Applications of Derivatives Test

Result

Applications of Derivatives Test - 3

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Question 1
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Let f be a real valued function defined on (0, 1) ∪(2, 4) such that f ‘(x) = 0 for every x, then

A
f is constant function if f 1/2 = f (3)

B
f is a constant function

C
f is a constant function if f 1/2 = 0

D
f is not a constant function

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.
Question 2
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Let f (x) = x⁴ –4x, then

A
f is increasing in [−∞,1)

B
f is decreasing in [1,∞)

C
f is increasing in [1,∞)

D
none of these.

Solution
Question 3
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The slope of the tangent to the curve x = a sin t, y = a at the point ‘t ’is

A
cot t

B
tan t

C
tan t/2

D
none of these.

Solution
Question 4
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A
relative maximum >relative minimum

B
relative minimum >relative maximum.

C
relative minimum does not exist

D
relative maximum does not exist

Solution

has a local maximum value = - 2 at x = - 1 and a local minimum value = 2 at x = 1.
Question 5
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The minimum value of (x) = sin x cos x is

A
-1/2

B
0

C
1/2

D
none of these.

Solution

Sinx cosx = 1/2 (sin2x) and minimum value of sin2x is –1 .
Question 6
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In case of strict decreasing functions, slope of tangent and hence derivative is

A
Negative

B
either negative or zero.

C
Zero

D
Positive

Solution

In the case of strictly decreasing functions, the slope of the tangent line is always negative. This implies that the derivative of a strictly decreasing function is always negative. The derivative represents the rate of change of the function, and if the function is strictly decreasing, the rate of change is negative.
Question 7
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Let f (x) = x³ −6x² +9x+18, then f (x) is strict decreasing in

A
[3,∞)

B
[−∞,1)

C
[1, 3]

D
(−∞,1]

Solution
Question 8
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Tangents to the curve y = x³ at the points (1, 1) and (–1, –1) are

A
parallel

B
intersecting but not at right angles

C
perpendicular

D
none of these

Solution

therefore , slopes of the tangents at (1,1) and (- 1 , -1)are equal. Hence, the two tangents in reference are parallel.
Question 9
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Let f(x) = x²⁵ (1 −x)⁷⁵ for all x ∈[0,1], then f (x) assumes its maximum value at

A
0

B
1/4

C
1/3

D
1/2

Solution
Question 10
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The stone projected vertically upwards moves under the action of gravity alone and its motion is described by x = 49 t –4.9 t² . It is at a maximum height when

A
t = 10

B
t = 5

C
t = 0

D
none of these

Solution
Question 11
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The function f (x) = 2 –3 x is

A
increasing

B
decreasing

C
neither decreasing nor increasing

D
none of these

Solution

f (x) = 2 –3x ⇒f ‘(x) = - 3 <0 for all x ∈ R . so, f is strictly decreasing function.
Question 12
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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) (x −a)² , then

A
f is decreasing at a if g (a) >

B
f is increasing at a if g (a) <0

C
f is increasing at a if g (a) >0

D
none of these

Solution

Since g is continuous at a , therefore , if g (a) >0 , then there is a nhd.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.
Question 13
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Minimum value of the function f(x) = x² +x+1 is

A
3

B

C
1

D
none of these

Solution
Question 14
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If the line y=x is a tangent to the parabola y=ax2+bx+c at the point (1,1) and the curve passes through (−1,0), then

Solution

The correct option is C
a=c=1/4, b=1/2
y=x is a tangent
∴slopes are equal.
dy/dx=2ax+b
⇒1=2a+b at (1,1)⋯(1)
Also, the parabola passes through (1,1)
⇒a+b+c=1 ⋯(2)
The parabola passes through (−1,0)
⇒0=a −b+c ⋯(3)
Solving (1),(2),(3), we get -
∴a=c=1/4
and b=1/2
Question 15
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The function f (x) = | x | has

A
only one minima

B
no maxima or minima

C
only one maxima

D
none of these

Solution

The modulus function has V-shaped graph,which means that it has only one minima.
Question 16
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At which point the line x/a + y/b = 1, touches the curve y = be-x/a

A
(0, 0)

B
(0, a)

C
(0, b)

D
(b, 0)

Solution