RD Sharma Class 12 Solutions - Maxima and Minima | Mathematics (Maths) Class 12 - JEE PDF Download

 Page 1


18. Maxima and Minima
Exercise 18.1
1. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = 4x
2
 – 4x + 4 on R
Answer
f(x) = 4x
2
 – 4x + 4 on R
= 4x
2
 – 4x + 1 + 3
= (2x – 1)
2
 + 3
Since, (2x – 1)
2
 =0
= (2x – 1)
2
 + 3 =3
= f(x) = f
Thus, the minimum value of f(x) is 3 at x = 
Since, f(x) can be made large. Therefore maximum value does not exist.
2. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = –(x – 1)
2
 + 2 on R
Answer
We have f(x) = – (x – 1)
2
 + 2
It can be observed that (x – 1)
2
=0 for every x?R
Therefore, f(x) = – (x – 1)
2
 + 2=2 for every x?R
The maximum value of f is attained when (x – 1) = 0
(x – 1)=0, x=1
Since, Maximum value of f = f(1) = – (1 – 1)
2
 + 2 = 2
Hence, function f does not have minimum value.
3. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = |x + 2| on R
Answer
|x + 2|=0 for x ? R
= f(x) = 0 for all x ? R
So the minimum value of f(x) is 0, which attains at x =2
Hence, f(x) = |x + 2| does not have the maximum value.
4. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
Page 2


18. Maxima and Minima
Exercise 18.1
1. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = 4x
2
 – 4x + 4 on R
Answer
f(x) = 4x
2
 – 4x + 4 on R
= 4x
2
 – 4x + 1 + 3
= (2x – 1)
2
 + 3
Since, (2x – 1)
2
 =0
= (2x – 1)
2
 + 3 =3
= f(x) = f
Thus, the minimum value of f(x) is 3 at x = 
Since, f(x) can be made large. Therefore maximum value does not exist.
2. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = –(x – 1)
2
 + 2 on R
Answer
We have f(x) = – (x – 1)
2
 + 2
It can be observed that (x – 1)
2
=0 for every x?R
Therefore, f(x) = – (x – 1)
2
 + 2=2 for every x?R
The maximum value of f is attained when (x – 1) = 0
(x – 1)=0, x=1
Since, Maximum value of f = f(1) = – (1 – 1)
2
 + 2 = 2
Hence, function f does not have minimum value.
3. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = |x + 2| on R
Answer
|x + 2|=0 for x ? R
= f(x) = 0 for all x ? R
So the minimum value of f(x) is 0, which attains at x =2
Hence, f(x) = |x + 2| does not have the maximum value.
4. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = sin 2x + 5 on R
Answer
We know that – 1 = sin2x = 1
= – 1 + 5 = sin2x + 5 = 1 + 5
= 4 = sin 2x + 5 = 6
Hence, the maximum and minimum value of h are 4 and 6 respectively.
5. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = |sin 4x + 3| on R
Answer
We know that – 1 = sin4x = 1
= 2 = sin4x + 3 = 4
= 2 = |sin 4x + 3| = 4
Hence, the maximum and minimum value of f are 4 and 2 respectively.
6. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = 2x
3
 + 5 on R
Answer
We have f(x) = 2x
3
 + 5 on R
Here, we observe that the values of f(x) increase when the values of x are increased and f(x) can be made
large,
So, f(x) does not have the maximum value
Similarly, f(x) can be made as small as we want by giving smaller values to x.
So, f(x) does not have the minimum value.
7. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = – |x + 1| + 3 on R
Answer
We know that – |x + 1| = 0 for every x ? R.
Therefore, g(x) = – |x + 1| + 3 = 3 for every x ? R.
The maximum value of g is attained when |x + 1| = 0
|x + 1| = 0
x = – 1
Since, Maximum Value of g = g( – 1) = – | – 1 + 1| + 3 = 3
Hence, function g does not have minimum value.
8. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = 16x
2
 –16x + 28 on R
Page 3


18. Maxima and Minima
Exercise 18.1
1. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = 4x
2
 – 4x + 4 on R
Answer
f(x) = 4x
2
 – 4x + 4 on R
= 4x
2
 – 4x + 1 + 3
= (2x – 1)
2
 + 3
Since, (2x – 1)
2
 =0
= (2x – 1)
2
 + 3 =3
= f(x) = f
Thus, the minimum value of f(x) is 3 at x = 
Since, f(x) can be made large. Therefore maximum value does not exist.
2. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = –(x – 1)
2
 + 2 on R
Answer
We have f(x) = – (x – 1)
2
 + 2
It can be observed that (x – 1)
2
=0 for every x?R
Therefore, f(x) = – (x – 1)
2
 + 2=2 for every x?R
The maximum value of f is attained when (x – 1) = 0
(x – 1)=0, x=1
Since, Maximum value of f = f(1) = – (1 – 1)
2
 + 2 = 2
Hence, function f does not have minimum value.
3. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = |x + 2| on R
Answer
|x + 2|=0 for x ? R
= f(x) = 0 for all x ? R
So the minimum value of f(x) is 0, which attains at x =2
Hence, f(x) = |x + 2| does not have the maximum value.
4. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = sin 2x + 5 on R
Answer
We know that – 1 = sin2x = 1
= – 1 + 5 = sin2x + 5 = 1 + 5
= 4 = sin 2x + 5 = 6
Hence, the maximum and minimum value of h are 4 and 6 respectively.
5. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = |sin 4x + 3| on R
Answer
We know that – 1 = sin4x = 1
= 2 = sin4x + 3 = 4
= 2 = |sin 4x + 3| = 4
Hence, the maximum and minimum value of f are 4 and 2 respectively.
6. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = 2x
3
 + 5 on R
Answer
We have f(x) = 2x
3
 + 5 on R
Here, we observe that the values of f(x) increase when the values of x are increased and f(x) can be made
large,
So, f(x) does not have the maximum value
Similarly, f(x) can be made as small as we want by giving smaller values to x.
So, f(x) does not have the minimum value.
7. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = – |x + 1| + 3 on R
Answer
We know that – |x + 1| = 0 for every x ? R.
Therefore, g(x) = – |x + 1| + 3 = 3 for every x ? R.
The maximum value of g is attained when |x + 1| = 0
|x + 1| = 0
x = – 1
Since, Maximum Value of g = g( – 1) = – | – 1 + 1| + 3 = 3
Hence, function g does not have minimum value.
8. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = 16x
2
 –16x + 28 on R
Answer
We have f(x) = 16x
2
 – 16x + 28 on R
= 16x
2
 – 16x + 4 + 24
= (4x – 2)
2
 + 24
Now, (4x – 2)
2
 = 0 for all x ? R
= (4x – 2)
2
 + 24= 24 for all x ? R
= f(x) = f
Thus, the minimum value of f(x) is 24 at x =
Hence, f(x) can be made large as possibly by giving difference value to x.
Thus, maximum values does not exist.
9. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = x
3
 – 1 on R
Answer
We have f(x) = x
3
 – 1 on R
Here, we observe that the values of f(x) increase when the values of x are increased, and f(x) can be made
large, by giving large value.
So, f(x) does not have the maximum value
Similarly, f(x) can be made as small as we want by giving smaller values to x.
So, f(x) does not have the minimum value.
Exercise 18.2
1. Question
Find the points of local maxima or local minima, if any, of the following functions, using the first derivative
test. Also, find the local maximum or local minimum values, as the case may be:
f(x) = (x – 5)
4
Answer
f(x) = (x – 5)
4
Differentiate w.r.t x
f ’(x) = 4(x – 5)
3
for local maxima and minima
f ‘ (x) = 0
= 4(x – 5)
3
 = 0
= x – 5 = 0
x = 5
f ‘ (x) changes from –ve to + ve as passes through 5.
So, x = 5 is the point of local minima
Page 4


18. Maxima and Minima
Exercise 18.1
1. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = 4x
2
 – 4x + 4 on R
Answer
f(x) = 4x
2
 – 4x + 4 on R
= 4x
2
 – 4x + 1 + 3
= (2x – 1)
2
 + 3
Since, (2x – 1)
2
 =0
= (2x – 1)
2
 + 3 =3
= f(x) = f
Thus, the minimum value of f(x) is 3 at x = 
Since, f(x) can be made large. Therefore maximum value does not exist.
2. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = –(x – 1)
2
 + 2 on R
Answer
We have f(x) = – (x – 1)
2
 + 2
It can be observed that (x – 1)
2
=0 for every x?R
Therefore, f(x) = – (x – 1)
2
 + 2=2 for every x?R
The maximum value of f is attained when (x – 1) = 0
(x – 1)=0, x=1
Since, Maximum value of f = f(1) = – (1 – 1)
2
 + 2 = 2
Hence, function f does not have minimum value.
3. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = |x + 2| on R
Answer
|x + 2|=0 for x ? R
= f(x) = 0 for all x ? R
So the minimum value of f(x) is 0, which attains at x =2
Hence, f(x) = |x + 2| does not have the maximum value.
4. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = sin 2x + 5 on R
Answer
We know that – 1 = sin2x = 1
= – 1 + 5 = sin2x + 5 = 1 + 5
= 4 = sin 2x + 5 = 6
Hence, the maximum and minimum value of h are 4 and 6 respectively.
5. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = |sin 4x + 3| on R
Answer
We know that – 1 = sin4x = 1
= 2 = sin4x + 3 = 4
= 2 = |sin 4x + 3| = 4
Hence, the maximum and minimum value of f are 4 and 2 respectively.
6. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = 2x
3
 + 5 on R
Answer
We have f(x) = 2x
3
 + 5 on R
Here, we observe that the values of f(x) increase when the values of x are increased and f(x) can be made
large,
So, f(x) does not have the maximum value
Similarly, f(x) can be made as small as we want by giving smaller values to x.
So, f(x) does not have the minimum value.
7. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = – |x + 1| + 3 on R
Answer
We know that – |x + 1| = 0 for every x ? R.
Therefore, g(x) = – |x + 1| + 3 = 3 for every x ? R.
The maximum value of g is attained when |x + 1| = 0
|x + 1| = 0
x = – 1
Since, Maximum Value of g = g( – 1) = – | – 1 + 1| + 3 = 3
Hence, function g does not have minimum value.
8. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = 16x
2
 –16x + 28 on R
Answer
We have f(x) = 16x
2
 – 16x + 28 on R
= 16x
2
 – 16x + 4 + 24
= (4x – 2)
2
 + 24
Now, (4x – 2)
2
 = 0 for all x ? R
= (4x – 2)
2
 + 24= 24 for all x ? R
= f(x) = f
Thus, the minimum value of f(x) is 24 at x =
Hence, f(x) can be made large as possibly by giving difference value to x.
Thus, maximum values does not exist.
9. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = x
3
 – 1 on R
Answer
We have f(x) = x
3
 – 1 on R
Here, we observe that the values of f(x) increase when the values of x are increased, and f(x) can be made
large, by giving large value.
So, f(x) does not have the maximum value
Similarly, f(x) can be made as small as we want by giving smaller values to x.
So, f(x) does not have the minimum value.
Exercise 18.2
1. Question
Find the points of local maxima or local minima, if any, of the following functions, using the first derivative
test. Also, find the local maximum or local minimum values, as the case may be:
f(x) = (x – 5)
4
Answer
f(x) = (x – 5)
4
Differentiate w.r.t x
f ’(x) = 4(x – 5)
3
for local maxima and minima
f ‘ (x) = 0
= 4(x – 5)
3
 = 0
= x – 5 = 0
x = 5
f ‘ (x) changes from –ve to + ve as passes through 5.
So, x = 5 is the point of local minima
Thus, local minima value is f(5) = 0
2. Question
Find the points of local maxima or local minima, if any, of the following functions, using the first derivative
test. Also, find the local maximum or local minimum values, as the case may be:
f(x) = x
3
 – 3x
Answer
We have, g (x) = x
3
 – 3x
Differentiate w.r.t x then we get,
g’ (x) = 3x
2
 – 3
Now, g‘(x) =0
= 3x
2
 = 3 ? x = ±1
Again differentiate g’(x) = 3x
2
 – 3
g’’(x)= 6x
g’’(1)= 6 > 0
g’’( – 1)= – 6>0
By second derivative test, x=1 is a point of local minima and local minimum value of g at
x =1 is g(1) = 1
3
 – 3 = 1 – 3 = – 2
However, x = – 1 is a point of local maxima and local maxima value of g at
x = – 1 is g( – 1) = ( – 1)
3
 – 3( – 1)
= – 1 + 3
= 2
Hence, The value of Minima is – 2 and Maxima is 2.
