CBSE Class 12

These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-6-ex-6-3/

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.3

Ex 6.3 Class 12 Question 1.
Find the slope of the tangent to the curve y = 3x⁴ – 4x at x = 4.
Solution:
y = 3x⁴ – 4x
Differentiating w.r.t. x,
∴ \(\frac { dy }{ dx }\) = 12x³ – 4
∴ Req. slope = \({ \left( \frac { dy }{ dx } \right) }_{ x=4 }\)
= 12(4³) – 4 = 764.

Exercise 6.3 Class 12 Question 2.
Find the slope of the tangent to the curve y = \(\frac { x-1 }{ x-2 }\), x ≠ 2 at x = 10.
Solution:
y = \(\frac { x-1 }{ x-2 }\), x ≠ 2
Differentiating w.r.t. x,

6.3 Class 12 Question 3.
Find the slope of the tangent to curve y = x³ – x + 1 at the point whose x-coordinate is 2.
Solution:
The curve is y = x³ – x + 1
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = 3x² – 1
∴ slope of tangent = \(\frac { dy }{ dx }\)
= 3x² – 1
= 3 x 2² – 1
= 11

Ex6.3 Class 12 Question 4.
Find the slope of the tangent to the curve y = x³ – 3x + 2 at the point whose x-coordinate is 3.
Solution:
The curve is y = x³ – 3x + 2
\(\frac { dy }{ dx }\) = 3x² – 3
∴ slope of tangent = \(\frac { dy }{ dx }\)
= 3x² – 3
= 3 x 3² – 3
= 24

Ex 6.3 Class 12 Maths Ncert Solutions Question 5.
Find the slope of the normal to the curve x = a cos³ θ, y = a sin³ θ at θ = \(\frac { \pi }{ 4 } \).
Solution:
y = a sin³θ and x = a cos³θ
Differentiating w.r.t. x,

Exercise 6.3 Class 12 Maths Question 6.
Find the slope of the normal to the curve x = 1 – a sin θ, y = b cos² θ at θ = \(\frac { \pi }{ 2 } \)
Solution:
x = 1 – a sin θ and y = b cos² θ
Differentiating w.r.t. x,

6.3 Maths Class 12 Question 7.
Find points at which the tangent to the curve y = x³ – 3x² – 9x + 7 is parallel to the x-axis.
Solution:
y = x³ – 3x² – 9x + 7
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = 3(x – 3)(x + 1)
Tangent is parallel to x-axis if the slope of tangent = 0
or \(\frac { dy }{ dx }=0\)
⇒ 3(x + 3)(x + 1) = 0
⇒ x = – 1, 3
when x = – 1, y = 12 & When x = 3, y = – 20
Hence the tangent to the given curve are parallel to x-axis at the points (- 1, – 12), (3, – 20)

Exercise 6.3 Class 12 Maths Ncert Solutions Question 8.
Find a point on the curve y = (x – 2)² at which the tangent is parallel to the chord joining the points (2,0) and (4,4).
Solution:
The equation of the curve is y = (x – 2)²
Differentiating w.r.t x
\(\frac { dy }{ dx }\) = 2(x – 2 )
The point A and B are (2,0) and (4,4) respectively.

Slope of AB = \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }\)
= \(\frac { 4-0 }{ 4-2 } =\frac { 4 }{ 2 } \) = 2 … (i)
Slope of the tangent = 2 (x – 2) ….(ii)
from (i) & (ii) 2 (x – 2) = 2
∴ x – 2 = 1 or x = 3
when x = 3,y = (3 – 2)² = 1
∴ The tangent is parallel to the chord AB at (3, 1)

6.3 Class 12 Maths Question 9.
Find the point on the curve y = x³ – 11x + 5 at which the tangent is y = x – 11.
Solution:
Here, y = x³ – 11x + 5
⇒ \(\frac { dy }{ dx }\) = 3x² – 11
The slope of tangent line y = x – 11 is 1
∴ 3x² – 11 = 1
⇒ 3x² = 12
⇒ x² = 4, x = ±2
When x = 2, y = – 9 & when x = -2,y = -13
But (-2, -13) does not lie on the curve
∴ y = x – 11 is the tangent at (2, -9)

Ncert Class 12 Maths Exercise 6.3 Solutions Question 10.
Find the equation of all lines having slope -1 that are tangents to the curve y = \(\frac { 1 }{ x-1 }\), x ≠ 1
Solution:
y = \(\frac { 1 }{ x-1 }\), x ≠ 1
Differentiating w.r.t x
\(\frac { dy }{ dx }\) = \(\frac{-1}{(x-1)^{2}}\)
Since the tangent have slope – 1,
\(\frac { dy }{ dx }\) = – 1 ⇒ \(\frac{-1}{(x-1)^{2}}\) = – 1
⇒ (x – 1)² = 1 ⇒ x – 1 = ± 1
⇒ x = 2 or x = 0
When x = 0, y = \(\frac { 1 }{ 0 – 1 }\) = – 1
When x = 2, y = \(\frac { 1 }{ 2 – 1 }\) = 1
∴ The required points are (0, – 1) and (2, 1)
Equation of the tangent at (0, – 1) is
y – 1 = – 1 (x – 0) ⇒ x + y + 1 = 0
Equation of the tangent at (2, 1) is
y – 1 = – 1(x – 2) ⇒ x + y – 3 = 0

Class 12 Ex 6.3 Question 11.
Find the equation of ail lines having slope 2 which are tangents to the curve y = \(\frac { 1 }{ x-3 }\), x ≠ 3.
Solution:
Here y = \(\frac { 1 }{ x-3 }\)
\(\frac { dy }{ dx } ={ (-1)(x-3) }^{ -2 }=\frac { -1 }{ { (x-3) }^{ 2 } } \)
∵ slope of tangent = 2
\(\frac { -1 }{ { (x-3) }^{ 2 } } =2\Rightarrow { (x-3) }^{ 2 }=-\frac { 1 }{ 2 } \)
Which is not possible as (x – 3)² > 0
Thus, no tangent to y = \(\frac { 1 }{ x-3 }\)
Hence there is no tangent to the curve with slope 2.

Class 12 Maths Ex 6.3 Question 12.
Find the equations of all lines having slope 0 which are tangent to the curve y = \(\frac { 1 }{ { x }^{ 2 }-2x+3 }\)
Solution:
y = \(\frac { 1 }{ { x }^{ 2 }-2x+3 }\)
Differentiating w.r.t x
∴ \(\frac { dy }{ dx }\) = \(\frac{-(2 x-2)}{\left(x^{2}-2 x+3\right)^{2}}\)
Since the slope of the tangent is zero,
\(\frac { dy }{ dx }\) = 0

Class 12 Maths Chapter 6 Exercise 6.3 Question 13.
Find points on the curve \(\frac { { x }^{ 2 } }{ 9 } + \frac { { y }^{ 2 } }{ 16 } \) = 1 at which the tangents are
i. parallel to x-axis
ii. parallel to y-axis
Solution:
The equation of the curve is \(\frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 16 }\) = 1 … (i)
Differentiating both sides w.r.t x
\(\frac { 2 }{ 9 }\)x + \(\frac { 2y }{ 16 }\)\(\frac { dy }{ dx }\) = 0
∴ \(\frac { dy }{ dx }\) = \(\frac { -16x }{ 9y }\)

i. Since the tangent is parallel to x axis,
\(\frac { dy }{ dx }\) = 0 ⇒ \(\frac{-16x}{9y}\)
Hence x = 0
When x = 0, we get \(\frac{0}{9}\) + \(\frac { { y }^{ 2 } }{ 16 }\) = 1
⇒ y² = 16 ⇒ y = ±4
∴ (0,4) and (0, – 4) are the points at which the tangents are parallel to the x axis

ii. Since the tangent is parallel to y axis, its slope is not defined, then the normal is parallel to x-axis whose slope is zero.

