Applications of Derivatives Test

Result

Applications of Derivatives Test - 2

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Weekly Quiz Competition

Question 1
1 / -0.25

The instantaneous rate of change at t = 1 for the function f (t) = te^−t + 9 is

A
2

B
0

C
9

D
–1

Solution
Question 2
1 / -0.25

A
an odd function

B
a decreasing function.

C
an increasing function

D
an even function

Solution
Question 3
1 / -0.25

The equation of the tangent to the curve y = e^2x at the point (0, 1) is

A
1 –y = 2 x

B
y –1 = 2 x

C
y + 1 = 2 x

D
none of these

Solution

Hence equation of tangent to the given curve at (0 , 1) is :
(y −1) = 2(x −0),i.e..y −1 = 2x
Question 4
1 / -0.25

The smallest value of the polynomial x³ −18x² +96 in the interval [0, 9] is

A
135

B
0

C
126

D
160

Solution

x³ - 18x² + 96x = x(x² - 18x + 96) = x[(x-9)² + 15]
=x(x-9)² +15 ≥0
Question 5
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Let f (x) = (x² −4)^1/3 , then f has a

A
local minima at x = 0

B
local maxima at x = 0

C
point of inflextion at x = 0

D
none of these

Solution

Also, for x <0 (slightly) , f ‘(x) <0 and for x >0 (slightly) f ‘(x) >0. Hence f has a local minima at x = 0 .
Question 6
1 / -0.25

If the graph of a differentiable function y = f (x) meets the lines y = –1 and y = 1, then the graph

A
meets the line y = 0 at least thrice

B
meets the line y = 0 at least once

C
meets the line y = 0 at least twice

D
does not meet the line y = 0.

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.
Question 7
1 / -0.25

Let f(x) = where x >0, then f is

A
an increasing function

B
neither increasing nor decreasing

C
a decreasing function

D
none of these

Solution
Question 8
1 / -0.25

The equation of the tangent to the curve y=(4 −x² )^2/3 at x = 2 is

A
x = 2

B
x = –2

C
y = –1.

D
y = 2

Solution

, which does not exist at x = 2 . However , we find that , at x = 2 . Hence , there is a vertical tangent to the given curve at x = 2 .The point on the curve corresponding to x = 2 is (2 , 0). Hence , the equation of the tangent at x = 2 is x = 2
Question 9
1 / -0.25

Given that f (x) = x^1/x , x >0, has the maximum value at x = e,then

A
e^π =π^e

B
e^π <π^e

C
e^π >π^e

D
e^π ⩽π^e

Solution
Question 10
1 / -0.25

The function f (x) = x³ has a

A
point of inflexion at = 0

B
local minima at x = 0

C
local maxima at x = 0

D
none of these

Solution

f ‘(0) = 0 , f ‘’(0) = 0 and f ‘’’(0) = 6 . So, f has a point of inflexion at 0.