NCERT Solutions for Maths Class 12 Exercise 6.5
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NCERT Solutions for Maths Class 12 Exercise 6.1
NCERT Solutions for Maths Class 12 Exercise 6.2
NCERT Solutions for Maths Class 12 Exercise 6.3
NCERT Solutions for Maths Class 12 Exercise 6.4
NCERT Solutions for Maths Class 12 Exercise 6.5
Maths Class
12 Ex 6.5 Question 1.
Find the maximum
and minimum values, if any, of the following functions given by
(i) f(x) = (2x – 1)² + 3
(ii) f(x) = 9x² + 12x + 2
(iii) f(x) = – (x – 1)² + 10
(iv) g(x) = x3 + 1
Solution:
(i) We have, f(x) = (2x – 1)² + 3
Minimum value of
(2x – 1)² is zero.
Minimum value of (2x – 1)² + 3 is 3.
Clearly, it does not have maximum value.
(ii) We have, f(x) = 9x² + 12x + 2
⇒ f(x) = (3x + 2)² – 2
Minimum value of (3 + 2)² is zero.
∴ Minimum value of (3x + 2)² – 2 is –2.
Clearly, f(x) does not have finite maximum value.
(iii) We have, f(x) = – (x – 1)² + 10
Maximum value of –(x – 1)² is zero.
Maximum value of f(x) = – (x – 1)² + 10 is 10.
Clearly, f(x) does not have finite minimum value.
(iv) We have, g(x) = x3 + 1
As x → ∞, g(x) → ∞;
Also, x → -∞, g(x) → -∞
Thus, there is no maximum or minimum value of f(x).
Maths Class
12 Ex 6.5 Question 2.
Find the maximum
and minimum values, if any, of the following functions given by
(i) f(x) = |x + 2| – 1
(ii) g(x) = –|x + 1| + 3
(iii) h(x) = sin 2x + 5
(iv) f(x) = |sin(4x + 3)|
(v) h(x) = x + 1, x ∈ (–1, 1)
Solution:
(i) We have, f(x) = |x + 2| – 1 ∀ x ∈ R
Now, |x + 2| ≥ 0 ∀ x ∈ R
|x + 2| – 1 ≥ – 1 ∀ x ∈ R,
So, –1 is the minimum value of f(x).
Now, f(x) = –1
⇒ |x + 2| – 1 = –1
⇒ |x + 2| = 0
⇒ x = – 2
(ii) We have, g(x) = = –|x + 1| + 3 ∀ x ∈ R
Now, |x + 1| ≥ 0 ∀ x ∈ R ⇒
–|x + 1| ≤ 0 ∀ x ∈ R
–|x + 1| + 3 ≤ 3 ∀ x ∈ R
So, 3 is the maximum value of g(x).
Now, g(x) = 3
⇒ –|x + 1| + 3 = 3
⇒ |x + 1| = 0
⇒ x = –1
(iii) We have, h(x) = sin 2x + 5
Since –1 ≤ sin 2x
≤ 1 ∀ x ∈ R
Adding 5 to all
terms, –1 + 5 ≤ sin 2x + 5 ≤ 1 + 5 ∀ x ∈ R
⇒ 4 ≤ sin 2x + 5 ≤ 6 ∀ x ∈ R
Thus, the maximum
value of f(x) is 6 and the minimum value is 4.
(iv) We have, f(x) = |sin 4x + 3|
Since –1 ≤ sin 4x ≤ 1 ∀ x ∈ R
Adding 3 to all
terms, –1 + 3 ≤ sin 2x + 3 ≤ 1 + 3 ∀ x ∈ R
⇒ 2 ≤ sin 2x + 3 ≤ 4 ∀ x ∈ R
Thus, the maximum
value of f(x) is 4 and the minimum value is 2.
(v) We have, h(x) = x + 1, x ∈ (–1, 1)
Since –1 < x
< 1
Adding 1 to all
terms, –1 + 1 ≤ x + 1 ≤ 1 + 1
⇒ 0 ≤ x + 1 ≤ 2
Thus, the maximum
value of h(x) is 2 and the minimum value is 0. But these values do not exist in
the interval (–1, 1).
Therefore, the
function h(x) has neither minimum value nor maximum value in (–1, 1).
Maths Class
12 Ex 6.5 Question 3.
Find the local
maxima and local minima, if any, of the following functions. Find also the
local maximum and the local minimum values, as the case may be:
(i) f(x) = x2
(ii) g(x) = x3 – 3x
(iii) h(x) = sin x + cos x, 0 < x
< π/2
(iv) f(x) = sin x – cos
x, 0 < x < 2Ï€
(v) f(x) = x3 – 6x2 +
9x + 15
(vi) g(x) = x/2 + 2/x, x > 0
(vii) g(x) = 1/(x2 + 2)
(viii) f(x) =
, x > 0
Solution:
(i) We have, f(x) = x² ⇒ f’(x) = 2x and f’’(x) = 2
Now, f'(x) = 0 ⇒ 2x = 0 i.e., x = 0
When x = 0, f’’(x) = 2 (+ve)
⇒ f(x) has a local minimum at x = 0 and
local minimum value f(0) = 0.
Prove that the
following functions do not have maxima or minima:
(i) f(x) = ex
(ii) f(x) = log x
(iii) h(x) = x3 + x2 +
x + 1
Solution:
(i) We have, f(x) = ex ⇒ f'(x) = ex;
Since f’(x) ≠ 0 for any value of x.
So, f(x) = ex does not have a maxima or minima.
(ii) We have, f(x) = log x ⇒ f’(x) = 1/x;
Clearly f’(x) ≠ 0
for any value of x.
So, f(x) = log x does not have a maxima or minima.
(iii) We have, f(x) = x3 + x2 +
x + 1
⇒ f’(x) = 3x2 + 2x + 1
Now, f’(x) = 0 ⇒ 3x2 + 2x + 1 = 0
i.e. f'(x) ≠ 0 for any real value of x.
Hence, there is neither maxima nor minima.
Maths Class
12 Ex 6.5 Question 5.
Find the absolute
maximum value and the absolute minimum value of the following functions in the
given intervals:
(i) f(x) = x3, x ∈ [–2, 2]
(ii) f(x) = sin x + cos x, x ∈ [0, Ï€]
(iii) f(x) = 4x – x2/2,
x ∈ [–2, 9/2]
(iv) f(x) = (x – 1)2
+ 3, x ∈ [–3, 1]
Solution:
(i) We have, f(x) = x3 in [ –2, 2]
∴ f'(x) = 3x²; Now, f’(x) = 0 at x = 0,
f(0) = 0
Now, f(–2) = (–2)3 = – 8; f(0) = (0)² = 0 and f(2) = (2)3
= 8
Hence, the absolute maximum value of f(x) is 8 which is attained at x = 2 and absolute
minimum value of f(x) is –8 which is attained at x = –2.
(ii) We have, f(x) = sin x + cos x in [0, π]
f’(x) = cos x – sin x for extreme values f’(x) = 0
Maths Class
12 Ex 6.5 Question 6.
Find the maximum
profit that a company can make, if the profit function is given by p(x) = 41 –
24x – 18x²
Solution:
Profit function is
given by p(x) = 41 – 24x – 18x²
∴ p'(x) = – 24 – 36x = –12(2 + 3x)
For maxima and minima, p'(x) = 0
Now, p'(x) = 0
⇒ –12(2 + 3x) = 0
⇒ x = –2/3
p'(x) changes sign from +ve to -ve.
⇒ p(x) has maximum value at x = –2/3
Maximum Profit = 41 + 16 – 8 = 49.
Maths Class
12 Ex 6.5 Question 7.
Find both the
maximum value and the minimum value of 3x4 – 8x3 +
12x2 – 48x + 25 on the interval [0, 3].
Solution:
Let f(x) = 3x4 –
8x3 + 12x2 – 48x + 25
∴ f'(x) = 12x3 – 24x2 +
24x – 48
= 12(x2 + 2)(x
– 2)
For maxima and minima, f'(x) = 0
⇒ 12(x2 + 2)(x – 2) = 0
⇒ x = 2
Now, we find f(x) at x = 0, 2 and 3, f(0) = 25,
f(2) = 3(24) – 8(23) + 12(22) – 48(2) + 25 =
–39
and f(3) = (34) – 8(33) + 12(32) – 48(3) + 25
= 243 – 216 + 108 – 144 +
25 = 16
Hence, at x = 0, maximum value = 25
and at x = 2, minimum value = –39.