∴ (3, 0) and (- 3, 0) are the points at which the tangents are parallel to they axis.

Class 12 Ch 6 Ex 6.3 Question 14.
Find the equations of the tangent and normal to the given curves at the indicated points:
(i) y = x⁴ – 6x³ + 13x² – 10x + 5 at (0,5)
(ii) y = x⁴ – 6x³ + 13x² – 10x + 5 at (1,3)
(iii) y = x³ at (1, 1)
(iv) y = x² at (0,0)
(v) x = cos t, y = sin t at t = \(\frac { \pi }{ 4 } \)
Solution:
(i) y = x⁴ – 6x³ + 13x² – 10x + 5
Differentiating w.r.t x

Equation of the tangent at (0, 5) is
(y- 5) = – 10(x- 0)
i.e., 10x+y-5 = 0
Equation of the normal at (0, 5) is
(y – 5) = – 10(x – 0)
i.e., x – 10y + 50 = 0

(ii) y = x⁴ – 6x³ + 13x² – 10x + 5
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = 4x³ – 18x² + 26x – 10
\(\frac { dy }{ dx }\)_{(1, 3)} = 4(1) – 18(1) + 26(1) – 10 = 2
At (1, 3), the slope of the tangent = 2
At (1, 3), the slope of the normal = \(\frac { – 1 }{ 2 }\)
Equation of the tangent at (1, 3) is y – 3 = 2( x – 1)
i.e., y = 2x + 1
Equation of the normal at (1, 3) is
y – 3 = y(x – 1)
i.e., x + 2y – 7 = 0

(iii) y = x³
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = 3x²
At(1, 1), \(\frac { dy }{ dx }\) = 3 x 1² = 3
At (1, 1), the slope of the tangent = 3
Equation of the tangent at(1, 1) is
y – 1 = 3(x – 1)
i.e., y = 3x – 2
At (1, 1), the slope o fthe normal = \(\frac { -1 }{ 3 }\)
Equation of the normal at (1, 1) is
y – 1 = \(\frac { -1 }{ 3 }\) (x – 1)
i.e., x + 3y – 4 = 0

(iv) y = x²
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = 2x
[ \(\frac { dy }{ dx }\) ]_{(0, 0)} = 0
At (0, 0), the slope of the tangent = 0
Equation of the tangent at (0, 0) is
y – 0 = 0(x – 0)
i.e., y = 0 or the x axis
Since the tangent is the x axis, the normal is parallel to y axis.
∴ The equation of the normal at (0, 0) is x = 0, since the normal passes through (0,0)

(v) y = sin t and x = cos t
Differentiating w.r.t. t, we get \(\frac { dy }{ dt }\) = cos t

Ncert Solutions For Class 12 Maths Chapter 6 Exercise 6.3 Question 15.
Find the equation of the tangent line to the curve y = x² – 2x + 7 which is
(i) parallel to the line 2x – y + 9 = 0
(ii) perpendicular to the line 5y – 15x = 13.
Solution:
y = x² – 2x + 7
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = 2x – 2
\(\frac { dy }{ dx }\) = 2(x – 1) … (1)
(i) Slope of the line 2x – y + 9 = 0 is 2
Since the tangent to the curve is parallel to the line 2x – y + 9 = 0, we get
\(\frac { dy }{ dx }\) = 2
From (1), we get 2(x – 1) = 2 ⇒ x = 2
When x = 2, y = 2² – 2 x 2 + 7 = 7
∴ At (2, 7) the tangent is parallel to 2x – y + 9 = 0
The equation of the tangent to the given curve at (2, 7) is y – 7 = 2(x – 2)
i.e., 2x – y + 3 = 0 or y – 2x – 3 = 0
Slope of the line – 15x = 13 is 3

(ii) Since the tangent to the curve is perpendicular to the line 5y – 15x = 13,
\(\frac { dy }{ dx }\) = \(\frac { -1 }{ 3 }\)
From (1), we get 2(x – 1) = \(\frac { -1 }{ 3 }\)

The tangent is perpendicular to the line 5y – 15x = 13
The equation of the tangent to the given

Question 16.
Show that the tangents to the curve y = 7x³ + 11 at the points where x = 2 and x = – 2 are parallel.
Solution:
Here, y = 7x³ + 11
⇒ x \(\frac { dy }{ dx }\) = 21 x²
Now m₁ = slope at x = 2 is
\({ \left( \frac { dy }{ dx } \right) }_{ x=2 }\)
= 21 x 2² = 84
and m₂ = slope at x = -2 is
\({ \left( \frac { dy }{ dx } \right) }_{ x=-2 }\)
= 21 x (- 2)² = 84
Hence, m₁ = m₂ Thus, the tangents to the given curve at the points where x = 2 and x = – 2 are parallel

Question 17.
Find the points on the curve y = x³ at which the slope of the tangent is equal to the y-coordinate of the point
Solution:
y = x³
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = 3x²
The slope of the tangent = 3x²
Since slope of the tangent = y coordinate,
3x² = y or 3x² = x³
= 3x² – x³ = 0 = x²(3 – x) = 0
⇒ x = 3 or x = 0
When x = 0, y = (0)³ = O
When x = 3, y= (3)³ = 27
∴ (0, 0) and (3, 27) are the points at which the slope of the tangent is equal to the y coordinate of the point.

Question 18.
For the curve y = 4x³ – 2x⁵, find all the points at which the tangent passes through the origin.
Solution:
Let, (h, k) be the point at which the tangent passes through the origin.
y = 4x³ – 2x⁵
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = 12x² – 10x⁴
At (h, k) the slope of the tangent = \(\frac { dy }{ dx }\)_(h,k) = 12h² – 10h⁴
Equation of the tangent at (h, k) is
y – k = (12h² – 10h⁴) (x – h)
Since the tangent passes through the origin,
(0 – k) = (12h² – 10h⁴) (0 – h)
⇒ k = 12h³ – 10h⁵ … (1)
(h, k) is a point on the curve y = 4x³ – 2x⁵
∴ k = 4h³ – 2h⁵ … (2)
From (1) & (2) we get
12h³ – 10h⁵ = 4h³ – 2h⁵
⇒ 8h³ – 8h⁵ = 0
⇒ 8h³ (1 – h²) = 0
⇒ 8h³ (1 – h) (1 +h) = 0
⇒ h = 0 or h= 1 or h = – 1
When h = 0, k = 0
When h = 1, k = 2
When h = – 1, k = – 2
∴ (0, 0), (1, 2) and (- 1, – 2) are the points at which the tangent passes through the origin.

