Question

For $a , b \in R -\{0\}$, let $f ( x )= ax ^2+ bx + a$ satisfies $f \left( x +\frac{7}{4}\right)= f \left(\frac{7}{4}- x \right) \forall x \in R$. Also the equation $f(x)=7 x+a$ has only one real and distinct solution.
The value of $( a + b )$ is equal to

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

Answer: (B)

Since $f \left( x +\frac{7}{4}\right)= f \left(\frac{7}{4}- x \right) \forall x \in R \Rightarrow f ( x )$ is symmetric about $x =\frac{7}{4}$
Hence $\frac{- b }{2 a }=\frac{7}{4} \Rightarrow \frac{- b }{ a }=\frac{7}{2}$....(1)
Also $f(x)=7 x+a$ has only one real solution, so $a x^2+b x+a=7 x+a \Rightarrow a x^2+x(b-7)=0$ has discriminant zero.
$\Rightarrow (b-7)^2-4(a)(0)=0 \Rightarrow b=7$
Put $ b=7$ in equation (1), we get $a=-2$
So, $ f(x)=-2 x^2+7 x-2$
Hence $(a+b)=(-2+7)=5$