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TS EAMCET 2017 Code A Solved Paper

Section: Mathematics

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Question : 36 of 160

Marks: +1, -0

Let

f(x)

be quadratic expression such that

f(0)+f(1)=0

f(-2)=0

, then

$f\left(\frac{-2}{5}\right)=0$
$f\left(\frac{2}{5}\right)=0$
$f\left(\frac{-3}{5}\right)=0$
$f\left(\frac{3}{5}\right)=0$

Solution:

Consider the function is,

f(x)=a x^{2}+b x+c

From the given condition,

f(0)+f(1)=0

c+a+b+c=0

a+b+2c=0\ldots\ldots(1)

Again,

f(-2)=0

a(-2)^{2}+b(-2)+c=0

4a-2b+c=0

4a-2b+c=0\ldots\ldots(2)

Solving equation (1) and (2), we get

\frac{a}{5}=\frac{b}{7}=\frac{c}{-6}=k

a=5k, b=7k, c=-6k

The function becomes

f(x)=k(5x^{2}+7x-6)

5x^{2}+7x-6=0

(x-2)(5x-3)=0

x=-2, \frac{3}{5}

Thus, the value of the function,

f\left(\frac{3}{5}\right)=0

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