Question

The graph of a polynomial $f(x)$ of degree 3 is as shown in the figure and slope of tangent at $Q(0,5)$ is 3 .

The equation of normal at the point where curve crosses $y$-axis, is

• A. $3x+y=15$
• B. $x+3y=15$
• C. $x+3y=5$
• D. $3x+y=5$

Answer 1

Answer: (B)

Let $f(x)=a{x}^3+b{x}^2+cx+d$
$\therefore f(0)=5\Rightarrow d=5$
So, $f(x)=a{x}^3+b{x}^2+cx+5$
$\therefore f^{\prime }(x)=3a{x}^2+2bx+c$
Now, $f^{\prime }(-2)=0\Rightarrow 12a-4b+c=0$....(1)
and $y=f(x)$ passes through $P(-2,0)$, so $0=-8a+4b-2c+5$...(2)
Also, $f^{\prime }(0)=3\Rightarrow c=3$
$\therefore$ On solving, we get $a=\frac{-1}{2},b=\frac{-3}{4}$
$\Rightarrow f(x)=\frac{-1}{2}{x}^3-\frac{3}{4}{x}^2+3x+5$
Equation of normal at $Q(0,5)$ is
$(y-5)=\frac{-1}{3}(x-0)\Rightarrow x+3y=15$