Applications of Derivatives Test

Result

Applications of Derivatives Test - 5

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Weekly Quiz Competition

Question 1
1 / -0.25

The function f (x) = x² –2 x is increasing in the interval

A
x ⩾−1

B
x ⩾1

C
x ≠−1

D
x ≠1

Solution
Question 2
1 / -0.25

The function f (x) = a x + b is strict increasing for all x ∈R iff

A
a <0

B
a = 0

C
a >0

D
none of these.

Solution

Since f ‘(x) = a , therefore , f (x) is strict increasing on R iff a >0.
Question 3
1 / -0.25

Tangents to the curve x² +y² = 2at the points (1, 1) and (–1, 1)

A
at right angles

B
parallel

C
intersecting but not at right angles

D
none of these

Solution

therefore , slope of tangent at (1,1) = - 1 and the slope of tangent at (- 1 ,1) = 1. Hence , the two tangents in reference are at right angles.
Question 4
1 / -0.25

If x be real, the minimum value of x² −8x+17 is

A
–1

B
2

C
1

D
0

Solution
Question 5
1 / -0.25

If a differentiable function f (x) has a relative minimum at x = 0, then the function y = f (x) + a x + b has a relative minimum at x = 0 for

A
all b if a = 0

B
all a >0.

C
all b >0

D
all a and all b

Solution

And y has a relative minimum at x = 0 if f ‘(0)+a = 0 . a =0.
Question 6
1 / -0.25

The function f (x) = x² , for all real x, is

A
neither decreasing nor increasing

B
Decreasing

C
Increasing

D
none of these

Solution

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 (- ∞,0].
Question 7
1 / -0.25

The function f (x) = a x + b is strict increasing for all x ∈R if

A
a = 0

B
a >0

C
a <0

D
none of these.

Solution

Since f ‘(x) = a , therefore , f (x) is strict increasing on R if f a >0.
Question 8
1 / -0.25

Equation of the tangent to the curve at the point (a, b) is

A

B

C

D
none of these

Solution

Hence ,slope of tangent at (a , b) = -b/a . Therefore , the equation of tangent at (a , b ) is ;
Question 9
1 / -0.25

a log | x | + bx² + x has its extreme values at x = –1 and x = 2, then

A

B

C
a = 2, b = –1

D
a = 2, b =

Solution

= 0, at x = -1 and x = 2 i.e a + 2b = 1 and 4b + a/2 = -1. ⇒ a = 2 , b = -1/2
Question 10
1 / -0.25

Let f (x) be differentiable in (0, 4) and f (2) = f (3) and S = {c : 2

A
S has exactly one point

B
S = { }

C
S has atleast one point

D
none of these

Solution

Conditions of Rolle ’s Theorem are satisfied by f(x) in [2,3].Hence there exist atleast one real c in (2, 3) s.t. f ‘(c) = 0 . Therefore , the set S contains atleast one element