QJEEAdv18Ma79f7: JEE Advanced 2018 Mathematics - Class 11 JEEAdv18Ma79f7 Solution: Let a, b, c three non-zero real numbers such that the equation…

Question

Verified Solution • 2025–26 Session moderate

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 ............

Accepted Answer

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

Step-by-Step Verified Solution