Question

The curve $y(x)=a{x}^3+b{x}^2+cx+5$ touches the $x$-axis at the point $P(-2,0)$ and cuts the $y$-axis at the point $Q$, where $y$ ' is equal to 3 . Then the local maximum value of $y(x)$ is :

• A. $\frac{27}{4}$
• B. $\frac{29}{4}$
• C. $\frac{37}{4}$
• D. $\frac{9}{2}$

Answer 1

Answer: (A)

$y(x)=a{x}^3+b{x}^2+cx+5$ is passing through $(-2,0)$ then $8a-4b+2c=5\dots \dots$ (1)
$y^{\prime }(x)=3a{x}^2+2bx+c$ touches $x$-axis at $(-2,0)$
$12a-4b+c=\mathrm{0.....}$(2)
again, for $x=0,y^{\prime }(x)=3$
$c=\mathrm{3.....}$(3)
Solving eq. (1), (2) \& (3) $a=-\frac{1}{2},b=-\frac{3}{4}$
$y^{\prime }(x)=-\frac{3}{2}{x}^2-\frac{3}{2}x+3$
$y(x)$ has local maxima at $x=1$
$y(1)=\frac{27}{4}$