Question

If for $f\left(x\right)={\mathrm{ax}}^2+\mathrm{bx}+12,f\text{'}\left(4\right)=15\mathrm{and}f\text{'}\left(2\right)=11,$ then a + b =

Answer 1

Given that,
$f\left(x\right)={\mathrm{ax}}^2+\mathrm{bx}+12.f\text{'}\left(x\right)=\frac{d\left\{f\right(x\left)\right\}}{\mathrm{dx}}=\frac{d}{\mathrm{dx}}\{{\mathrm{ax}}^2+\mathrm{bx}+12\}\Rightarrow f\text{'}\left(x\right)=2\mathrm{ax}+bf\text{'}\left(4\right)=15\Rightarrow 8a+b=15....\left(1\right]f\text{'}\left(2\right)=11\Rightarrow 4a+b=11....\left(2\right]\mathrm{Solving}\left(1\right]\mathrm{and}\left(2\right]\mathrm{we}\mathrm{get},a=1\mathrm{and}b=7a+b=8$