3. Question
Find the points of local maxima or local minima, if any, of the following functions, using the first derivative
test. Also, find the local maximum or local minimum values, as the case may be:
f(x) = x
3
 (x – 1)
2
Answer
We have, f(x) = x
3
(x – 1)
2
Differentiate w.r.t x, we get,
f ‘(x) = 3x
2
(x – 1)
2
 + 2x
3
(x – 1)
= (x – 1)(3x
2
(x – 1) + 2x
3
)
= (x – 1)(3x
3
 – 3x
2
 + 2x
3
)
= (x – 1)(5x
3
 – 3x
2
)
= x
2
 (x – 1)(5x – 3)
For all maxima and minima,
f ’(x) = 0
= x
2
(x – 1)(5x – 3) = 0
Page 5


18. Maxima and Minima
Exercise 18.1
1. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = 4x
2
 – 4x + 4 on R
Answer
f(x) = 4x
2
 – 4x + 4 on R
= 4x
2
 – 4x + 1 + 3
= (2x – 1)
2
 + 3
Since, (2x – 1)
2
 =0
= (2x – 1)
2
 + 3 =3
= f(x) = f
Thus, the minimum value of f(x) is 3 at x = 
Since, f(x) can be made large. Therefore maximum value does not exist.
2. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = –(x – 1)
2
 + 2 on R
Answer
We have f(x) = – (x – 1)
2
 + 2
It can be observed that (x – 1)
2
=0 for every x?R
Therefore, f(x) = – (x – 1)
2
 + 2=2 for every x?R
The maximum value of f is attained when (x – 1) = 0
(x – 1)=0, x=1
Since, Maximum value of f = f(1) = – (1 – 1)
2
 + 2 = 2
Hence, function f does not have minimum value.
3. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = |x + 2| on R
Answer
|x + 2|=0 for x ? R
= f(x) = 0 for all x ? R
So the minimum value of f(x) is 0, which attains at x =2
Hence, f(x) = |x + 2| does not have the maximum value.
4. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = sin 2x + 5 on R
Answer
We know that – 1 = sin2x = 1
= – 1 + 5 = sin2x + 5 = 1 + 5
= 4 = sin 2x + 5 = 6
Hence, the maximum and minimum value of h are 4 and 6 respectively.
5. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = |sin 4x + 3| on R
Answer
We know that – 1 = sin4x = 1
= 2 = sin4x + 3 = 4
= 2 = |sin 4x + 3| = 4
Hence, the maximum and minimum value of f are 4 and 2 respectively.
6. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = 2x
3
 + 5 on R
Answer
We have f(x) = 2x
3
 + 5 on R
Here, we observe that the values of f(x) increase when the values of x are increased and f(x) can be made
large,
So, f(x) does not have the maximum value
Similarly, f(x) can be made as small as we want by giving smaller values to x.
So, f(x) does not have the minimum value.
7. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = – |x + 1| + 3 on R
Answer
We know that – |x + 1| = 0 for every x ? R.
Therefore, g(x) = – |x + 1| + 3 = 3 for every x ? R.
The maximum value of g is attained when |x + 1| = 0
|x + 1| = 0
x = – 1
Since, Maximum Value of g = g( – 1) = – | – 1 + 1| + 3 = 3
Hence, function g does not have minimum value.
8. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = 16x
2
 –16x + 28 on R
Answer
We have f(x) = 16x
2
 – 16x + 28 on R
= 16x
2
 – 16x + 4 + 24
= (4x – 2)
2
 + 24
Now, (4x – 2)
2
 = 0 for all x ? R
= (4x – 2)
2
 + 24= 24 for all x ? R
= f(x) = f
Thus, the minimum value of f(x) is 24 at x =
Hence, f(x) can be made large as possibly by giving difference value to x.
Thus, maximum values does not exist.
9. Question
Find the maximum and the minimum values, if any, without using derivatives of the following functions:
f(x) = x
3
 – 1 on R
Answer
We have f(x) = x
3
 – 1 on R
Here, we observe that the values of f(x) increase when the values of x are increased, and f(x) can be made
large, by giving large value.