Question 19.
Find the points on the curve x² + y² – 2x – 3 = 0 at which the tangents are parallel to the x-axis.
Solution:
Here, x² + y² – 2x – 3 = 0
⇒ \(\frac { dy }{ dx } =\frac { 1-x }{ y }\)
Tangent is parallel to x-axis, if \(\frac { dy }{ dx }\) = 0 i.e.
if 1 – x = 0
⇒ x = 1
Putting x = 1 in (i)
⇒ y = ±2
Hence, the required points are (1,2), (1, -2) i.e. (1, ±2).

Question 20.
Find the equation of the normal at the point (am², am³ ) for the curve ay² = x³ .
Solution:
ay² = x³
Differentiating w.r.t. x,

Question 21.
Find the equation of the normal’s to the curve y = x³ + 2x + 6 which are parallel to the line x + 14y + 4 = 0.
Solution:
y = x³ + 2x + 6
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = 3x² + 2
Slope of the normal = \(\frac{-1}{3 x^{2}+2}\)
Slope of the line x + 14y + 4 = 0 is \(\frac { -1 }{ 14 }\)
Since the normal is parallel to x + 14y + 4 = 0
\(\frac{-1}{3 x^{2}+2}\) = \(\frac { -1 }{ 14 }\) ⇒ 3x² + 2 = 14
⇒ 3x² = 12 ⇒ x² = 4 ⇒ x = ± 2
When x = 2, y = 18 and when x = – 2, y = – 6
∴ At (2, 18) and (- 2, – 6), the normals are parallel to x + 14y + 4 = 0.
Slope of the normal = \(\frac { -1 }{ 14 }\)
Equation of the normal at (2, 18) is
y – 8 = \(\frac { -1 }{ 14 }\)(x – 2)
i.e., x + 14y – 254 = 0
Equation of the normal at ( -2, -6) is
y + 6 = \(\frac { -1 }{ 14 }\) (x + 2) i.e., x + 14y + 86 = 0

Question 22.
Find the equations of the tangent and normal to the parabola y² = 4ax at the point (at²,2at).
Solution:
y² = 4ax
Differentiating w.r.t. x,
2y\(\frac { dy }{ dx }\) = 4a

Equation of the tangent at (at², 2at) is
y – 2at = \(\frac { 1 }{ t }\)(x – at²)
yt – 2at² = x – at²
x – yt + at² = 0 or ty = x + at²
Slope of the normal at (at², 2at) is – t.
Equation of the normal at (at², 2at) is
y – 2at = – t(x – at²)
xt + y = at³ + 2at
y = – tx + 2at + at³

Question 23.
Prove that the curves x = y² and xy = k cut at right angles if 8k² = 1.
Solution:
The curve are
x = y² … (1)
xy = k … (2)
Let P be the point of intersection
Substituting x = y² in (2),

Since the two curves cut at right angles at P, the product of their slopes at P is – 1.

Taking the cubes we get k² = \(\frac { 1 }{ 8 }\) or 8k² = I

Question 24.
Find the equations of the tangent and normal to the hyperbola \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } }\) = 1 at the point (x₀, y₀).
Solution:

Question 25.
Find the equation of the tangent to the curve y = \(\sqrt { 3x – 2 } \) which is parallel to the line 4x – 2y + 5 = 0.
Solution:
y = \(\sqrt { 3x – 2 } \)
Differentiating w.r.t. x,
\(\frac { dy }{ dx }\) = \(\frac{3}{2 \sqrt{3 x-2}}\)
Slope of the tangent = \(\frac{3}{2 \sqrt{3 x-2}}\)
Slope of the line 4x – 2y + 5 = 0 is 2
Since the tangent is parallel to the line dy
4x – 2y + 5 = 0, we get \(\frac { dy }{ dx }\) = 2

∴ At(\(\frac { 41 }{ 48 }\), \(\frac { 3 }{ 4 }\)), the tangent is parallel to the line 4x – 2y + 5 = 0
Equation of the tangent at (\(\frac { 41 }{ 48 }\), \(\frac { 3 }{ 4 }\)) is
y – \(\frac { 3 }{ 4 }\) = 2(x – \(\frac { 41 }{ 48 }\)
24y – 18 = 48x – 41
i.e., 48x – 24y = 23

Question 26.
The slope of the normal to the curve y = 2x² + 3 sin x at x = 0 is
(a) 3
(b) \(\frac { 1 }{ 3 }\)
(c) -3
(d) \(-\frac { 1 }{ 3 }\)
Solution:
(d) ∵ y = 2x² + 3sinx
∴ \(\frac { dy }{ dx }\) = 4x+3cosx
at
x = 0, \(\frac { dy }{ dx }\) = 3
∴ slope = 3
⇒ slope of normal is = \(\frac { 1 }{ 3 }\)

Question 27.
The line y = x + 1 is a tangent to the curve y² = 4x at the point
(a) (1,2)
(b) (2,1)
(c) (1,-2)
(d) (-1,2)
Solution:
(a) The curve is y² = 4x,
∴ \(\frac { dy }{ dx } =\frac { 4 }{ 2y } =\frac { 2 }{ y } \)
Slope of the given line y = x + 1 is 1
∴ \(\frac { 2 }{ y }\) = 1
y = 2 Putting y = 2 in
⇒ y² = 4x
⇒ 2² = 4x
⇒ x = 1
∴ The Point of contact is (1, 2)

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) = e^x
(ii) f(x) = log x
(iii) h(x) = x³ + x² + x + 1
Solution:
(i) f'(x) = e^x;
Since f’ (x) ≠ 0 for any value of x.
So f(x) = e^x 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 = \({ \left[ x\left( x-1 \right) +1 \right] }^{ \frac { 1 }{ 3 } },0\le x\le 1\)

These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.4 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-6-ex-6-4/

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.4

Ex 6.4 Class 12 NCERT Solutions Question 1.
Using differentials, find the approximate value of each of the following up to 3 places of decimal.
i. \(\sqrt { 25.3 } \)
ii. \(\sqrt { 49.5 } \)
iii. \(\sqrt { 0.6 } \)
iv. \({ \left( 0.009 \right) }^{ \frac { 1 }{ 3 } }\)
v. \({ \left( 0.999 \right) }^{ \frac { 1 }{ 10 } }\)
vi. \({ \left( 15 \right) }^{ \frac { 1 }{ 4 } }\)
vii. \({ \left( 26 \right) }^{ \frac { 1 }{ 3 } }\)
viii. \({ \left( 255 \right) }^{ \frac { 1 }{ 4 } }\)
ix. \({ \left( 82 \right) }^{ \frac { 1 }{ 4 } }\)
x. \({ \left( 401 \right) }^{ \frac { 1 }{ 2 } }\)
xi. \({ \left( 0.0037 \right) }^{ \frac { 1 }{ 2 } }\)
xii. \({ \left( 26.57 \right) }^{ \frac { 1 }{ 3 } }\)
xiii. \({ \left( 81.5 \right) }^{ \frac { 1 }{ 4 } }\)
xiv. \({ \left( 3.968 \right) }^{ \frac { 3 }{ 2 } }\)
xv. \({ \left( 32.15 \right) }^{ \frac { 1 }{ 5 } }\)
Solution:

iv. \({ \left( 0.009 \right) }^{ \frac { 1 }{ 3 } }\)

v. \({ \left( 0.999 \right) }^{ \frac { 1 }{ 10 } }\)

