Application of Derivatives - Online Test
Q1. The instantaneous rate of change at t = 1 for the function f (t) = is
Answer : Option C
Explaination / Solution:
⇒f′(1)=−e−1+e−1=0
Q2. If the graph of a differentiable function y = f (x) meets the lines y = – 1 and y = 1, then the graph
Answer : Option C
Explaination / Solution:
Since the graph cuts the lines y = -1 and y = 1 , therefore ,it must cut y = 0 atleast once as the graph is a continuous curve in this case.
Q3. Let f be a real valued function defined on (0, 1) ∪(2, 4) such that f ‘ (x) = 0 for every x, then
Answer : Option D
Explaination / Solution:
f ‘ (x) =0 f (x) is constant in ( 0 , 1 ) and also in ( 2, 4 ). But this does not mean that f ( x) has the same value in both the intervals . However , if f ( c ) =f ( d ) , where c ( 0 , 1 ) and d ( 2, 4) then f ( x ) assumes the same value at all x ( 0 ,1 ) U (2, 4 ) and hence f is a constant function.
Q4. In case of strict decreasing functions, slope of tangent and hence derivative is
Answer : Option B
Explaination / Solution:
In case of a strict decreasing function , slope of tangent and hence derivative is either negative or zero because decreasing function change sign from positive to negative and making obtuse angle measured clockwise.Hence negative and will be zero at peak.
Q5. The function f (x) = 2 – 3 x is
Answer : Option C
Explaination / Solution:
f (x) = 2 – 3x f ‘ (x) = - 3 < 0 for all x R . so, f is strictly decreasing function.
Q6. The function f (x) = – 2 x is increasing in the interval
Answer : Option A
Explaination / Solution:
f(x) = x2- 2x
f'(x) = 2x - 2 = 2(x - 1)
So , f( x) is increasing if 2(x-1) 0 , i.e.if x 1
f'(x) = 2x - 2 = 2(x - 1)
So , f( x) is increasing if 2(x-1) 0 , i.e.if x 1
Q7.
The function f (x) = , for all real x, is
Answer : Option A
Explaination / Solution:
f(x) = x2
f'(x) = 2x for all x in R.
Since f ‘(x) = 2x > 0 for x >0, and f ‘ (x) = 2x< 0 for x < 0 ,therefore on R , f is neither increasing nor decreasing . Infact , f is strict increasing on [ 0 , ) and strict decreasing on (- ].
f'(x) = 2x for all x in R.
Since f ‘(x) = 2x > 0 for x >0, and f ‘ (x) = 2x< 0 for x < 0 ,therefore on R , f is neither increasing nor decreasing . Infact , f is strict increasing on [ 0 , ) and strict decreasing on (- ].
Q8. The function f (x) = m x + c where m, c are constants, is a strict decreasing function for all x∈R , if
Answer : Option D
Explaination / Solution:
f ( x) = mx + c is strict decreasing on R
if f ‘ ( x) < 0 i.e. if m < 0 .
Q9. The function f(x) = is
Answer : Option A
Explaination / Solution:
f′(x)=ddx(tan−1x)
Therefore , f is strictly increasing on R
Q10. Let f (x) = then f (x) is decreasing in
Answer : Option B
Explaination / Solution:
Here ,
