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

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Aicharya · Accepted Answer

$\begin{aligned} & y=\mathrm{a} x^3+\mathrm{b} x^2+\mathrm{c} x+5 \text { touches X-axis at } (-2,0) \\ & \Rightarrow 0=-8 \mathrm{a}+4 \mathrm{~b}-2 \mathrm{c}+5 \\ & \Rightarrow 8 \mathrm{a}-4 \mathrm{~b}+2 \mathrm{c}=5 \end{aligned}$
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 $a=\frac{-1}{2}, b=\frac{-3}{4}$ $\begin{aligned} \therefore \quad a+b+c & =\frac{-1}{2}+\frac{-3}{4}+3 \\ & =\frac{-2-3+12}{4} \\ & =\frac{7}{4} \end{aligned}$

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…

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