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

The curve \(y=a x^3+b x^2+c x+5\) touches \(X\)-axis at \(P(-2,0)\) \(\therefore 0=\mathrm{a}(-2)^3+\mathrm{b}(-2)^2+\mathrm{c}(-2)+5\)
\(\therefore 8 a-4 b+2 c=5\quad\ldots(1)\)
\(\frac{d y}{d x}=3 a x^2+2 b x+c\) and at point \(Q\) on \(Y\) axis, we have \(\frac{d y}{d x}=3\).
Let \(Q \equiv(0, k)\)
\(\therefore 3=3 a(0)+2(0)+c \Rightarrow c=3\quad\ldots(2)\)
The equation (1) becomes \(8 a-4 b+6=5\) i.e. \(8 a-4 b=-1\)
\(\Rightarrow 2 \mathrm{a}-\mathrm{b}=\frac{-1}{4}\)
At \(\mathrm{P}(-2,0)\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=0\)
\(12 a-4 b+c=0\quad\ldots(3)\)
From (1), (2) & (3)
\(\mathrm{a}=\frac{-1}{2}, \mathrm{~b}=\frac{-3}{4}, \mathrm{c}=3\)