∆y is approximately equal to dy
∴ (1) ⇒ \({ \left( 0.999 \right) }^{ \frac { 1 }{ 10 } }\) ≈ 1 – 0.0001 ≈ 0.9999
i.e., \({ \left( 0.999 \right) }^{ \frac { 1 }{ 10 } }\) ≈ 0.9999

vi. \({ \left( 15 \right) }^{ \frac { 1 }{ 4 } }\)

vii. \({ \left( 26 \right) }^{ \frac { 1 }{ 3 } }\)

viii. \({ \left( 255 \right) }^{ \frac { 1 }{ 4 } }\)

ix. \({ \left( 82 \right) }^{ \frac { 1 }{ 4 } }\)

x. \({ \left( 401 \right) }^{ \frac { 1 }{ 2 } }\)

xi. \({ \left( 0.0037 \right) }^{ \frac { 1 }{ 2 } }\)

xii. \({ \left( 26.57 \right) }^{ \frac { 1 }{ 3 } }\)

xiii. \({ \left( 81.5 \right) }^{ \frac { 1 }{ 4 } }\)

xv. \({ \left( 32.15 \right) }^{ \frac { 1 }{ 5 } }\)

Exercise 6.4 Class 12 NCERT Solutions Question 2.
Find the approximate value of f (2.01), where f (x) = 4x² + 5x + 2
Solution:
f(x+∆x) = f(2.01), f(x) = f (2) = 4.2² + 5.2 + 2 = 28,
f’ (x) = 8x + 5 Now, f(x + ∆x) = f(x) + ∆f(x)
= f(x) + f’ (x) . ∆x = 28 + (8x + 5) ∆x
= 28 + (16 + 5) x 0.01
= 28 + 21 x 0.01
= 28 + 0.21
Hence, f(2.01) ≈ 28 x 21.

6.4 Class 12 NCERT Solutions Question 3.
Find the approximate value of f (5.001), where f(x) = x³ – 7x² +15.
Solution:
Let x + ∆x = 5.001, x = 5 and ∆x = 0.001,
f(x) = f(5) = – 35
f(x + ∆x) = f(x) + ∆f(x) = f(x) + f'(x).∆x
= (x³ – 7x² + 15) + (3x² – 14x) × ∆x
f(5.001) = – 35 + (3 × 5² – 14 × 5) × 0.001
⇒ f (5.001) = – 35 + 0.005
= – 34.995.

Exercise 6.4 Class 12 Maths Question 4.
Find the approximate change in the volume V of a cube of side x metres caused by increasing the side by 1%.
Solution:
The side of the cube = x meters.
Increase in side = 1% = 0.01 × x = 0.01 x
Volume of cube V = x³
∴ ∆v = \(\frac { dv }{ dx }\) × ∆x
= 3x² × 0.01 x
= 0.03 x³ m³

Class 12 Maths Ex 6.4 Solutions Question 5.
Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%.
Solution:
The side of the cube = x m;
Decrease in side = 1% = 0.01 x
Increase in side = ∆x = – 0.01 x
Surface area of cube = 6x² m² = S
∴ \(\frac { ds }{ dx }\) × ∆x = 12x × (- 0.01 x)
= – 0.12 x² m².

Ex 6.4 Class 12 Maths Ncert Solutions Question 6.
If the radius of a sphere is measured as 7m with an error of 0.02 m, then find the approximate error in calculating its volume.
Solution:
Radius of the sphere = 7m : ∆r = 0.02 m.
Volume of the sphere V = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)
\(\Delta V=\frac { dV }{ dr } \times \Delta r=\frac { 4 }{ 3 } .\pi .3{ r }^{ 2 }\times \Delta r\)
= 4π × 7² × 0.02
= 3.92 πm³

Ex6.4 Class 12 NCERT Solutions Question 7.
If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area.
Solution:
Radius of the sphere = 9 m: ∆r = 0.03m
Surface area of sphere S = 4πr²
∆s = \(\frac { ds }{ dr }\) × ∆r
= 8πr × ∆r
= 8π × 9 × 0.03
= 2.16 πm².

Question 8.
If f (x) = 3x² + 15x + 5, then the approximate value of f (3.02) is
(a) 47.66
(b) 57.66
(c) 67.66
(d) 77.66
Solution:
(d) x + ∆x = 3.02, where x = 30, ∆x = 0.2,
∆f(x) = f(x + ∆x) – f(x)
⇒ f(x + ∆x) = f(x) + ∆f(x) = f(x) + f'(x) ∆x
Now f(x) = 3×2 + 15x + 5; f(3) = 77, f’ (x) = 6x + 15
f’ (3) = 33
∴ f (3.02) = 87 + 33 x 0.02 = 77.66

Question 9.
The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is
(a) 0.06 x³ m³
(b) 0.6 x³ m³
(c) 0.09 x³ m³
(d) 0.9 x³ m³
Solution:
(c) Side of a cube = x meters
Volume of cube = x³,
for ∆x ⇒ 3% of x = 0.03 x
Let ∆v be the change in v0l. ∆v = \(\frac { dv }{ dx }\) x ∆x = 3x² × ∆x
But, ∆x = 0.03 x
⇒ ∆v = 3x² x 0.03 x
= 0.09 x³m³

These NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-7-ex-7-6/

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.6

Question 1.
x sinx
Solution:
By part integration
∫x sinx dx = x(-cosx) – ∫1(-cosx)dx
= – x cosx + ∫cosxdx
= – x cosx + sinx + c

Question 2.
x sin 3x
Solution:
∫xsin 3x dx = x ∫sin 3 dx – ∫(\(\frac { d }{ dx }\)(x) ∫sin 3x dx)dx
= \(=x\left(\frac{-\cos 3 x}{3}\right)-\int 1\left(\frac{-\cos 3 x}{3}\right) d x\)
= \(\frac{-1}{3} x \cos 3 x+\frac{1}{9} \sin 3 x+C\)

Question 3.
\({ x }^{ 2 }{ e }^{ x }\)
Solution:
\(\int { { x }^{ 2 }{ e }^{ x } } dx={ x }^{ 2 }{ e }^{ x }-2{ x }{ e }^{ x }+2{ e }^{ x }+c\)
\(={ e }^{ x }\left( { x }^{ 2 }-2x+2 \right) +c\)

Question 4.
x logx
Solution:

Question 5.
x log 2x
Solution:

