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…

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
  1. $3,-\frac{1}{2},-\frac{3}{4}$
  2. $-\frac{3}{4},-\frac{1}{2}, 3$
  3. $-\frac{1}{2},-\frac{3}{4}, 3$
  4. $-\frac{1}{2}, 3,-\frac{3}{4}$

Solution

$\begin{aligned} & y=a x^3+b x^2+c x+5 \text { touches X-axis at }(-2,0) \\ & \Rightarrow 0=-8 a+4 b-2 c+5 \\ & \Rightarrow 8 \mathrm{a}-4 b+2 c=5...(i) \end{aligned}$
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

Asked in: MHT CET 2024 (11 May Shift 2)

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