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

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