Question 6.
\({ x }^{ 2 }logx\)
Solution:

Question 7.
x sin^-1 x
Solution:

Question 8.
x tan^-1 x
Solution:
\(I=x\quad { tan}^{ -1 }x.\left( \frac { { x }^{ 2 } }{ 2 } \right) -\int { \frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } } .\frac { { x }^{ 2 } }{ 2 } dx\)
\(=\frac { { x }^{ 2 } }{ 2 } { tan }^{ -1 }x-\frac { 1 }{ 2 } \int { \left( 1-\frac { 1 }{ 1+{ x }^{ 2 } } \right) dx } \)
\(=\frac { { x }^{ 2 } }{ 2 } { tan }^{ -1 }x-\frac { 1 }{ 2 } x+\frac { 1 }{ 2 } { tan }^{ -1 }\)

Question 9.
x cos^-1 x
Solution:

Question 10.
(sin^-1 x)²
Solution:
Method I
Take 1 as the second function and integrate, w.r.t. x

Method II

Method III

Question 11.
\(\frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}}\)
Solution:

Question 12.
x sec²x
Solution:
∫x sec²x dx
= \(x \int \sec ^{2} x d x-\int\left(\frac{d}{d x}(x) \int \sec ^{2} x d x\right) d x\)
= x tan x -∫tan x dx
= x tan x – log |sec x| + C
= x tan x + log |cos x| + C

Question 13.
tan^-1 x
Solution:
\(\int { { tan }^{ -1 }xdx } =x{ tan }^{ -1 }x-\frac { 1 }{ 2 } \int { \frac { 2x }{ 1+{ x }^{ 2 } } dx } \)
\(=x{ tan }^{ -1 }x-\frac { 1 }{ 2 } log|1+{ x }^{ 2 }|+C \)

Question 14.
x(logx)²
Solution:

Question 15.
(x² + 1) log x
Solution:

Question 16.
\({ e }^{ x }(sinx+cosx)\)
Solution:

Question 17.
\(\frac { { xe }^{ x } }{ { (1+x) }^{ 2 } } \)
Solution:

Question 18.
\(\frac { { e }^{ x }(1+sinx) }{ 1+cosx } \)
Solution:

Question 19.
\({ e }^{ x }\left( \frac { 1 }{ x } -\frac { 1 }{ { x }^{ 2 } } \right) \)
Solution:

Question 20.
\(\frac { { (x-2)e }^{ x } }{ { (x-1) }^{ 3 } } \)
Solution:

Question 21.
\({ e }^{ 2x }sinx\)
Solution:

Question 22.
\(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\)
Solution:

Question 23.
\(\int { { x }^{ 2 }{ e }^{ { x }^{ 3 } } } dx\quad equals\)
(a) \(\frac { 1 }{ 3 } { e }^{ { x }^{ 3 } }+C\)
(b) \(\frac { 1 }{ 3 } +{ e }^{ { x }^{ 2 } }+C\)
(c) \(\frac { 1 }{ 2 } { e }^{ { x }^{ 3 } }+C\)
(d) \(\frac { 1 }{ 2 } { e }^{ { x }^{ 2 } }+C\)
Solution:
(a) \(\frac { 1 }{ 3 } { e }^{ { x }^{ 3 } }+C\)

Question 24.
\(\int { { e }^{ x }secx(1+tanx) } dx\quad equals\)
(a) \({ e }^{ x }cosx+C\)
(b) \({ e }^{ x }secx+C\)
(c) \({ e }^{ x }sinx+C\)
(d) \({ e }^{ x }tanx+C\)
Solution:
(b) \({ e }^{ x }secx+c\)

These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-6-ex-6-2/

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.2

Ex 6.2 Class 12 Question 1.
Show that the function given by f(x) = 3x + 17 is strictly increasing on R.
Solution:
f(x) = 3x + 17
Since f(x) is a polynomial function, it is continuous and differentiable in R
∴ f'(x) = 3 > 0 for all x ∈ R
Hence f'(x) is strictly increasing on R

Exercise 6.2 Class 12 NCERT Solutions Question 2.
Show that the function given by f (x) = e^2x is strictly increasing on R.
Solution:
f (x) = e^2x
Since f(x) is an exponential function, it is continuous and differentiable in R
∴ f (x) = e^2x > 0 for all x ∈ R
Hence f(x) is strictly increasing on R.

6.2 Class 12 NCERT Solutions Question 3.
Show that the function given by f (x) = sin x is
(a) strictly increasing in \(\left( 0,\frac { \pi }{ 2 } \right) \)
(b) strictly decreasing in \(\left( \frac { \pi }{ 2 } ,\pi \right) \)
(c) neither increasing nor decreasing in (0, π)
Solution:
We have f(x) = sinx
∴ f’ (x) = cosx
(a) f’ (x) = cos x is + ve in the interval \(\left( 0,\frac { \pi }{ 2 } \right) \)
⇒ f(x) is strictly increasing on \(\left( 0,\frac { \pi }{ 2 } \right) \)

(b) f’ (x) = cos x is a -ve in the interval \(\left( \frac { \pi }{ 2 } ,\pi \right) \)
⇒ f (x) is strictly decreasing in \(\left( \frac { \pi }{ 2 } ,\pi \right) \)

(c) f’ (x) = cos x is +ve in the interval \(\left( 0,\frac { \pi }{ 2 } \right) \)
while f’ (x) is -ve in the interval \(\left( \frac { \pi }{ 2 } ,\pi \right) \)
∴ f(x) is neither increasing nor decreasing in (0, π)

Ex 6.2 Class 12 Ncert Solutions Question 4.
Find the intervals in which the function f given by f(x) = 2x² – 3x is
(a) strictly increasing
(b) strictly decreasing
Solution:
f(x) = 2x² – 3x
⇒ f’ (x) = 4x – 3
⇒ f’ (x) = 0 at x = \(\frac { 3 }{ 4 }\)
The point \(x=\frac { 3 }{ 4 }\) divides the real

(i) f(x) is strictly increasing in (\(\frac { 3 }{ 4 }\), ∞)
(ii) f(x) is strictly increasing in (- ∞, \(\frac { 3 }{ 4 }\))

Ex6.2 Class 12 Question 5.
Find the intervals in which the function f given by f (x) = 2x³ – 3x² – 36x + 7 is
(a) strictly increasing
(b) strictly decreasing
Solution:
f (x) = 2x³ – 3x² – 36x + 7
Differentiating w.r.t. x,
f (x) = 6 (x – 3) (x + 2)
⇒ f’ (x) = 0 at x = 3 and x = – 2
The points x = 3, x = – 2, divide the real line into three disjoint intervals viz. (-∞, -2), (-2, 3), (3, ∞)
Now f’ (x) is +ve in the intervals (-∞, -2) and (3, ∞). Since in the interval (-∞, -2) each factor x – 3, x + 2 is -ve.
⇒ f’ (x) = + ve.
(a) f is strictly increasing in (-∞, -2)∪(3, ∞)

(b) In the interval (-2, 3), x+2 is +ve and x-3 is -ve.
f (x) = 6(x – 3)(x + 2) = + x – = -ve
∴ f is strictly decreasing in the interval (-2, 3).

