Question

For $a,b\in R-\{0\}$, let $f(x)=a{x}^2+bx+a$ satisfies $f\left(x+\frac{7}{4}\right)=f\left(\frac{7}{4}-x\right)\forall x\in R$. Also the equation $f(x)=7x+a$ 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 $\frac{-b}{2a}=\frac{7}{4}\Rightarrow \frac{-b}{a}=\frac{7}{2}$
Also $f(x)=7x+a$ has only one real solution, so
$a{x}^2+bx+a=7x+a\Rightarrow a{x}^2+x(b-7)=0$ 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, $f(x)=-2x^2+7x-2$
Hence $(a+b)=(-2+7)=5$