So, f(x) does not have the maximum value
Similarly, f(x) can be made as small as we want by giving smaller values to x.
So, f(x) does not have the minimum value.
Exercise 18.2
1. Question
Find the points of local maxima or local minima, if any, of the following functions, using the first derivative
test. Also, find the local maximum or local minimum values, as the case may be:
f(x) = (x – 5)
4
Answer
f(x) = (x – 5)
4
Differentiate w.r.t x
f ’(x) = 4(x – 5)
3
for local maxima and minima
f ‘ (x) = 0
= 4(x – 5)
3
 = 0
= x – 5 = 0
x = 5
f ‘ (x) changes from –ve to + ve as passes through 5.
So, x = 5 is the point of local minima
Thus, local minima value is f(5) = 0
2. Question
Find the points of local maxima or local minima, if any, of the following functions, using the first derivative
test. Also, find the local maximum or local minimum values, as the case may be:
f(x) = x
3
 – 3x
Answer
We have, g (x) = x
3
 – 3x
Differentiate w.r.t x then we get,
g’ (x) = 3x
2
 – 3
Now, g‘(x) =0
= 3x
2
 = 3 ? x = ±1
Again differentiate g’(x) = 3x
2
 – 3
g’’(x)= 6x
g’’(1)= 6 > 0
g’’( – 1)= – 6>0
By second derivative test, x=1 is a point of local minima and local minimum value of g at
x =1 is g(1) = 1
3
 – 3 = 1 – 3 = – 2
However, x = – 1 is a point of local maxima and local maxima value of g at
x = – 1 is g( – 1) = ( – 1)
3
 – 3( – 1)
= – 1 + 3
= 2
Hence, The value of Minima is – 2 and Maxima is 2.
3. Question
Find the points of local maxima or local minima, if any, of the following functions, using the first derivative
test. Also, find the local maximum or local minimum values, as the case may be:
f(x) = x
3
 (x – 1)
2
Answer
We have, f(x) = x
3
(x – 1)
2
Differentiate w.r.t x, we get,
f ‘(x) = 3x
2
(x – 1)
2
 + 2x
3
(x – 1)
= (x – 1)(3x
2
(x – 1) + 2x
3
)
= (x – 1)(3x
3
 – 3x
2
 + 2x
3
)
= (x – 1)(5x
3
 – 3x
2
)
= x
2
 (x – 1)(5x – 3)
For all maxima and minima,
f ’(x) = 0
= x
2
(x – 1)(5x – 3) = 0
= x =0, 1, 
At x =  f ’(x) changes from –ve to + ve
Since, x =  is a point of Minima
At x =1 f ‘ (x) changes from –ve to + ve
Since, x =1 is point of maxima.
4. Question
Find the points of local maxima or local minima, if any, of the following functions, using the first derivative
test. Also, find the local maximum or local minimum values, as the case may be:
f(x) = (x – 1) (x + 2)
2
Answer
We have, f(x) = (x – 1)(x + 2)
2
Differentiate w.r.t x, we get,
f ‘(x) = (x + 2)
2
 + 2(x – 1)(x + 2)
= (x + 2)(x + 2 + 2x – 2)
= (x + 2)(3x)
For all maxima and minima,
f ’(x) = 0
= (x + 2)(3x) = 0
= x =0, – 2
At x = – 2 f ’(x) changes from –ve to + ve
Since, x = – 2 is a point of Maxima
At x =0 f ‘ (x) changes from –ve to + ve
Since, x =0 is point of Minima.
Hence, local min value = f(0) = – 4
local max value = f( – 2) = 0.
5. Question
Find the points of local maxima or local minima, if any, of the following functions, using the first derivative
test. Also, find the local maximum or local minimum values, as the case may be:
Answer
We have, f(x) = (x – 1)
3
(x + 1)
2
Differentiate w.r.t x, we get,
f ‘(x) = 3(x – 1)
2
(x + 1)
2
 + 2(x – 1)
3
(x + 1)
= (x – 1)
2
(x + 1) (3 (x + 1) + 2(x – 1)
= (x – 1)
2
(x + 1)(5x + 1)