Exercise 6.2 Class 12 Maths Ncert Solutions Question 6.
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
Solution:
(a) Let f(x) = x² + 2x – 5
Differentiating w.r.t. x,
f'(x) = 2x + 2 = 2(x + 1)
f'(x) = 0 ⇒ 2(x + 1) = 0 = – 1
x = – 1 divides R into disjoint open intervals as (-∞, -1) and (-1, ∞)

∴ f is strictly decreasing in (-∞, -1)
f is strictly increasing in (-1, ∞)

(b) Let f(x) = 10 – 6x – 2x²
Differentiating w.r.t. x,
f'(x) = – 6 – 4x = – 2(2x + 3)
f'(x) = 0 ⇒ – 2(2x + 3) = 0

∴ f is strictly decreasing in (-∞, \(\frac { -3 }{ 2 }\)) and
f is strictly increasing in (\(\frac { 3 }{ 2 }\), ∞)

(c) Let f(x) = – 2x³ – 9x² – 12x + 1
∴ f’ (x) = – 6x² – 18x – 12
= – 6(x² + 3x + 2)
f'(x) = – 6(x + 1)(x + 2), f’ (x) = 0 gives x = -1 or x = -2
The points x = – 2 and x = – 1 divide the real line into three disjoint intervals namely ( – ∞, – 2) ( – 2, – 1) and( – 1, ∞).

f (x) is increasing in (-2, -1)
In the interval (-1, ∞) i.e.,-1 < x < ∞,(x + 1) and (x + 2) are both positive. f’ (x) = (-) (+) (+) = -ve.
⇒ f (x) is decreasing in (-1, ∞)
Hence, f (x) is increasing for – 2 < x < – 1 and decreasing for x < – 2 and x > – 1.

(d) Let f(x) = 6 – 9x – x²
Differentiating w.r.t. x,

(e) Let f(x) = (x + 1)³(x – 3)³
Differentiating w.r.t. x,
f'(x) = 3 (x² – 2x – 3)² (2x – 2)
= 6 [(x + 1) (x – 3)]² (x – 1)
= 6 (x + 1)² (x – 1 ) (x – 3)²
f'(x) = 0 ⇒ x = – 1, 1 or 3
x = – 1, x = 1 or x = 3 divide R into disjoint open intervals (-∞, -1), (-1, 1), (1, 3) and (-3, – ∞)

Thus f is strictly decreasing in (- ∞, – 1) and (- 1, 1) and f is strictly increasing in (1, 3) and (3, ∞)

Exercise 6.2 Class 12 Maths Question 7.
Show that y = log(1 + x) – \(\frac { 2x }{ 2 + x }\), x > – 1 is an increasing function of x throughout its domain.
Solution:

When x > – 1, we get y’ = (+)(+) = (+) i.e.,
y’ is positive for x > – 1
∴ When x > – 1, y = log(1 + x) – \(\frac { 2x }{ 2 + x }\) an increasing function.

Class 12 Maths Ex 6.2 NCERT Solutions Question 8.
Find the values of x for which y = [x (x – 2)]² is an increasing function.
Solution:
y = x⁴ – 4x³ + 4x²
∴ \(\frac { dy }{ dx }\) = 4x³ – 12x² + 8x
For the function to be increasing \(\frac { dy }{ dx }\) >0
4x³ – 12x² + 8x>0
⇒ 4x(x – 1)(x – 2)>0

Thus, the function is increasing for 0 < x < 1 and x > 2.

Ex 6.2 Class 12 Maths Ncert Solutions Question 9.
Prove that y = \(\frac { 4sin\theta }{ (2+cos\theta ) } -\theta \) is an increasing function of θ in \(\left[ 0,\frac { \pi }{ 2 } \right]\)
Solution:

∴ \(\frac { dy }{ dθ }\) > 0 since (4 – cos θ) is positive.
Hence y is strictly increasing on (0, \(\frac { π }{ 2 }\)).
Also y is continuous at θ = 0 and θ = \(\frac { π }{ 2 }\).
Hence y is increasing on [0, \(\frac { π }{ 2 }\)]

Class 12 Maths Chapter 6 Exercise 6.2 Question 10.
Prove that the logarithmic function is strictly increasing on (0, ∞).
Solution:
Let f (x) = log x.
The domain of f is (0, ∞).
Now, f’ (x) = \(\frac { 1 }{ x }\)
When takes the values x > 0
\(\frac { 1 }{ x }\) > 0, when x > 0,
∵ f’ (x) > 0
Hence, f (x) is an increasing function for x > 0 i.e

Another method:
From the graph, we can observe that as x increases f(x) also increases. Hence logx is an increasing function on (0, ∞).

Ex 6.2 Class 12 Maths  Question 11.
Prove that the function f given by f (x) = x² – x + 1 is neither strictly increasing nor strictly decreasing on (- 1, 1).
Solution:
let f (x) = x² – x + 1
Differentiating both sides w.r.t. x,
f'(x) = 2x – 1
f'(x) = 2(x – \(\frac { 1 }{ 2 }\)) … (1)
f'(x) = 0 ⇒ 2(x – \(\frac { 1 }{ 2 }\)) = 0 ⇒ x = \(\frac { 1 }{ 2 }\)
∴ The domain of f(x) is (-1, 1)

∴ f(x) is strictly increasing in (\(\frac { 1 }{ 2 }\), 1) and strictly decreasing in (- 1, \(\frac { 1 }{ 2 }\))
In the interval (- 1, 1) the given function is neither strictly increasing nor strictly decreasing.

Class 12 Ex 6.2 NCERT Solutions Question 12.
Which of the following functions are strictly decreasing on \(\left[ 0,\frac { \pi }{ 2 } \right] \)
i. cos x
ii. cos 2x
iii. cos 3x
iv. tan x
Solution:

sin 3x can be positive or negative.
∴ f'(x) can be positive or negative.
Hence cos 3x is neither increasing nor decreasing on (0, (\(\frac { π }{ 2 }\)).

iv. Let f(x) = tan x, x ∈ (0, \(\frac { π }{ 2 }\))
Differentiating w.r.t. x,
f'(x) = sec²x is positive
tan x is strictly increasing on (0, \(\frac { π }{ 2 }\))
Thus cos x and cos 2x are strictly decreasing on (0, \(\frac { π }{ 2 }\))

6.2 Class 12 Maths Question 13.
On which of the following intervals is the function f given by f (x) = x¹⁰⁰ + sin x – 1 strictly decreasing ?
i. (0, 1)
ii. \(\left[ \frac { \pi }{ 2 } ,\pi \right] \)
iii. \(\left[ 0,\frac { \pi }{ 2 } \right] \)
iv. none of these
Solution:
f(x) = x¹⁰⁰ + sin x – 1
Differentiating w.r.t x,
∴ f’ (x)= 100x⁹⁹ + cos x

i. When x ∈ (0, 1), x⁹⁹ > 0 and cosx > 0
∴ f'(x) is positive
i.e.,/(x) is strictly increasing on (0, 1)

ii. When x ∈ (\(\frac { π }{ 2 }\), π), 100x⁹⁹ > 100 and -1 < cosx < 0
f'(x)= 100x⁹⁹ + cosx > 100 + a negative number greater than – 1.
f'(x) is positive
∴ Hence f(x) is strictly increasing on (\(\frac { π }{ 2 }\), π)

iii. When x ∈ (0, \(\frac { π }{ 2 }\)), 100x⁹⁹ > 0 and cosx > 0
Hence f'(x) is positive
f(x) is strictly increasing on (0, \(\frac { π }{ 2 }\))
∴ f(x) is not strictly decreasing in any of the intervals (0, 1), (\(\frac { π }{ 2 }\), π) and (0, \(\frac { π }{ 2 }\)).

