Question

For $a, b \in R-\{0\}$, let $f(x)=a x^2+b x+a$ 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 $\frac{- b }{2 a }=\frac{7}{4} \Rightarrow \frac{- b }{ a }=\frac{7}{2}$
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 $( a + b )=(-2+7)=5$