Question

The curve y(x) = ax 3 + bx 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)

$\frac{27}{4}$

Answer 2

f(x) = y = ax3 + bx2 + cx + 5 …(i)
$\frac{dy}{dx}$=3ax2+2bx+c……(ii)
Touches x -axis at P(–2, 0)
⇒Y|x=−2=0⇒−8a+4b−2c+5=0…(iii)
Touches x -axis at P(–2, 0) also implies
$\frac{dy}{dx}$|x=−2=0⇒12a−4b+c=0…(iv)
y = f(x) cuts the y -axis at (0, 5)
Given,
$\frac{dy}{dx}$|x=0=c=3…(v)
From (iii), (iv) and (v)
a=−$\frac{1}{2}$,b=−$\frac{3}{4}$,c=3
⇒f(x)=−$\frac{x^2}{2}$−$\frac{3}{4}$x2+3x+5
f′(x)=−$\frac{3}{2}$x2−$\frac{3}{4}$x+3
=−$\frac{3}{2}$(x+2)(x−1)
f ′(x) = 0 at x = –2 and x = 1
By first derivative test x = 1 in point of local maximum
Hence local maximum value of f(x) is f(1)