Question

The curve y = ax³ + bx² + cx + 8 touches x- axis at P(–2, 0) and cuts the y-axis at a point Q where its gradient is 3. The values of a, b, c are respectively

• A. – $\frac{1}{2}$, –$\frac{3}{4}$, 3
• B. 3, – $\frac{1}{2}$, – 4
• C. –$\frac{1}{2}$, –$\frac{7}{4}$, 2
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

$\frac{dy}{dx}$= 3ax² + 2bx + c

Since the curve touches x- axis at (–2, 0) so

$\left. \ \frac{dy}{dx} \right|_{( - 2,0)}$= 0

⇒ 12a – 4b + c = 0 ... (i)

The curve cut the y- axis at (0, 8) so

$\left. \ \frac{dy}{dx} \right|_{(0,8)}$= 3

⇒ c = 3

Also the curve passes through (–2, 0) so

0 = –8a + 4b –2c + 8 ⇒ –8a + 4b –2 = 0 (ii)

Solving (i) and (ii) a = – $\frac{1}{4}$, b = 0