Question

The curve y(x) = ax³ + bx² + 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

The Correct Option is A

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)