NCERT Solutions: Exercise 6.2 - Application of Derivative

# Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE PDF Download

Q1: Show that the function given by f(x) = 3x  17 is strictly increasing on R.
Ans: Let  be any two numbers in R.
Then, we have:
Let x1 be x2 any two numbers in R.
Then, we have:

Hence, f is strictly increasing on R.

Q2: Show that the function given by/(x) = e2x is strictly increasing on R.
Ans:
Let x1 and x2 be any twro numbers in R.
Then, we have:
Hence, f is strictly increasing on R.

Q3: Show that the function given by f(x) = sin x is
(a) strictly increasing in
(b) strictly decreasing in
(c) neither increasing nor decreasing in (0, π)

Ans: The given function is f(x) = sinx.

(a) Since for each   we have
Hence,/is strictly increasing in
(b) Since for each , we have
Hence,f is strictly decreasing in
(c) From the results obtained in (a) & (b), it is clear that f is neither increasing nor decreasing in (0, π)

Q4: Find the intervals in which the function f given by f(x) = 2x2 − 3x is
(a) strictly increasing
(b) strictly decreasing
Ans: The given function is f(x) = 2x2 - 3x.

Now. the point 3/4 divides the real line into two disjoint intervals i.e..

Hence, the given function (f) is strictly decreasing in interval
In interval
Hence, the given function (J) is strictly increasing in interval

Q5: Find the intervals in which the function/given by/x) = 2x3 - 3x2 - 36x+ 7 is
(a) strictly increasing
(b) strictly decreasing
Ans:
The given function is

The points x = -2 andx = 3 divide the real line into three disjoint intervals i.e..

In intervals (-∞, -2) and (3, ∞) , f'(x) is positive while in interval (-2,3), f'(x) is negative.
Hence, the given function (f) is strictly increasing in intervals (-∞, -2) and (3, ∞) , while function if) is strictly decreasing in interval (-2,3).

Q6: Find the intervals in which the following functions are strictly increasing or decreasing:
(a) x2 + 2x- 5
(b) 10 - 6x— 2x2
(c) -2x3 - 9x2 - 12x+ 1
(d) 6 - 9x -x2
(e) (x + 1)3 (x — 3)3
Ans: (a) We have,
f(x) = x2 +2x-5
∴ f'(x) = 2x+2
Now,
f'(x) = 0 ⇒ x = -1
Point x = -1 divides the real line into two disioint. intervals i.e.. (-∞, -1) and (-1, ∞)
In interval
∴ f is strictly decreasing in interval (-∞, -1).
Thus, f is strictly decreasing for x < -1.
In interval
∴ f is strictly increasing in interval (-1, ∞)
Thus, f is strictly deere asing for x < - 1.
(b) We have,

The point  x = -3/2 divides the real line into two disjoint intervals i.e., (-∞, -3/2) and (-3/2, ∞)
In interval  (-∞, -3/2) i.e , when x < -3/2, f'(x) = -6-4x<0
∴ f is strictly increasing for x< -3/2
In interval  (-3/2, ∞) i.e , when x > -3/2, f'(x) = -6-4x<0
∴ f is strictly decreasing for  x> -3/2
(c) We have
f(x) = -2x3 - 9x2 - 12x +1
∴ f'(x) = -6x2 - 18x - 12 = -6(x2+3x+2) = -6(x+1)(x+2)
Now,
f'(x) = 0 ⇒ x = -1 and x = -2
Points x = - 1 and x = -2 divide the real line into three disjoint intervals i.e...
(-∞, -2),(-2, -1) and (-1, ∞)
In intervals  (-∞, -2) and (-1, ∞) he., when x < - 2 and x > - 1,
f'(x) = -6(x+1)(x+2) < 0
∴ f is strictly de ere as ing for x < - 2 and x > -1.
Now. in interval (-2, -1) i.e.. when - 2 < x < - 1.
f'(x) = -6(x+1)(x+2) > 0
∴ f is strictly increasing for 2 < x < 1
(d) We have.
f(x) = 6-9x = x2
∴  f'(x) = -9 -2x
Now, f'
(x) = 0 gives x = -9/2
The point  x = -9/2 divides the real line into two disjoint intervals i.e., (-∞, -9/2) and (-9/2,∞ )
In interval  (-∞, -9/2) i.e., for
∴ f is strictly increasing for x < -9/2
In interval  (-9/2,∞ ) i.e. for
∴ f is strictly decreasing for
(e) We have.
f(x) = (x+1)3 (x-3)3
f'(x) = 3(x+1)2(x-3)3 +3(x-3)(x+1)3
= 3(x+1)2(x-3)2 [x-3+x+1]
=3(x+1)2(x-3)2(2x-2)
=6 (x+1)2(x-3)(x-1)
The points x = -1,x = 1. and x = 3 divide the real line into four disjoint intervals i.e., (-∞, -1) (-1,1), (1,3), and (3, ∞)
In intervals  (-∞, -1) and (-1,1), f'(x) = 6(x+1)2(x-3)2(x-1) <0
∴ f is strictly decreasing in intervals  (-∞, -1) and (-1,1)
In intervals
∴ f is strictly increasing in intervals  (1,3), and (3, ∞)

Q7: Show that  , is an increasing function of x throughout its domain.
Ans:
We have.

Since x > - 1. point x =0 divides the domain (-1 ∞) in two disjoint intervals i.e.. -1 < x < 0 & x > 0.
When -1 < x < 0. we have:

Hence, function f is increasing throughout this domain.

Q8: Find the values of x for which  y = [x(x-2)]2 is an increasing function.
Ans:
We have.

