The curve \(y=\mathrm{a} x^3+\mathrm{b} x^2+\mathrm{c} x+5\) touches the x -axis at \((-2,0)\) and cuts the \(y\)-axis at a point Q where its gradient is 3 , then the value of \(\mathrm{a}+\mathrm{b}+\mathrm{c}\) is

\(\begin{aligned}
& y=\mathrm{a} x^3+\mathrm{b} x^2+\mathrm{c} x+5 \text { touches X-axis at } (-2,0) \\
& \Rightarrow 0=-8 \mathrm{a}+4 \mathrm{~b}-2 \mathrm{c}+5 \\
& \Rightarrow 8 \mathrm{a}-4 \mathrm{~b}+2 \mathrm{c}=5
\end{aligned}\)
Also, it cuts Y -axis at a point Q
\(\therefore \quad\) Put \(x=0\) in the equation of curve, we get
\(\begin{aligned}
& y=5 \\
& y=\mathrm{a} x^3+b x^2+\mathrm{c} x+5 \\
\therefore \quad & \frac{\mathrm{~d} y}{\mathrm{~d} x}=3 \mathrm{a} x^2+2 \mathrm{~b} x+\mathrm{c} \\
& \left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)_{\mathrm{Q}(0,5)}=3 \\
& \Rightarrow 3 \mathrm{a}(0)^2+2 \mathrm{~b}(0)+\mathrm{c}=3 \\
& \Rightarrow \mathrm{c}=3
\end{aligned}\)
Equation (i) becomes,
\(\begin{aligned}
& 8 a-4 b+6=5 \\
& \Rightarrow 8 a-4 b+1=0...(ii)
\end{aligned}\)
Also, X -axis is tangent to curve
\(\begin{aligned}
& \left(\frac{d y}{d x}\right)_{x=-2}=0 \\
& 3 a(-2)^2+2 b(-2)+3=0 \\
& \Rightarrow 12 a-4 b+3=0...(iii)
\end{aligned}\)
Solving (ii), (iii) we get
\(a=\frac{-1}{2}, b=\frac{-3}{4}\)
\(\begin{aligned}
\therefore \quad a+b+c & =\frac{-1}{2}+\frac{-3}{4}+3 \\
& =\frac{-2-3+12}{4} \\
& =\frac{7}{4}
\end{aligned}\)