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Comprehension Based Questions: Applications of Derivatives - Free MCQ Practice


MCQ Practice Test & Solutions: Test: Comprehension Based Questions: Applications of Derivatives (7 Questions)

You can prepare effectively for JEE Crack JEE with 35 Years of Previous Year Solved Papers with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Comprehension Based Questions: Applications of Derivatives". These 7 questions have been designed by the experts with the latest curriculum of JEE 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 10 minutes
  • - Number of Questions: 7

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Test: Comprehension Based Questions: Applications of Derivatives - Question 1

PASSAGE - 1

If a continuous function f defined on the real line R, assumes positive and negative values in R then the equation f(x) = 0 has a root in R. For example, if it is known that a continuous function f on R is positive at some point and its minimum value is negative then the equation f(x) = 0 has a root in R.

Consider f(x) = kex – x for all real x where k is a real constant.

Q. The line y = x meets y = kex for k < 0 at

Detailed Solution: Question 1

For k = 0, line y = x meets y = 0, i.e., x-axis only at one point.
For k < 0, y = kex meets y = x only once as shown in the graph.

Test: Comprehension Based Questions: Applications of Derivatives - Question 2

PASSAGE - 1

If a continuous function f defined on the real line R, assumes positive and negative values in R then the equation f(x) = 0 has a root in R. For example, if it is known that a continuous function f on R is positive at some point and its minimum value is negative then the equation f(x) = 0 has a root in R.

Consider f(x) = kex – x for all real x where k is a real constant.

Q. The positive value of k for which kex – x = 0 has only one root is

Detailed Solution: Question 2

Let f (x) = kex – x

Now for f (x) = 0 to have only one root means the line y = x must be tangential to the curve y = kex.
Let it be so at (x1, y1) then

Test: Comprehension Based Questions: Applications of Derivatives - Question 3

PASSAGE - 1

If a continuous function f defined on the real line R, assumes positive and negative values in R then the equation f(x) = 0 has a root in R. For example, if it is known that a continuous function f on R is positive at some point and its minimum value is negative then the equation f(x) = 0 has a root in R.

Consider f(x) = kex – x for all real x where k is a real constant.

Q. For k > 0, the set of all values of k for which kex – x = 0 has two distinct roots is

Detailed Solution: Question 3

∵ For y = x to be tangent to the curve y = kex, k = 1/e
∴ For y = kex to meet y = x at two points we should 

Test: Comprehension Based Questions: Applications of Derivatives - Question 4

PASSAGE - 2

Let f (x) = (1 – x)2 sin2x + x2 for all x ∈ I R and let 

Q.

Consider the statements:

P : Th er e exists some x ∈ R such th at f (x) + 2 x =  2(1 + x2)

Q : There exists some x ∈ R such that 2 f (x) + 1 = 2x (1 + x) Then

Detailed Solution: Question 4

For the statement P
f(x) + 2x = 2 (1 + x2)
⇒ (1 – x)2 sin2x + x2 + 2x = 2(1+ x2)
⇒ (1 – x)2 sin2x = x2 – 2x + 1 + 1
⇒ (1 – x)2 sin2x = (1 – x)2 + 1
⇒ (1 – x)2 cos2x = –1

Which is not possible for any real value of x.
∴ P is not true.
Also let H(x) = 2f(x) + 1 – 2x (1 + x)
H(0) = 2f(0) + 1 – 0 = 1
and H(1) = 2f(1) + 1 – 4 = – 3
⇒ H(x) has a solution in (0, 1)
∴ Q is true.

Test: Comprehension Based Questions: Applications of Derivatives - Question 5

PASSAGE - 2

Let f (x) = (1 – x)2 sin2x + x2 for all x ∈ I R and let 

Q. Which of the following is true?

Detailed Solution: Question 5



∴ h(x) is decreasing function.

Test: Comprehension Based Questions: Applications of Derivatives - Question 6

PASSAGE - 3

(the set of all real numbers) be a function. Suppose the function f is twice differentiable, f(0) = f(1) = 0 and satisfies 

Q. Which of the following is true for 0 < x < 1?

Detailed Solution: Question 6

We have f ''(x) – 2f '(x) +f(x) > ex


Let  g(x) = e–xf(x)
Then we have g ''(x) > 1> 0
So g is concave upward.
Also g(0) = g(1) = 0

Test: Comprehension Based Questions: Applications of Derivatives - Question 7

PASSAGE - 3

(the set of all real numbers) be a function. Suppose the function f is twice differentiable, f(0) = f(1) = 0 and satisfies 

Q. If the function e–x f(x) assumes its minimum in the interval   which of the following is true? 

Detailed Solution: Question 7

g(x) = e–xf(x)
⇒ g'(x) = e–xf '(x) – e–xf(x) = e–x (f '(x) – f(x))
As = 1/4 is point of local minima in [0, 1]

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