Application of Derivatives 1 Question 18
####18. The curve $y=a x^{3}+b x^{2}+c x+5$, touches the $X$-axis at $P(-2,0)$ and cuts the $Y$-axis at a point $Q$, where its gradient is 3 . Find $a, b, c$.
(1994, 5M)
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Solution:
- Given, $y=a x^{3}+b x^{2}+c x+5$ touches $X$-axis at $P(-2,0)$ which implies that $X$-axis is tangent at $(-2,0)$ and the curve is also passes through $(-2,0)$.
The curve cuts $Y$-axis at $(0,5)$ and gradient at this point is given 3 , therefore at $(0,5)$ slope of the tangent is 3.
Now,
$ \frac{d y}{d x}=3 a x^{2}+2 b x+c $
Since, $X$-axis is tangent at $(-2,0)$.
$ \begin{aligned} & \therefore \quad \frac{d y}{d x}{ }_{x=-2}=0 \\ & \Rightarrow \quad 0=3 a(-2)^{2}+2 b(-2)+c \\ & \Rightarrow \quad 0=12 a-4 b+c \end{aligned} $
Again, slope of tangent at $(0,5)$ is 3 .
$ \begin{aligned} & \therefore \quad\left|\frac{d y}{d x}\right|_{(0,5)}=3 \\ & \Rightarrow \quad 3=3 a(0)^{2}+2 b(0)+c \\ & \Rightarrow \quad 3=c \end{aligned} $
Since, the curve passes through $(-2,0)$.
$ \begin{array}{rlrl} & & 0 & =a(-2)^{3}+b(-2)^{2}+c(-2)+5 \\ \Rightarrow & 0 & =-8 a+4 b-2 c+5 \end{array} $
From Eqs. (i) and (ii),
$ 12 a-4 b=-3 $
From Eqs. (ii) and (iii),
$ -8 a+4 b=1 $
On adding Eqs. (iv) and (v), we get
$ 4 a=-2 \Rightarrow a=-1 / 2 $
On putting $a=-1 / 2$ in Eq. (iv), we get
$ \begin{array}{rlrl} & & 12(-1 / 2)-4 b & =-3 \\ \Rightarrow & -6-4 b & =-3 \\ \Rightarrow & & -3 & =4 b \\ \Rightarrow & & b & =-3 / 4 \\ & \therefore & a=-1 / 2, b & =-3 / 4 \text { and } c=3 \end{array} $
