 10 ) = 0 and α − β = b − a \alpha-\beta=b-a α − β = b − a , then α \alpha α is equal to :
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sin − 1 sin θ − ( π 2 − sin − 1 sin θ ) > 0 \sin ^{-1} \sin \theta-\left(\frac{\pi}{2}-\sin ^{-1} \sin \theta\right)>0 sin − 1 sin θ − ( 2 π − sin − 1 sin θ ) > 0
⇒ sin − 1 sin θ > π 4 \Rightarrow \sin ^{-1} \sin \theta>\frac{\pi}{4} ⇒ sin − 1 sin θ > 4 π
⇒ sin θ > 1 2 \Rightarrow \sin \theta>\frac{1}{\sqrt{2}} ⇒ sin θ > 2 1
So, θ ∈ ( π 4 , 3 π 4 ) \theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right) θ ∈ ( 4 π , 4 3 π )
θ ∈ ( π 4 , 3 π 4 ) = ( a , b ) \theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)=(\mathrm{a}, \mathrm{b}) θ ∈ ( 4 π , 4 3 π ) = ( a , b )
b − a = π 2 = α − β b-a=\frac{\pi}{2}=\alpha-\beta b − a = 2 π = α − β
⇒ β = α − π 2 \Rightarrow \beta=\alpha-\frac{\pi}{2} ⇒ β = α − 2 π
⇒ α x 2 + β x + sin − 1 [ ( x − 3 ) 2 + 1 ] + cos − 1 [ ( x − 3 ) 2 + 1 ] = 0 \Rightarrow \alpha x^{2}+\beta \mathrm{x}+\sin ^{-1}\left[(\mathrm{x}-3)^{2}+1\right]+\cos ^{-1}\left[(\mathrm{x}-3)^{2}+1\right]=0 ⇒ α x 2 + β x + sin − 1 [ ( x − 3 ) 2 + 1 ] + cos − 1 [ ( x − 3 ) 2 + 1 ] = 0
x = 3 , 9 α + 3 β + π 2 + 0 = 0 x=3,9 \alpha+3 \beta+\frac{\pi}{2}+0=0 x = 3 , 9 α + 3 β + 2 π + 0 = 0
\[
⇒ 9 α + 3 ( α − π 2 ) + π 2 = 0 ⇒ 12 α − π = 0 α = π 12 \begin{aligned}
& \Rightarrow 9 \alpha+3\left(\alpha-\frac{\pi}{2}\right)+\frac{\pi}{2}=0 \\\\
& \Rightarrow 12 \alpha-\pi=0 \\\\
& \alpha=\frac{\pi}{12}
\end{aligned} ⇒ 9 α + 3 ( α − 2 π ) + 2 π = 0 ⇒ 12 α − π = 0 α = 12 π
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