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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:
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
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:  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Hence, f is strictly increasing on R.

Q3: Show that the function given by f(x) = sin x is
(a) strictly increasing in Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE 
(b) strictly decreasing in Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
(c) neither increasing nor decreasing in (0, π)

Ans: The given function is f(x) = sinx.
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
(a) Since for each  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE we have  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Hence,/is strictly increasing in Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
(b) Since for each Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE, we have Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Hence,f is strictly decreasing in Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
(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.
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Now. the point 3/4 divides the real line into two disjoint intervals i.e.. Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Hence, the given function (f) is strictly decreasing in interval Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
In interval  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Hence, the given function (J) is strictly increasing in interval Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE

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  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
The points x = -2 andx = 3 divide the real line into three disjoint intervals i.e..
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
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  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
∴ f is strictly decreasing in interval (-∞, -1).
Thus, f is strictly decreasing for x < -1.
In interval Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
∴ f is strictly increasing in interval (-1, ∞)
Thus, f is strictly deere asing for x < - 1.
(b) We have,
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
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  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE 
∴ f is strictly increasing for x < -9/2
In interval  (-9/2,∞ ) i.e. for  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE 
∴ f is strictly decreasing for Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
(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 Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
∴ f is strictly increasing in intervals  (1,3), and (3, ∞)

Q7: Show that Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE , is an increasing function of x throughout its domain.
Ans:
 We have.
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
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:
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
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.
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
The points x = 0.x = 1. andx = 2 divide the real line into four disjoint intervals
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
∴ 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  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE is an increasing function of θ in [0, π/2]
Ans: We have.
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
In interval  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE we have cos Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
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.
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
The point 1/2 divides the interval (-1. 1) into two disjoint intervals i.e.. Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Now. in interval  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Therefore, f is strictly decreasing in interval  (-1, 1/2)
However, in interval  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
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:
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE is strictly decreasing in interval Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE is strictly decreasing in interval Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE 
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
The point  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE divides the interval Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE 1 into two disjoint intervals i.e., 0 Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Now, in interval  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
f3 is strictly decreasing in interval Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
However, in interval  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
∴ f3 is strictly increasing in interval Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Hence, fis neither increasing nor decreasing in interval Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE.
  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
∴ f4 is strictly increasing in interval Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Therefore, functions cos x and cos 2.x are strictly decreasing in Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Hence, the correct answers are A and B.

Question 13: On which of the following intervals is the function f given by Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE strictly decreasing?
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
(D) ) None of these
Ans: 
We have,
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
In interval  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Thus, function fis strictly increasing in interval (0,1).
In interval  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Thus, function f is strictly increasing in interval  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
∴ f is strictly increasing in intervalExercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE .
Hence, function f is strictly decreasing in none of the intervals.
The correct answer is D.

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
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Now, function f will be increasing in  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Therefore, we have to find the least value of a such that
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
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,
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
The points x = 1 andx = -1 divide the real line in three disjoint intervals i.e.,  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
In interval (-1,1). it is observed that:
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
In intervals  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE , it is observed that:
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
∴ 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:-
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
In interval   Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
∴ f  is strictly decreasing in Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE

Q17: Prove that the function f given by f (x) = log cos x is strictly decreasing on  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE  and strictly increasing on Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Ans: We have,
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
∴ f is strictly decreasing on Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
In interval  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
∴ f is strictly increasing on Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE

Q18: Prove that the function given by Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE is increasing in R.
ANS: We have,
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
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.
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
The points x = 0 and x = 2 divide the real line into three disjoint intervals i.e.;
Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
In intervals  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE is always positive.
∴ f is decreasing on (-∞,0) and (2, ∞)
In interval (0,2),  Exercise 6.2 - Application of Derivative NCERT Solutions | Mathematics (Maths) Class 12 - JEE
∴ f is strictly increasing on (0.2).
Hence f is strictly increasing in interval (0,2).
The correct answer is D.

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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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.
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