6.2 Maths Class 12 NCERT Solutions Question 14.
Find the least value of a such that the function f given by f (x) = x² + ax + 1 is strictly increasing on (1, 2).
Solution:
We have f (x) = x² + ax + 1
∴ f’ (x) = 2x + a.
Since f (x) is an increasing function on (1, 2)
f’ (x) > 0 for all 1 < x < 2 Now, f” (x) = 2 for all x ∈ (1, 2) ⇒ f” (x) > 0 for all x ∈ (1, 2)
⇒ f’ (x) is an increasing function on (1, 2)
⇒ f’ (x) is the least value of f’ (x) on (1, 2)
But f’ (x)>0 ∀ x∈ (1, 2)
∴ f’ (1)>0 ⇒ 2 + a > 0 ⇒ a > – 2
∴ Thus, the least value of ‘a’ is – 2.

Class 12 Maths Exercise 6.2 Question 15.
Let I be any interval disjoint from (- 1, 1). Prove that the function f given by f(x) = x + \(\frac { 1 }{ x } \) is strictly increasing on I.
Solution:
f(x) = x + \(\frac { 1 }{ x } \) and I = R – (-1, 1)
Differentiating w.r.t. x,
f'(x) = 1 – \(\frac{1}{x^{2}}\) = \(\frac{x^{2}-1}{x^{2}}\)
When x² > 1, f'(x) > 0
When x ∈ R – (- 1, 1), x² > 1
Hence f'(x) is positive.
i.e., when x ∈ 1, f(x) is strictly increasing

Question 16.
Prove that the function f given by f (x) = log sin x is strictly increasing on (0, \(\frac { π }{ 2 }\)) and strictly decreasing on (\(\frac { π }{ 2 }\), π)
Solution:
f'(x) = \(\frac { 1 }{ sin\quad x }\)cos x cot x
when 0 < x < \(\frac { \pi }{ 2 }\), f’ (x) is +ve; i.e., increasing
When \(\frac { \pi }{ 2 }\) < x < π, f’ (x) is – ve; i.e., decreasing,
∴ f (x) is decreasing. Hence, f is increasing on (0, π/2) and strictly decreasing on (π/2, π).

Question 17.
Prove that the function f given by f(x) = log cos x is strictly decreasing on (0, \(\frac { π }{ 2 }\))and strictly increasing on (\(\frac { π }{ 2 }\), π)
Solution:
Given f(x) = log cos x
f'(x) = \(\frac { 1 }{ cosx }\) (-sinx) = – tanx
In the interval \(\left( 0,\frac { \pi }{ 2 } \right)\), f’ (x) = -ve
∴ f is strictly decreasing.
In the interval \(\left( \frac { \pi }{ 2 } ,\pi \right)\), f’ (x) is + ve.
∴ f is strictly increasing in the interval.

Question 18.
Prove that the function given by
f (x) = x³ – 3x² + 3x -100 is increasing in R.
Solution:
f’ (x) = 3x² – 6x + 3
= 3 (x² – 2x + 1)
= 3 (x -1 )²
Now x ∈ R, f'(x) = (x – 1)² ≥ 0
i.e. f'(x) ≥ 0 ∀ x ∈ R; hence, f(x) is increasing on R.

Question 19.
The interval in which y = x² e^-x is increasing is
(a) (-∞, ∞)
(b) (-2 0)
(c) (2, ∞)
(d) (0, 2)
Solution:
y = x² e^-x
Differentiating w.r.t. x, we get

x = 0 and x = 2 divide the domain R into disjoint open intervals as (-∞, 0), (0, 2) and (2, ∞).
∴ f is strictly increasing on (0, 2)

These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-6-ex-6-1/

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.1

Question 1.
Find the rate of change of the area of a circle with respect to its radius r when (a) r = 3 cm (b) r = 4 cm
Solution:
Let A be the area of the circle with radius r.
A = πr²
Differentiating w.r.t. r, \(\frac { dA }{ dr }\) = 2πr
i. r = 3cm, \(\frac { dA }{ dr }\) = 2π(3) = 6πcm²/cm
ii. r = 4cm, \(\frac { dA }{ dr }\) = 2π(4) = 8πcm² /cm

Question 2.
The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm?
Solution:
Let x be the length of the cube, V the volume and S the surface area at time l
∴ V = x³ and S = 6x².
\(\frac { dV }{ dt }\) = 8cm³/s
V = x³
Differentiating both sides, w.r.t. t

When the edge of the cube is 12 cm, area increases at the rate of \(\frac { 8 }{ 3 }\) cm²/s.

Question 3.
The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
Solution:
Let r be the radius of the circle.
Area of circle = πr² = A
also \(\frac { dr }{ dt }\) = 3 cm/ sec.
r = 10 m
Differentiating A w.r.t. t, we get
\(\frac { dA }{ dt }\) = 2πr.\(\frac { dr }{ dt }\)
= 2π(10)(3) = 60π cm²/sec

Question 4.
An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?
Solution:
Let V be the volume and x be the edge of the cube at time ‘t’.
V = x³
Given \(\frac { dx }{ dt }\) = 3 cm/s
Differentiating both sides, w.r.t. t
\(\frac { dV }{ dt }\) = 3x².\(\frac { dx }{ dt }\) = 3x²(3) = 9x²
When x = 10 cm, \(\frac { dV }{ dt }\) = 9 x (10)² = 900 cm³/s

Question 5.
A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
Solution:
Let r be the radius of a wave circle:
\(\frac { dx }{ dt }\) = 5cm/sec.
A = πr²
Differentiating w.r.t. t
\(\frac { dA }{ dt }\) = 2πr\(\frac { dr }{ dt }\) = 2πr(5) = 10πr
When r = 8, \(\frac { dA }{ dt }\) = 10π(8) = 80π cm²/s

Question 6.
The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?
Solution:
The rate of change of circle w.r.t time t is given to be 0.7 cm/sec. i.e. \(\frac { dr }{ dt }\) = 0.7 cm/sec.
Now, circumference of the circle is c = 2πr
\(\frac{d \mathrm{C}}{d t}=2 \pi \cdot \frac{d r}{d t}\) = 2π(0.7) = 1.4πcm/s
The rate of increase of circumference is 1.4 π cm/s.

