Test: Application of Derivatives - 1 - JEE MCQ

# Test: Application of Derivatives - 1 - JEE MCQ

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## 26 Questions MCQ Test - Test: Application of Derivatives - 1

Test: Application of Derivatives - 1 for JEE 2024 is part of JEE preparation. The Test: Application of Derivatives - 1 questions and answers have been prepared according to the JEE exam syllabus.The Test: Application of Derivatives - 1 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Application of Derivatives - 1 below.
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Test: Application of Derivatives - 1 - Question 1

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

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Test: Application of Derivatives - 1 - Question 2
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Test: Application of Derivatives - 1 - Question 3

### The equation of the tangent to the curve y = e2x at the point (0, 1) is

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Hence equation of tangent to the given curve at (0 , 1) is :

(y−1) = 2(x−0),i.e..y−1 = 2x

Test: Application of Derivatives - 1 - Question 4

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

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x3 - 18x2 + 96x = x(x2 - 18x + 96) = x[(x-9)2 + 15]
=x(x-9)+15 ≥ 0

Test: Application of Derivatives - 1 - Question 5

Let f (x) = (x2−4)1/3, then f has a

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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 .

Test: Application of Derivatives - 1 - Question 6

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

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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.

Test: Application of Derivatives - 1 - Question 7

Let f(x)  =  where x>0, then f is

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Test: Application of Derivatives - 1 - Question 8

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

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, 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

Test: Application of Derivatives - 1 - Question 9

Given that f (x) = x1/x , x>0, has the maximum value at x = e,then

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Test: Application of Derivatives - 1 - Question 10

The function f (x) = x3 has a

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

Test: Application of Derivatives - 1 - Question 11

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

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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.

Test: Application of Derivatives - 1 - Question 12

Let f (x) = x4 – 4x, then

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Test: Application of Derivatives - 1 - Question 13

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

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Test: Application of Derivatives - 1 - Question 14

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has a local maximum value = - 2 at x = - 1 and a local minimum value = 2 at x = 1.

Test: Application of Derivatives - 1 - Question 15

The minimum value of (x) = sin x cos x is

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Sinx cosx = 1/2 (sin2x) and minimum value of sin2x is – 1 .

Test: Application of Derivatives - 1 - Question 16

In case of strict decreasing functions, slope of tangent and hence derivative is

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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.

Test: Application of Derivatives - 1 - Question 17

Let f (x) = x3−6x2+9x+18, then f (x) is strict decreasing in

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Test: Application of Derivatives - 1 - Question 18

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

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therefore , slopes of the tangents at (1,1) and (- 1 , -1)are equal. Hence, the two tangents in reference are parallel.

Test: Application of Derivatives - 1 - Question 19

Let f(x) = x25(1−x)75 for all x ∈ [0,1], then f (x) assumes its maximum value at

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Test: Application of Derivatives - 1 - Question 20

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

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Test: Application of Derivatives - 1 - Question 21

The function f (x) = 2 – 3 x is

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

Test: Application of Derivatives - 1 - Question 22

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

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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.

Test: Application of Derivatives - 1 - Question 23

Minimum value of the function f(x) = x2+x+1 is

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Test: Application of Derivatives - 1 - Question 24

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

Detailed Solution for Test: Application of Derivatives - 1 - Question 24

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

Test: Application of Derivatives - 1 - Question 25

The function f (x) = | x | has

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The modulus function has V-shaped graph,which means that it has only one minima.

Test: Application of Derivatives - 1 - Question 26

At which point the line x/a + y/b = 1,  touches the curve  y = be-x/a

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