1. If both the critical points of f(x) = ax^3 + bx^2 + cx + d are −ve, then

Question

While drawing the graph of y = f(x), f(x) is polynomial function, coefficient of highest degree term dominates, for example, if f(x) = ax² + bx + c then if a is +ve, its graph will be upward parabola, otherwise it will be downward. In the same way if f(x) = ax³ + bx2 + cx + d, the graph of f(x) will be either

as a is −ve or +ve, respectively. The point where f(x) cuts the x-axis is called its roots. The maximum number of +ve real roots which f(x) = 0 can have, are equal to the change of sign in the expression f(x). The maximum number of −ve real roots which f(x) = 0 can have, are equal to the change of sign in the expression f(−x). For example, f(x) = x³ − x + 1 have two change of sign this implies f(x) = 0 can have maximum of two positive real roots.

1. If both the critical points of f(x) = ax³ + bx² + cx + d are −ve, then 

(A) ac < 0

(B) bc > 0 

(C) ab < 0

(D) None of these

2. The least number of imaginary roots of f(x) = ax⁶ + bx⁴ - cx² + dx - c, where a, b, c and d are same as in Question 1. 

(A)  Four 

(B)  Two 

(C)  Six 

(D)  0

3 . If critical points are c₁ and c₂ and f(c₁) f(c₂) < 0, then f(x) = 0 have (here f(x) is same as in Question 1) 

(A)  Only one real root which is positive

(B) Three real roots in which at least two are negative 

(C) Only one real root which is negative

(D)  None of these

Answer 1

Correct option  1. (B) 2. (B) 3. (B)

1. f′(x) = 3ax² + 2bx + c 

As both the roots are negative, we have

-2b/3a < 0 and < 0

⇒ a and b have same signs.

⇒ a and c have different signs.

⇒ bc < 0

2.  Case I: a is positive

Case I(a): d > 0

f(x) have 3 signs of change and f(−x) have 1 sign of change.

Case I(b): d < 0

f(x) have 1 sign of changes and f(−x) have 3 signs of change.

Case II: a < 0

Case II(a): d > 0 

f(x) have 1 sign of changes and f(−x) have 3 signs of change.

Case II(b): d < 0

f(x) have 3 sign of change and f(−x) have 1 signs of change.

So,

⇒ Maximum number of real roots = 4

⇒ Least number of imaginary root = 2

3.  There are only two cases Fig.

Here, C₁ and C₂ are negative. Hence, a and b are negative.