Question 7.
The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of
(a) the perimeter, and
(b) the area of the rectangle.
Solution:
The length x of a rectangle is decreasing at dx the rate of 5cm/min. ⇒ \(\frac { dx }{ dt }\) = – 5cm min … (i)
The width y is increasing at the rate of 4cm/min.
\(\frac { dy }{ dt }\) = 4 cm/minute
(a) p = 2x + 2y
Differentiating both sides w.r.t., t,
\(\frac { dP }{ dt }\) = 2\(\frac { dx }{ dt }\) + 2\(\frac { dy }{ dt }\)
= 2(-5) + 2(4) = – 2cm/minute
Perimeter is decreasing at the rate of 2cm/ minute

(b) A = xy
Differentiating both sides w.r.t., t,
\(\frac { dA }{ dt }\) = x.\(\frac { dy }{ dt }\) + y.\(\frac { dx }{ dt }\)
When x = 8 cm and y = 6 cm
= 32 – 30 = 2 cm²/minute
Area is increasing at the rate of 2 cm²/minute

Question 8.
A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
Solution:
Let V be the volume and r be the radius of the balloon at time t.
Given \(\frac { dV }{ dt }\) = 900
V = \(\frac { 4 }{ 3 }\) πr³
Differentiating both sides w.r.t. t
\(\frac { dV }{ dt }\) = \(\frac { d }{ dt }\)(\(\frac { 4 }{ 3 }\) πr³) = \(\frac { d }{ dr }\)(\(\frac { 4 }{ 3 }\) πr³) \(\frac { d }{ dr }\)
900 = 4πr².\(\frac { dr }{ dt }\)
\(\frac { dr }{ dt }\) = \(\frac{900}{4 \pi r^{2}}\) =\(\frac{225}{\pi r^{2}}\)
when r = 15 cm
\(\frac { dr }{ dt }\) = \(\frac{225}{\pi(15)^{2}}\) = \(\frac{1}{\pi}\) cm/sec.
When radius is 15 cm, the rate at which the radius increases is \(\frac { 1 }{ π }\) cm/sec.

Question 9.
A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.
Solution:
Let r be the variable radius of the balloon which is in the form of sphere Vol. of the sphere
Differentiating w.r.t. r, \(\frac { dV }{ dr }\) = 4π²
When r = 10, \(\frac { dV }{ dr }\) = 4π(10)² = 400πcm³/cm
i.e., The rate at which volume of the bal¬loon is increasing with respect to the radius is 400πcm³/cm

Question 10.
A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4m away from die wall ?
Solution:

Let AB be the ladder and OB be the wall. At an instant,
let OA = x, OB = y,
x² + y² = 25 …(i)
Differentiating both sides w.r.t., t,
2x.\(\frac { dx }{ dt }\) + 2y.\(\frac { dy }{ dt }\) = 0 … (1)
x² + y² = 25 ⇒ y = \(\sqrt{25-x^{2}}\)
When x = 4m, y = \(\sqrt{25-16}\) = 3m
When x = 400 cm, y = 300 cm
(1) ⇒ (2 x 400 x 2) + (2 x 300 x \(\frac { dy }{ dt }\)) = 0
1600 + 600\(\frac { dy }{ dt }\) = 0
∴\(\frac { dy }{ dt }\) = \(\frac { -16 }{ 6 }\) = \(\frac { -8 }{ 3 }\) cm/s
∴ Height of the ladder on the wall is de-creasing at the rate of \(\frac { 8 }{ 3 }\) cm/s.

Question 11.
A particle moves along the curve 6y = x³ + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.
Solution:
We have
6y = x³ + 2 … (i)
Differentiating both sides w.r.t., t,

Hence the required points on the curve are (4, 11) and (- 4, \(\frac { -31 }{ 3 }\))

Question 12.
The radius of an air bubble is increasing at the rate of \(\frac { 1 }{ 2 }\) cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?
Solution:
Let r be the radius then V = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }\)
\(\frac { dr }{ dt } =\frac { 1 }{ 2 } \)cm/sec
\(\frac { dv }{ dt } =\frac { d }{ dt } \left( \frac { 4 }{ 3 } \pi { r }^{ 3 } \right) =\frac { 4 }{ 3 } { \pi .3r }^{ 2 }.\frac { dr }{ dt } \) = 2πr²
Hence, the rate of increase of volume when radius is 1 cm = 2π x 1² = 2π cm³/sec.

Question 13.
A balloon, which always remains spherical, has a variable diameter \(\frac { 3 }{ 2 } \)(2x+1). Find the rate of change of its volume with respect to x.
Solution:
Let r be the radius and V be the volume at time t.

Question 14.
Sand is pouring from a pipe at the rate of 12 cm³/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?
Solution:
Let V be the volume, r be the base radius and h be the height of the sand cone at time t.

When h = 4 cm, \(\frac { dh }{ dt }\) = \(\frac { 1 }{ 48π }\) cm/sec
When the height is 4 cm, the rate at which the height of the sand cone increases is \(\frac { 1 }{ 48π }\) cm/sec.

Question 15.
The total cost C (x) in Rupees associated with the production of x units of an item is given by C (x) = 0.007 x³ – 0.003 x² + 15x + 4000. Find the marginal cost when 17 units are produced.
Solution:
C(x) = 0.007x³ – 0.003x² + 15x + 4000
Marginal cost, MC = \(\frac { d }{ dx }\) C(x)
= 3(0.007)x² – 2(0.003)x + 15
= 0.021 x² – 0.006x + 15
When x = 17
MC = 0.021(17²) – (0.006)17 + 15
= 6.069 – 0.102 + 15 = 20.967
When 17 units are produced,
Marginal cost = ₹ 20.967
∴ The required marginal cost ₹ 30.02 (approx)

Question 16.
The total revenue in rupees received from the sale of JC units of a product is given by
R (x) = 13x² + 26x + 15.
Find the marginal revenue when x = 7.
Solution:
Marginal Revenue (MR)
= Rate of change of total revenue w.r.t. the
number of items sold at an instant = \(\frac { dR }{ dx } \)
We know R(x) = 13x² + 26x + 15,
MR = \(\frac { dR }{ dx } \) = 26x + 26 = 26(x+1)
Now x = 7, MR = 26 (x + 1) = 26 (7 + 1) = 208
Hence, Marginal Revenue = Rs 208.

Question 17.
The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(a) 10π
(b) 12π
(c) 8π
(d) 11π
Solution:
(b) ∵ A = πr² ⇒ \(\frac { dA }{ dr } \) = 2π x 6 = 12π cm²/radius

Question 18.
The total revenue in Rupees received from the sale of x units of a product is given by R (x) = 3x² + 36x + 5. The marginal revenue, when x = 15 is
(a) 116
(b) 96
(c) 90
(d) 126
Solution:
(d) R(x) = 3x² + 36x + 5 ,
MR = \(\frac { dR }{ dx } \) = 6x + 36 ,
At x = 15; \(\frac { dR }{ dx } \) = 6 x 15 + 36 = 90 + 36 = 126