The points x = 0.x = 1. andx = 2 divide the real line into four disjoint intervals

∴ y is strictly decreasing in intervals  (-∞ , 0) and (1,2)
However, in intervals (0. 1) and (2, ∞), dy/dx > 0
∴ y is strictly increasing in intervals (0,1) and (2, ∞).
∴ y is strictly increasing for 0 < x < 1 and x > 2.

Q9: Prove that   is an increasing function of θ in [0, π/2]
Ans: We have.

In interval   we have cos

Tli ere fore.}' is strictly increasing in interval (0, π/2)
Also, the given function is continuous at  x=0 and x = π/2
Hence, is increasing in interval [0, π/2].

Q10: Prove that the logarithmic function is strictly increasing on (0, ∞).
Ans: The given function is f (x) = log x.
∴ f'(x) = 1/x
It is clear that for x > 0,  f'(x) = 1/x >0 .
Hence f(x) = log x is strictly increasing in interval (0;∞).

Q11: Prove that the function/given by f(x) = x2 - x + 1 is neither strictly increasing nor strictly decreasing on (-1. 1).
Ans: The given function is f(x) = x2 - x + 1.

The point 1/2 divides the interval (-1. 1) into two disjoint intervals i.e..
Now. in interval
Therefore, f is strictly decreasing in interval  (-1, 1/2)
However, in interval
Therefore, f is strictly increasing in interval  (1/2,1)
Hence. f is neither strictly increasing nor decreasing in interval (-1.1).

Q12: Which of the following functions are strictly decreasing on (0,π/2) ?
(A) cos x
(B) cos2x
(C) cos 3x
(D) tan x

Ans:

is strictly decreasing in interval

is strictly decreasing in interval

The point   divides the interval  1 into two disjoint intervals i.e., 0
Now, in interval
f3 is strictly decreasing in interval
However, in interval
∴ f3 is strictly increasing in interval
Hence, fis neither increasing nor decreasing in interval .

∴ f4 is strictly increasing in interval
Therefore, functions cos x and cos 2.x are strictly decreasing in
Hence, the correct answers are A and B.

Question 13: On which of the following intervals is the function f given by  strictly decreasing?

(D) ) None of these
Ans:
We have,

In interval

Thus, function fis strictly increasing in interval (0,1).
In interval

Thus, function f is strictly increasing in interval

∴ f is strictly increasing in interval .
Hence, function f is strictly decreasing in none of the intervals.

Q14: Find the least value of a such that the function f given f(x) = x2  + ax+ 1  is strictly increasing on (1, 2).
Ans: We have

Now, function f will be increasing in

Therefore, we have to find the least value of a such that

Thus, the least value of a for/to be increasing on (1? 2) is given by.
-a/2 =1
-a/2 =1 ⇒ a = -2
Hence, the required value of a is -2.

Q15: Let I be any interval disjoint from (−1, 1). Prove that the function f given by f(x) = x 1/x.  is strictly increasing on I.
Ans: We have,

The points x = 1 andx = -1 divide the real line in three disjoint intervals i.e.,
In interval (-1,1). it is observed that:

In intervals   , it is observed that:

∴ f is strictly increasing on  (-∞, 1) and (1, ∞)

Q16: Prove that the function f given by f(x) = log sin x is strictly increasing on (0,π/2)  and strictly decreasing on (π/2, π)
Ans:-

In interval
∴ f  is strictly decreasing in

Q17: Prove that the function f given by f (x) = log cos x is strictly decreasing on    and strictly increasing on
Ans: We have,

∴ f is strictly decreasing on
In interval

∴ f is strictly increasing on

Q18: Prove that the function given by  is increasing in R.
ANS: We have,

Thus, f'(x) is always positive in R.
Hence, the given function (f) is increasing in R.

Q19: The interval in which y= x2 e-x is increasing is
(A) (-∞,∞)
(B) (-2,0)
(C) (2,∞)
(D) (0,2)

Ans: We have.

The points x = 0 and x = 2 divide the real line into three disjoint intervals i.e.;

In intervals   is always positive.
∴ f is decreasing on (-∞,0) and (2, ∞)
In interval (0,2),
∴ f is strictly increasing on (0.2).
Hence f is strictly increasing in interval (0,2).

The document Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE is a part of the JEE Course Mathematics (Maths) Class 12.
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## Mathematics (Maths) Class 12

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## FAQs on Exercise 6.2 - Application of Derivative NCERT Solutions - Mathematics (Maths) Class 12 - JEE

 1. What is the application of derivatives in real life?
Ans. The application of derivatives in real life includes determining maximum and minimum values, finding rates of change, optimizing functions, and analyzing motion and growth.
 2. How can derivatives be used to find the maximum or minimum value of a function?
Ans. Derivatives can be used to find the maximum or minimum value of a function by setting the derivative equal to zero and solving for critical points, then using the second derivative test or analyzing the behavior of the function around those points.
 3. How are derivatives used in economics?
Ans. Derivatives are used in economics to analyze the marginal cost, revenue, and profit functions of a company, optimize production levels, and determine the elasticity of demand for a product.
 4. Can derivatives be used to analyze the speed of a moving object?
Ans. Yes, derivatives can be used to analyze the speed of a moving object by finding the derivative of the position function with respect to time, which gives the velocity function, and then finding the derivative of the velocity function to get the acceleration function.
 5. In what ways can derivatives be applied in engineering?
Ans. Derivatives can be applied in engineering to analyze the slope of a curve, optimize the design of structures or systems, determine the rate of change of a variable, and model the behavior of physical systems.

## Mathematics (Maths) Class 12

204 videos|288 docs|139 tests

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