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AP EAMCET Engineering 2012 Solved Paper

Section: Mathematics

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Question : 128 of 160

Marks: +1, -0

In

\Delta PQR, \angle R = \frac{\pi}{4}, \tan\left(\frac{P}{3}\right), \tan\left(\frac{Q}{3}\right)

are the roots of the equation

a x^{2}+b x+c=0

, then

$a+b=c$
$b+c=0$
$a+c=0$
$b=c$

Solution:

Given,

R=\frac{\pi}{4}

Also,

P+Q+R=\pi

⇒

P+Q+\frac{\pi}{4}=\pi

⇒

P+Q=\frac{3\pi}{4}

⇒

\frac{P}{3}+\frac{Q}{3}=\frac{\pi}{4}

⇒

\tan\left(\frac{P}{3}+\frac{Q}{3}\right)=\tan\left(\frac{\pi}{4}\right)

⇒

\frac{\tan\frac{P}{3}+\tan\frac{Q}{3}}{1-\tan\frac{P}{3}\tan\frac{Q}{3}}=1

......(i)

Since,

\tan\frac{P}{3}

and

\tan\frac{Q}{3}

are the roots of the equation

a x^{2}+b x+c=0

\therefore\tan\frac{P}{3}+\tan\frac{Q}{3}=-\frac{b}{a}

and

\tan\frac{P}{3}\cdot\tan\frac{Q}{3}=\frac{c}{a}

\therefore

From Eq. (i),

\frac{-\frac{b}{a}}{1-\frac{c}{a}}=1

⇒

\frac{-b}{a-c}=1

⇒

a+b=c

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