Question

If the function f (x) = | x2 + a | x | +b| has exactly three points of non-differentiability, then which of the following can be true?

Solution

Correct option is

b = 0, a < 0

 has three points of non-differentiability.

∴  f (x) is non differentiable at x = 0, x1x2  

⇒ x2 + ax + b = 0 has one root zero and other positive root. 

 b = 0 and a < 0, is one of the case.

 

 

SIMILAR QUESTIONS

Q2

In [0, 1] lagrange’s mean value theorem is not applicable to

Q3

Let f (x) satisfy the requirement of lagrange’s mean value theorem in [0, 2]. If f (0) and   

 

Q4

Let f : [2, 7] and [0, ) be a continuous and differentiable function.

Then the value of (f (7) – f (2))  is (where c Ïµ (2, 7))

Q5

The equation sin x + x cos x = 0 has at least one root in the interval

Q6

Let f (x) = ax5 + bx4 + cx3 + dx2 + ex, where abcde Ïµ R and f (x) = 0 has a positive root α, then 

Q7

Between any two real roots of the equation ex sin x – 1 = 0, the equation excos x + 1 = 0 has

Q8

f (x) is a polynomial of degree 4 with real coefficients such that f (x) = 0 is satisfied by x = 1, 2, 3 only, then f’(1). f’(2). f’(3) is equal to:

Q9

If f (x) is a polynomial of degree 5 with real coefficients such that  has 8 real roots then f (x) = 0 has:

Q10

If f (x) = loge x and g(x) = x2 and c Ïµ (4, 5), then  is equal to: