These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-6-ex-6-5/
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.5
Ex 6.5 Class 12 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:
Method I:
(i) f(x) = (2x – 1)² + 3
∴ f(x) ≥ 3 since (2x – 1)² ≥ 0 for x ∈ R
Hence the minimum value of f is 3, when
2x – 1 or x = \(\frac { 1 }{ 2 }\)
There is no maximum value since
f(x) → ∞ as x → ∞ and x → – ∞
Method II:
f(x) = (2x – 1)² + 3
Then f'(x) = 2(2x – 1)
∴ f'(x) = 0 ⇒ 2(2x – 1) = 0 ⇒ \(\frac { 1 }{ 2 }\)
Sign of f'(x) = 2(2x – 1)
f'(x) = 0 ⇒ 6(3x + 2) = 0 ⇒ x = \(\frac { – 2 }{ 3 }\)
Sign of f'(x) = 6(3x + 2)
f has no maximum value
Method III
f(x) – 9x² + 12x + 2
f'(x) = 18x + 12 and f”'(x) = 18
f'(x) = 0 ⇒ 18x + 12 = 0
⇒ 6(3x + 2) = 0
⇒ x = \(\frac { – 2 }{ 3 }\)
f has no maximum value
iii.
Method I
f(x) = – (x – 1)² + 10 = 10 – (x – 1)²
∴ f'(x) ≤ 10 since (x – 1)² ≥ 0 for all x ∈ R
The maximum value of is 10, when x – 1 = 0
of x = 1
f has no minimum value as f(x) → – ∞ as x → ∞ and x → – ∞.
Method II
f(x) = – (x – 1)² + 10, f'(x) = – 2(x – 1)
There is no maximum value
Method III (Second derivative test)
There is no maximum value
ii.
Method I
f(x) = 9x² + 12x + 2 = (3x + 2)² – 2
f(x) ≥ – 2 since (3x + 2)² ≥ 0 for all x ∈ R
The minimum value of f is – 2, when
(3x + 2) = 0 or when x = \(\frac { – 2 }{ 3 }\)
f has no maximum value as f(x) → ∞ as x → ∞ and x → ∞
Method II
f(x) = 9x² + 12x + 2
f'(x) = 18x + 12 = 6(3x + 2)
f'(x) = 0 ⇒ – 2(x – 1) = 0 ⇒ x = 1
Sign of f'(x) = – 2(x – 1)
∴ f has a maximum at x – 1 and the maxi-mum value = f(1) = -(1 – 1)² + 10 = 10
There is no minimum value
Method III
f(x) = – (x – 1)² + 10
f'(x) = – 2 (x – 1) and f”(x) = – 2
f'(x) = 0 ⇒ – 2 (x – 1) = 0 ⇒ x = 1
f”(1) = – 2, is negative
∴ f has a maximum value at x = 1
The maximum value = f(1) = – (1 – 1)² + 10 = 10
f has no minimum value
iv. Method I
g(x) = x³ + 1
When x → ∞, g(x) → ∞ and
When x → – ∞, g(x) → – ∞
∴ g does not attain a maximum value or a minimum value.
Method II
g(x) = x³ + 1 g'(x) = 3x²
g'(x) = 0 ⇒ 3x² = 0 ⇒ x = 0
Sign of g'(x) = 3x²
g does not attain a maximum value or a minimum value
Method III
g(x) = x³ + 1
g'(x) = 3x² and g”(x) = 6x
g'(x) = 0 ⇒ 3x² = 0 ⇒ x = 0
g'(0) = 6(0) = b
Hence the second derivative test fails.
So use the first derivative test (Method II).
Exercise 6.5 Class 12 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 min. value of f(x)
now f(x) = – 1
⇒ |x + 2|- 1
⇒ |x + 2| = 0
⇒ x = – 2
(ii) We have g(x) = – |x + 1| + 3 ∀x ∈ R
Now | x + 1| ≥ 0 ∀x ∈ R
-|x + 1| + 3 ≤3 ∀x ∈ R
So 3 is the minimum value of f(x).
Now f(x) = 3
⇒ -|x + 1| + 3
⇒ |x + 1| = 0
⇒ x = – 1.
(iii) Thus maximum value of f(x) is 6 and minimum value is 4.
(iv) Let f(x) = |sin 4x + 3|
Maximum value of sin 4x is 1
∴ Maximum value of |sin(4x+3)| is |1+3| = 4
Minimum value of sin 4x is -1
∴ Minimum value of f(x) is |-1+3| = |2|= 2
(v) h(x) = x + 1 x ∈ (- 1, 1)
i.e., – 1 < x < 1
∴ – 1 + 1 < x + 1 < 2
i.e., 0 < x + 1 < 2
i.e., 0 < h(x) < 2
Hence h(x) does not attain a maximum value or a minimum value
Another Method
h(x) = x + 1
∴ h'(x) = 1
h'(x) ≠ 0 for x ∈ (- 1, 1)
∴ h(x) does not attain a maximum value or a minimum value in ( – 1, 1)
6.5 Class 12 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) = sinx + cosx, 0 < x < \(\frac { \pi }{ 2 }\)
(iv) f(x) = sin x – cos x, 0 < x < 2π
(v) f(x) = x3 – 6x2 + 9x + 15
(vi) g(x) = \(\frac { x }{ 2 } +\frac { 2 }{ x }\), x > 0
(vii) g(x) = \(\frac { 1 }{ { x }^{ 2 }+2 }\)
(viii) f(x) = \(x\sqrt { 1-x } \), x < 1
Solution:
(i) f(x) = x2
Now f'(x) = 0 ⇒ 2x = 0 i.e., x = 0
At x = 0; When x is slightly < 0, f’ (x) is -ve When x is slightly > 0, f(x) is +ve
∴ f(x) changes sign from -ve to +ve as x increases through 0.
⇒ f’ (x) has a local minimum at x = 0 local minimum value f(0) = 0.
(ii) g(x) = x3 – 3x
g'(x) = 3x² – 3 = 3(x² – 1)
g'(x) = 0 ⇒ 3(x² – 1) = 0
⇒ 3(x + 1) (x – 1) = 0
⇒ x = – 1 or x = 1
g”(x) = 6x
g”(-1) = 6(- 1) = – 6 < 0 g”(1) = 6(1) = 6 > 0
∴ By the second derivative test,
x = – 1 is a point of local maximum,
x = 1 is a point of local minimum
The local maximum value = g(-1)
= (-1)3 – 3(- 1) = 2
The local minimum value = g(1)
= (1)3 – 3(1) = – 2
(iii) h(x) = sinx + cosx, 0 < x < \(\frac { \pi }{ 2 }\)
h(x) = cos x – sin x
h(x) = 0 ⇒ cos x – sin x = 0
Therefore, by second derivative test, x = \(\frac { \pi }{ 4 }\)
is a local maximum point
The local maximum value = h\(\frac { \pi }{ 4 }\)
= \(\sin \frac{\pi}{4}+\cos \frac{\pi}{4}\)
= \(=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
(iv) f(x) = sin x – cos x, 0 < x < 2π
f'(x) = cosx + sinx
f”(x) = – sinx + cosx
f'(x) = 0 ⇒ cosx + sinx 0
⇒ sinx = – cosx
(v) f(x) = x3 – 6x2 + 9x + 15
f'(x) = 3x² – 12x + 9 = 3(x² – 4x + 3)
f”(x) = 6x – 12
f'(x) = 0 ⇒ 3(x² – 4x + 3) = 0
⇒ 3(x – 1) (x – 3) = 0
⇒ x = 1 or x = 3
At x = 1, f”(x) = 6(1) – 12 = 6 – 12 < 0 At x = 3, f”(x) = 6(3) – 12 = 18 – 12 >0
f has a local maxima at x = 1 and a local minima at x = 3
The local maximum value = f(1)
= (1)³ – 6 (1)² + 9(1) + 15 = 19
The local minimum value = f(3)
= (3)³ – 6(3)² + 9(3) +15 = 15
(vi) g(x) = \(\frac { x }{ 2 } +\frac { 2 }{ x }\), x > 0
(vii) g(x) = \(\frac { 1 }{ { x }^{ 2 }+2 }\)
(viii) f(x) = \(x\sqrt { 1-x } \), x < 1
Ex6.5 Class 12 Question 4.
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) f'(x) = ex;
Since f’ (x) ≠ 0 for any value of x.
So f(x) = ex does not have a max. or min.
(ii) f’ (x) = \(\frac { 1 }{ x }\); Clearly f’ (x) ≠ 0 for any value of x.
So, f’ (x) = log x does not have a maximum or a minimum.
(iii) We have f(x) = x3 + x2 + x + 1
⇒ f’ (x) = 3x2 + 2x + 1
Now, f’ (x) = 0 = > 3x2 + 2x + 1 = 0
\(x=\frac { -2\pm \sqrt { 4-12 } }{ 6 } =\frac { -1+\sqrt { -2 } }{ 3 } \)
i.e. f'(x) = 0 at imaginary points
i.e. f'(x) ≠ 0 for any real value of x
Hence, there is neither max. nor min.
Ex 6.5 Class 12 Maths Ncert Solutions 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, π]
(iii) f(x) = \(4x-\frac { 1 }{ 2 } { x }^{ 2 },x\in \left[ -2,\frac { 9 }{ 2 } \right] \)
(iv) f(x) = \({ (x-1) }^{ 2 }+3,x\in \left[ -3,1 \right] \)
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)³ = 0
and f(0) = (2) = 8
Hence, the absolute maximum value of f (x) is 8 which it attained at x = 2 and absolute minimum value of f(x) = – 8 which is attained at x = -2.
(ii)
(iii) f(x) = \(4x-\frac { 1 }{ 2 } { x }^{ 2 },x\in \left[ -2,\frac { 9 }{ 2 } \right] \)
f'(x) = 4 – x
f'(x) = 0 ⇒ 4 – x = 0 ⇒ x = 4
f(- 2) = 4(-2) – \(\frac { 1 }{ 2 }\)(-2)² = – 8 – 2 = – 10
f(4) = 4(4) – \(\frac { 1 }{ 2 }\)(4)² = 16 – 8 = 8
(iv) f(x) = \({ (x-1) }^{ 2 }+3,x\in \left[ -3,1 \right] \)
f'(x) = 2(x – 1)
f'(x) = 0 ⇒ 2(x – 1) = 0 ⇒ x = 1
f(- 3) = (- 3 – 1)² + 3 = 16 + 3 = 19
f(1) = (1 – 1)² + 3 = 0 + 3 = 3
∴ Absolute minimum value = Min{19, 3} = 3
Absolute maximum value = Max{19, 3} = 19
Exercise 6.5 Class 12 Maths 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 in 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 = \(-\frac { 2 }{ 3 }\),
p'(x) changes sign from +ve to -ve.
⇒ p (x) has maximum value at x = \(-\frac { 2 }{ 3 }\)
Maximum Profit = 41 + 16 – 8 = ₹ 49.
Ex 6.5 Class 12 Maths 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
at x = 2, Minimum value = – 39.
Class 12 Ex 6.5 Question 8.
At what points in the interval [0, 2π], does the function sin 2x attain its maximum value?
Solution:
We have f (x) = sin 2x in [0, 2π], f'(x) = 2 cos 2 x
For maxima and minima f’ (x) = 0 ⇒ cos 2 x = 0
Maximum value of f =
Max{1, – 1, 1, – 1, 0} = 1
∴ sin 2x attains the maximum value at x = \(\frac { π }{ 4 }\), x = \(\frac { 5π }{ 4 }\)
6.5 Maths Class 12 Question 9.
What is the maximum value of the function sin x + cos x?
Solution:
∴ The maximum value of f(x) is \(\sqrt{2}\)
Class 12 Maths 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 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
∴ max.f(x) = 139 at x = – 2.
6.5 Class 12 Maths 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:
∵ f(x) = x4 – 62x2 + ax + 9, x ∈ [0, 2]
∴ 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 at x = 1 when a = 120.
Class 12 Maths Ncert Solutions Chapter 6 Exercise 6.5 Question 12.
Find the maximum and minimum values of x + sin 2x on [0, 2π]
Solution:
∴ f(x) = x + sin2x ∈ [0, 2π]
∴ f’ (x) = 1 + 2 cos2x
For maxima and minima f'(x) = 0
Exercise 6.5 Class 12 Ncert Solutions Question 13.
Find two numbers whose sum is 24 and whose product is as large as possible.
Solution:
Let the required numbers hex and (24 – x)
∴ Their product, p = x(24 – x) = 24x – x²
Now \(\frac { dp }{ dx }\) = 0 ⇒24 – 2x = 0 ⇒ x = 12
Also \(\frac { { d }^{ 2 }p }{ { dx }^{ 2 } } \) = – 2 < 0:
⇒ p is max at x = 12
Hence, the required numbers are 12 and (24 – 12) i.e. 12.
Ncert Solutions For Class 12 Maths Chapter 6 Exercise 6.5 Question 14.
Find two positive numbers x and y such that x + y = 60 and xy³ is maximum.
Solution:
i. x + y = 60
y = 60 – x
z = xy³
∴ z = x(60 – x)³
ii. Differentiating z with respect to x, dz
\(\frac { dz }{ dx }\) = x(3(60 – x)²(- 1)) + (60 – x)³
= (60 – x)²(- 3x + 60 – x)
\(\frac { dz }{ dx }\) = (60 – x)²(60 – 4x)
\(\frac{d^{2} z}{d x^{2}}\) = (60 – x)² (- 4) + (60 – 4x)(2(60 – x)(-1))
= -[4(60 – x)² + 2(60 – x)(60 – 4x)]
For maxima, \(\frac { dz }{ dx }\) = 0 dx
(60 – x)²(60 – 4x) = 0
∴ x = 60 or x = 15
When x = 60, y = 0
When x = 15, y = 45
At x = 60,
\(\frac{d^{2} z}{d x^{2}}\) = – [4(60 – 60)² + 2(60 – 60)(60 – 240)] = 0
At x = 15,
\(\frac{d^{2} z}{d x^{2}}\) = – [4(60 – 15)² + 2(60 – 15)(60 – 60)]
= – [4(45)²] < 0
∴ z is maximum at x = 15
Hence the numbers are 15 and 45.
Ex 6.5 Class 12 Ncert Solutions Question 15.
Find two positive numbers x and y such that their sum is 35 and the product x2 y5 is a maximum.
Solution:
Given x + y = 35
∴ y = 35 – x
Let P = x²y5
P = x²(35 – x)5
Differentiating w.r.t. x,
\(\frac { dP }{ dx }\) = x²(35 – x)4(- 1) + (35 – x)5 2x
= x (35 – x)4[- 5x + 70 – 2x]
\(\frac { dP }{ dx }\) = x(35 – x)4 (70 – 7x)
\(\frac{d^{2} P}{d x^{2}}\) = x(35 – x)4(- 7) + x(70 – 7x)4(35 – x)³(-1) dx + (35 – x)4(70 – 7x)(1)
For maxima, \(\frac { dP }{ dx }\) = 0
⇒ x(70 – 7x) (35 – x)4 = 0
⇒ x = 0, x = 10, x = 35
x = 0 and x = 35 are not possible
∴ x = 10
When x = 10, y = 35 – 10 = 25
At x = 10, \(\frac{d^{2} P}{d x^{2}}\) < 0
∴ P is maximum at x = 10
Hence the x²y is maximum when x = 10
and y = 25
Ncert Maths Class 12 Exercise 6.5 Solutions 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
Hence, the required numbers are 8 and (16 – 8) i.e. 8 and 8.
Class 12 Maths Chapter 6 Exercise 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
x² – 12 x + 27 = 0 ⇒ (x – 9)(x – 3) = 0
x = 9 or x = 3,
but x = 9 is not possible.
∴ x = 3
When x = 3, \(\frac{d^{2} V}{d x^{2}}\) = 4[6(3) – 36] < 0
∴ Volume is maximum when x = 3
∴ Side of the square to be cut off = 3 cm.
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 comer 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 x be the side of the square cut off and
V be the volume of the box.
Length of the box = 45 – 2x
Breadth of the box = 24 – 2x
Height of the box = x
Volume V = (45 – 2x) (24 – 2x)x
= (1080 – 90x – 48x + 4x²)x
V = 4x³ – 138x² + 1080x
\(\frac { dV }{ dx }\) = 12x² – 276x + 1080
\(\frac{d^{2} V}{d x^{2}}\) = 24x – 276 dx
For maxima, \(\frac { dV }{ dx }\) = 0
⇒ 12x² – 276x + 1080 = 0
⇒ x² – 23x + 90 = 0
(x – 18)(x – 5) = 0
x = 18 or x = 5 But x = 18 is not possible
∴ x = 5
At x = 5, \(\frac{d^{2} V}{d x^{2}}\) = 24(5) – 276
= 120 – 276 < 0
By second derivative test, V is maximum when x = 5.
Hence the volume of the box will be maxi¬mum when 5 cm is cut off from the side of the square.
Question 19.
Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
Solution:
Let x and y be the length and breadth of the rectangle inscribed in a circle of radius r.
∴ x² + y² = (2a)² ⇒ x² + y² = 4a² … (i)
∴ Perimeter = 2 (x + y)
∴ The area is maximum when the rectangle is a square.
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 x be the radius, y be the height, S be the surface area and V be the volume of the right circular cylinder.
S = 2πxy + 2πx² … (1)
∴ y = 2x
i.e., height = 2x i.e., radius = diameter
∴ Volume is maximum, when height of the cylinder is equal to the diameter of the cylinder.
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.
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 one part be of length x, then the other part = 28 – x
Let the part of the length x be converted into a circle of radius r.
Let r be the radius of the circle
Circumference of the circle is 28 – x
∴ 28 – x = 2πr,
Total area of the square and circle is given by
Thus A is minimum
Length of the square piece is \(\frac { 112 }{ π+4 }\) the circular piece = 28 – \(\frac { 112 }{ π+4 }\) = \(\frac { 28 }{ π+4 }\)
Question 23.
Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is \(\frac { 8 }{ 27 }\) of the volume of the sphere.
Solution:
Let h, r be the height and base radius of the inscirbed cone with volume V.
From the figure, (h – R)² + r² = R²
h² – 2RA + R² + r² = R²
r² = 2Rh – h² … (i)
For maxima V’ = 0
V’ = 0 ⇒ \(\frac { 1 }{ 3 }\)π(4Rh – 3h²) = 0
i.e., h (4R – 3h) = 0
⇒ h = 0 or h = \(\frac { 4R }{ 3 }\)
Since h ≠ 0, h = \(\frac { 4R }{ 3 }\)
When h = \(\frac { 4R }{ 3 }\) , V” = \(\frac { -4πR }{ 3 }\) < 0
V is maximum when h = \(\frac { 4R }{ 3 }\)
From (ii), Maximum volume
∴ Volume of the largest cone = \(\frac { 8 }{ 27 }\) of the volume of the sphere.
Question 24.
Show that die right circular cone of least curved surface and given volume has an altitude equal to \(\sqrt{2}\) time the radius of the base.
Solution:
Let r and h be the radius and height of the cone
Hence h = \(\sqrt{2}\)
The curved surface area of the cone is minimum when altitude equals to \(\sqrt{2}\) times the radius of the base.
Question 25.
Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan-1\(\sqrt{2}\).
Solution:
Let v be the volume, l be the slant height and 0 be the semi vertical angle of a cone.
Let h, l, r be the height, slant height and radius of the cone and a be the semivertical angle. Given l is a constant.
From the figure, r² = l² – h²
Volume, v = \(\frac { 1 }{ 3 }\)πr²h
Question 26.
Show that semi-vertical angle of right circular cone of given surface area and maximum volume is \({ sin }^{ -1 }\left( \frac { 1 }{ 3 } \right) \)
Solution:
Let r be the radius, h be the altitude, l be the slant height, V be the volume and S be the surface area and 0 be the semi -vertical angle of a right circular cone.
S = πr² + πrl = constant
∴ θ = \({ sin }^{ -1 }\left( \frac { 1 }{ 3 } \right) \)
i.e., For a given surface area, the volume of a right circular cone is maximum when the semi-vertical angle is \({ sin }^{ -1 }\left( \frac { 1 }{ 3 } \right) \)
Question 27.
The point on die curve x² = 2y which is nearest to the point (0, 5) is
(a) (2 \(\sqrt{2}\), 4)
(b) (2 \(\sqrt{2}\), 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]
Question 28.
For all real values of x, the minimum value of \(\frac { 1-x+{ x }^{ 2 } }{ 1+x+{ x }^{ 2 } } \)
(a) 0
(b) 1
(c) 3
(d) \(\frac { 1 }{ 3 }\)
Solution:
(d) Let \(y=\frac { 1-x+{ x }^{ 2 } }{ 1+x+{ x }^{ 2 } } \)
Question 29.
The maximum value of \({ \left[ x\left( x-1 \right) +1 \right] }^{ \frac { 1 }{ 3 } },0\le x\le 1\) is
(a) \({ \left( \frac { 1 }{ 3 } \right) }^{ \frac { 1 }{ 3 } } \)
(b) \(\frac { 1 }{ 2 } \)
(c) 1
(d) 0
Solution:
(c) Let y = \({ \left[ x\left( x-1 \right) +1 \right] }^{ \frac { 1 }{ 3 } },0\le x\le 1\)