Maths Class
12 Ex 6.5 Question 8.
At what points in
the interval [0, 2Ï€] does the function sin 2x attain its maximum value?
Solution:
We have, f(x) =
sin 2x in [0, 2Ï€] ⇒ f’(x) = 2 cos 2x
For maxima and minima, f’(x) = 0 ⇒ cos 2x = 0
Maths Class
12 Ex 6.5 Question 9.
What is the
maximum value of the function sin x + cos x?
Solution:
Consider the
interval [0, 2Ï€].
Let f(x) = sin x + cos x,
f’(x) = cos x – sin x
For maxima and minima, f’(x) = 0
Maths Class
12 Ex 6.5 Question 10.
Find the maximum
value of 2x3 – 24x + 107 in the interval [1, 3]. Find the
maximum value of the same function in [–3, –1].
Solution:
∵ f(x) = 2x3 – 24x +
107
∴ f’(x) = 6x2 – 24,
For maxima and minima, f'(x) = 0 ⇒ x = ±2
For the interval [ 1, 3], we find the values of f(x) at x = 1, 2, 3;
f(1) = 85, f(2) =
75, f(3) = 89
Hence, maximum value of f(x) = 89 at x = 3.
For the interval [–3, –1], we find the values of f(x) at x = – 3, – 2, – 1;
f(–3) = 125;
f(–2) = 139
f(–1) = 129
∴ Maximum value of f(x) = 139 at x = –2.
Maths Class
12 Ex 6.5 Question 11.
It is given that
at x = 1, the function x4 – 62x2 + ax + 9
attains its maximum value, on the interval [0, 2]. Find the value of a.
Solution:
We have, f(x) = x4 –
62x2 + ax + 9
∴ f’(x) = 4x3 – 124x +
a
Now, f’(x) = 0 at x = 1
⇒ 4 – 124 + a = 0
⇒ a = 120
Now, f”(x) = 12x2 – 124
At x = 1, f”(1) = 12 – 124 = – 112 < 0
⇒ f(x) has a maximum value at x = 1
when a = 120.
Maths Class
12 Ex 6.5 Question 12.
Find the maximum
and minimum values of x + sin 2x on [0, 2Ï€].
Solution: We have, f(x) = x + sin 2x on [0, 2Ï€]
∴ f’(x) = 1 + 2 cos 2x
For maxima and minima f’(x) = 0
Maths Class
12 Ex 6.5 Question 13.
Find two numbers
whose sum is 24 and whose product is as large as possible.
Solution:
Let the required
numbers are x and (24 – x).
∴
Their product, p
= x(24 – x) = 24x – x²
Now, dp/dx = 0 ⇒24 – 2x = 0 ⇒ x = 12
Also, d2p/dx2 = –2 < 0 ⇒ p is maximum at x = 12.
Hence, the required numbers are 12 and (24 – 12) i.e., 12.
Maths Class
12 Ex 6.5 Question 14.
Find two positive
numbers x and y such that x + y = 60 and xy3 is maximum.
Solution:
We have, x + y =
60
⇒ y = 60 – x … (i)
Let z = xy3
⇒ z = x(60 – x)3
⇒ dz/dx = x[3(60 – x)2(–1)]
+ (60 – x)3
= (60 – x)2[–3x + 60
– x]
= (60 – x)2(60 – 4x)
Maths Class
12 Ex 6.5 Question 15.
Find two positive
numbers x and y such that their sum is 35 and the product x2y5 is
a maximum.
Solution:
We have, x + y =
35 ⇒ y = 35 – x
Product, P = x2y5
= x2(35 –
x)5
Maths Class
12 Ex 6.5 Question 16.
Find two positive
numbers whose sum is 16 and the sum of whose cubes is minimum.
Solution:
Let two numbers
be x and 16 – x.
Maths Class
12 Ex 6.5 Question 17.
A square piece of
tin of side 18 cm is to be made into a box without top, by cutting a square
from each corner and folding up the flaps to form the box. What should be the side
of the square to be cut off so that the volume of the box is the maximum
possible?
Solution:
Let each side of
the square to be cut off be x cm.
∴ For the box, length = 18 – 2x;
breadth = 18 – 2x and height = x
Maths Class
12 Ex 6.5 Question 18.
A rectangular
sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting
off square from each corner and folding up the flaps. What should be the side
of the square to be cut off so that the volume of the box is maximum?
Solution:
Let each side of the
square cut off from each corner be x cm.
∴ Sides of the rectangular box are (45
– 2x), (24 – 2x) and x cm.
Maths Class
12 Ex 6.5 Question 19.
Show that of all
the rectangles inscribed in a given fixed circle, the square has the maximum
area.
Solution:
Let the length
and the breadth of the rectangle inscribed in a circle of radius a be x and y respectively.
∴ x² + y² = (2a)² ⇒ x² + y² = 4a² … (i)
∴ Perimeter = 2(x + y)
Maths Class
12 Ex 6.5 Question 20.
Show that the
right circular cylinder of given surface and maximum volume is such that its
height is equal to the diameter of the base.
Solution:
Let S be the
given surface area of the closed cylinder whose radius is r and height is h. Let
V be its Volume. Then
Maths Class
12 Ex 6.5 Question 21.
Of all the closed
cylindrical cans (right circular), of a given volume of 100 cubic centimetres,
find the dimensions of the can which has the minimum surface area?
Solution:
Let r be the
radius and h be the height of cylindrical can.
Maths Class
12 Ex 6.5 Question 22.
A wire of length
28 m is to be cut into two pieces. One of the pieces is to be made into a
square and the other into a circle. What should be the length of the two pieces
so that the combined area of the square and the circle is minimum?
Solution:
Let the length of
one part be x m, then the other part = (28 – x) m.
Let the part of the length x m be converted into a circle of radius r.
Prove that the
volume of the largest cone that can be inscribed in a sphere of radius R
is 8/27 of the volume of the sphere.
Solution:
Let a cone VAB of
greatest volume be inscribed in the sphere. Let ∠AOC = θ.
∴ AC, the radius of the base of the
cone = R sin θ and VC = VO + OC = R(1 + cos θ)
= R + R cos θ = height of the cone
Maths Class
12 Ex 6.5 Question 24.
Show that the
right circular cone of least curved surface and given volume has an altitude
equal to √2 time the radius of the base.
Solution:
Let r and h be
the radius and the height of the cone.
Maths Class
12 Ex 6.5 Question 25.
Show that the
semi-vertical angle of the cone of the maximum volume and of given slant height
is tan-1 √2.
Solution:
Let v be the
volume, l be the slant height and a
be the semi-vertical angle of a cone.
Maths Class
12 Ex 6.5 Question 26.
Show that
semi-vertical angle of right circular cone of given surface area and maximum
volume is sin-1 (1/3).
Solution:
Let r be the radius,
l be the slant height and h be the height of the cone of given surface area s. Then,
Choose the correct
answer in the Exercises 27 and 29.
Maths Class
12 Ex 6.5 Question 27.
The point on the
curve x² = 2y which is nearest to the point (0, 5) is:
(A) (2√2, 4)
(B) (2√2, 0)
(C) (0, 0)
(D) (2, 2)
Solution:
(A) Let P(x, y) be a point on the curve. The
other point is A(0, 5).
z = PA² = x² + y² + 25 – 10y [∵ x² = 2y]
Maths Class
12 Ex 6.5 Question 28.
For all real
values of x, the minimum value of (1 – x + x2)/(1 + x + x2) is:
(A) 0
(B) 1
(C) 3
(D) 1/3
Solution:
The maximum value
of [x(x – 1) + 1]1/3, 0 ≤ x ≤ 1 is:
(A) (1/3)1/3
(B) 1/2
(C) 1
(D) 0
Solution:
NCERT Solutions for Maths Class 12 Exercise 6.1
NCERT Solutions for Maths Class 12 Exercise 6.2