JEE Advance - Mathematics (2018 - Paper 1 Offline - No. 13)
Let a, b, c three non-zero real numbers such that the equation $$\sqrt 3 a\cos x + 2b\sin x = c,x \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]$$, has two distinct real roots $$\alpha $$ and $$\beta $$ with $$\alpha + \beta = {\pi \over 3}$$. Then, the value of $${b \over a}$$ is ............
Answer
0.5
Explanation
We have, $$\alpha $$, $$\beta $$ are the roots of
$$\sqrt 3 a\cos x + 2b\sin x = c$$
$$ \therefore $$ $$\sqrt 3 a\cos \alpha + 2b\sin \alpha = c$$ ... (i)
and $$\sqrt 3 a\cos \beta + 2b\sin \beta = c$$ ... (ii)
On subtracting Eq. (ii) from Eq. (i), we get
$$\sqrt 3 a(\cos \alpha - \cos \beta ) + 2b(\sin \alpha - \sin \beta ) = 0$$
$$ \Rightarrow \sqrt 3 a\left( { - 2\sin \left( {{{\alpha + \beta } \over 2}} \right)} \right)\sin \left( {{{\alpha - \beta } \over 2}} \right) + 2b\left( {2\cos \left( {{{\alpha + \beta } \over 2}} \right)} \right)\sin \left( {{{\alpha - \beta } \over 2}} \right) = 0$$
$$ \Rightarrow \sqrt 3 a\sin \left( {{{\alpha + \beta } \over 2}} \right) = 2b\cos \left( {{{\alpha + \beta } \over 2}} \right)$$
$$ \Rightarrow \tan \left( {{{\alpha + \beta } \over 2}} \right) = {{2b} \over {\sqrt 3 a}}$$
$$ \Rightarrow \tan \left( {{\pi \over 6}} \right) = {{2b} \over {\sqrt 3 a}}$$ [$$ \because $$ $$\alpha $$ + $$\beta $$ = $${\pi \over 3}$$, given]
$$ \Rightarrow {1 \over {\sqrt 3 }} = {{2b} \over {\sqrt 3 a}} \Rightarrow {b \over a} = {1 \over 2}$$
$$ \Rightarrow {b \over a} = 0.5$$
$$\sqrt 3 a\cos x + 2b\sin x = c$$
$$ \therefore $$ $$\sqrt 3 a\cos \alpha + 2b\sin \alpha = c$$ ... (i)
and $$\sqrt 3 a\cos \beta + 2b\sin \beta = c$$ ... (ii)
On subtracting Eq. (ii) from Eq. (i), we get
$$\sqrt 3 a(\cos \alpha - \cos \beta ) + 2b(\sin \alpha - \sin \beta ) = 0$$
$$ \Rightarrow \sqrt 3 a\left( { - 2\sin \left( {{{\alpha + \beta } \over 2}} \right)} \right)\sin \left( {{{\alpha - \beta } \over 2}} \right) + 2b\left( {2\cos \left( {{{\alpha + \beta } \over 2}} \right)} \right)\sin \left( {{{\alpha - \beta } \over 2}} \right) = 0$$
$$ \Rightarrow \sqrt 3 a\sin \left( {{{\alpha + \beta } \over 2}} \right) = 2b\cos \left( {{{\alpha + \beta } \over 2}} \right)$$
$$ \Rightarrow \tan \left( {{{\alpha + \beta } \over 2}} \right) = {{2b} \over {\sqrt 3 a}}$$
$$ \Rightarrow \tan \left( {{\pi \over 6}} \right) = {{2b} \over {\sqrt 3 a}}$$ [$$ \because $$ $$\alpha $$ + $$\beta $$ = $${\pi \over 3}$$, given]
$$ \Rightarrow {1 \over {\sqrt 3 }} = {{2b} \over {\sqrt 3 a}} \Rightarrow {b \over a} = {1 \over 2}$$
$$ \Rightarrow {b \over a} = 0.5$$
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