Application of Derivatives - Online Test

Q1.

The function f (x) =  strictly increases on


Answer : Option A
Explaination / Solution:

Here , 



i.e. if x(x – 2 )<0 i.e. if 0 < x < 2. Hence f is strictly increasing on [0,2]

Q2.

At (0, 0) the curve 


Answer : Option D
Explaination / Solution:

 and hence the tangent to the curve at ( 0 , 0 ) makes an angle of  with +ve X-axis.

Q3. Let f (x) = x – cos x, x∈R, then f is
Answer : Option D
Explaination / Solution:



Hence an increasing function.


Q4. In  the function f (x) =  is
Answer : Option D
Explaination / Solution:


Hence an increasing function.


Q5. Let f (x) = tan x – 4x, then in the interval is
Answer : Option C
Explaination / Solution:




Q6.

Let f (x) =  then f (x) is strictly decreasing in


Answer : Option D
Explaination / Solution:




Hence , f is strictly increasing on [1,3].

Q7. Let f (x) =  then f (x) is
Answer : Option C
Explaination / Solution:




Hence an increasing function.


Q8. Let f (x) – log(1 + x), where x > 0, then f is
Answer : Option C
Explaination / Solution:


Hence decreasing function.


Q9. Let f (x) =  – 4x, then
Answer : Option B
Explaination / Solution:

f (x) = x4 – 4x

f'(x) = 4(x3) - 4 = 4(x 3- 1) = 4 {(x2) + x + 1)} (x-1) 
= f ‘(x) > 0,if,(x - 1) > 0,
and f ‘(x) <0,  if (x -1 ) < 0.
So,f is decreasing on ( - ,1]

and f is increasing on [1,)


Q10. Let g (x) be continuous in a neighbourhood of ‘a’ and g (a) ≠ 0. Let f be a function such that f ‘ (x) = g(x)  then
Answer : Option A
Explaination / Solution:

Since g is continuous at a , therefore , if g ( a ) > 0 , then there is a neighbourhood of a, say ( a-e , a+ e ) in which g ( x ) is positive .This means that f ‘ (x)>0 in this nhd of a and hence    f ( x ) is increasing at a.