NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise
NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise contains all the questions with detailed solutions. Students are advised to solve these questions for better understanding of the concepts in chapter 8.
Krati Saraswat2 Feb, 2024
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NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise (Applications of Integrals)
NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise Applications of Integrals is prepared by academic team of Physics Wallah. We have prepared NCERT Solutions for all exercise of chapter-8. Given below is step by step solutions of all questions given in NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise.NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1
NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise Overview
NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Excercise addresses these significant subjects. In order to fully comprehend the concepts presented in the chapter and make effective use of the provided solutions, it is recommended that students go over each topic in great detail. The intention is for students to effortlessly achieve excellent exam scores after reviewing and practicing these responses.NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise
Solve The Following Questions of NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise Question 1. Find the area under the given curves and given lines:(i) y = x 2 , x = 1, x = 2 and x -axis
(ii) y = x 4 , x = 1, x = 5 and x –axis
Solution : (i)The required area is represented by the shaded area ADCBA as
(ii)The required area is represented by the shaded area ADCBA as
NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2
Question 2. Find the area between the curves y = x and y = x 2 Solution : The required area is represented by the shaded area OBAO as
The points of intersection of the curves,
y
=
x
and
y
=
x
2
, is A (1, 1).
We draw AC perpendicular to
x
-axis.
∴ Area (OBAO) = Area (ΔOCA) – Area (OCABO) … (1)
Question 3. Find the area of the region lying in the first quadrant and bounded by
y
= 4
x
2
,
x
= 0,
y
= 1 and
y
= 4
Solution :
The area in the first quadrant bounded by
y
= 4
x
2
,
x
= 0,
y
= 1, and
y
= 4 is represented by the shaded area ABCDA as
Question 4. Sketch the graph of y = |x + 3| and evaluate
Solution :
The given equation is y = |x + 3|
The corresponding values of
x
and
y
are given in the following table.
On plotting these points, we obtain the graph of y = |x + 3| as follows.
Question 5. Find the area bounded by the curve
y
= sin
x
between
x
= 0 and
x
= 2π
Solution :
The graph of
y
= sin
x
can be drawn as
∴ Required area = Area OABO + Area BCDB
Question 6. Find the area enclosed between the parabola
y
2
= 4
ax
and the line
y
=
mx
Solution :
The area enclosed between the parabola,
y
2
= 4
ax
, and the line,
y
=
mx
, is represented by the shaded area OABO as
The points of intersection of both the curves are (0, 0) and
.
We draw AC perpendicular to
x
-axis.
∴ Area OABO = Area OCABO – Area (ΔOCA)
Question 7. Find the area enclosed by the parabola 4
y
= 3
x
2
and the line 2
y
= 3
x
+ 12
Solution :
The area enclosed between the parabola, 4
y
= 3
x
2
, and the line, 2
y
= 3
x
+ 12, is represented by the shaded area OBAO as
The points of intersection of the given curves are A (–2, 3) and (4, 12).
We draw AC and BD perpendicular to
x-
axis.
∴ Area OBAO = Area CDBA – (Area ODBO + Area OACO)
Question 8. Find the area of the smaller region bounded by the ellipse
and the line
Solution :
The area of the smaller region bounded by the ellipse,
= 1
, and the line,
, is represented by the shaded region BCAB as
∴ Area BCAB = Area (OBCAO) – Area (OBAO)
Question 9. Find the area of the smaller region bounded by the ellipse
and the line
Solution :
The area of the smaller region bounded by the ellipse,
, and the line,
, is represented by the shaded region BCAB as
∴ Area BCAB = Area (OBCAO) – Area (OBAO)
Question 10. Find the area of the region enclosed by the parabola
x
2
=
y
, the line
y
=
x
+ 2 and
x
-axis
Solution :
The area of the region enclosed by the parabola,
x
2
=
y
, the line,
y
=
x
+ 2, and
x
-axis is represented by the shaded region OACO as
The point of intersection of the parabola,
x
2
=
y
, and the line,
y
=
x
+ 2, is A (–1, 1) and C(2, 4).
Question 11.Using the method of integration, find the area enclosed by the curve |x| + |y| = 1
[Hint: the required region is bounded by lines x + y = 1, x – y = 1, – x + y = 1 and – x – y = 11]
Solution : The area bounded by the curve, |x| + |y| = 1, is represented by the shaded region ADCB as
The curve intersects the axes at points A (0, 1), B (1, 0), C (0, –1), and D (–1, 0).
It can be observed that the given curve is symmetrical about
x
-axis and
y
-axis.
∴ Area ADCB = 4 × Area OBAO
Question 12. Find the area bounded by curves {(x, y) : y ≥ x
2
and y = |x|}.
Solution :
The area bounded by the curves, {(x, y) : y ≥ x
2
and y = |x|}
.
, is represented by the shaded region as
It can be observed that the required area is symmetrical about
y
-axis.
Question 13.Using the method of integration, find the area of the triangle whose vertices are A (2, 0), B (4, 5) and C (6, 3).
Solution :
Vertices of the given triangle are A (2, 0), B (4, 5) and C (6, 3).
Equation of side AB is
Equation of side BC
Equation of side CA is
Area (ΔABC) = Area (ABLA) + Area (BLMCB) – Area (ACMA)
Question 14.Using the method of integration, find the area of the region bounded by the lines: 2
x
+
y
= 4, 3
x
– 2
y
= 6 and
x
– 3
y
+ 5 = 0
Solution :
The given equations of lines are
2
x
+
y
= 4 … (1)
3
x
– 2
y
= 6 … (2)
And,
x
– 3
y
+ 5 = 0 … (3)
The area of the region bounded by the lines is the area of ΔABC. AL and CM are the perpendiculars on
x
-axis.
Area (ΔABC) = Area (ALMCA) – Area (ALB) – Area (CMB)
Question 15. Find the area of the region {(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9}.
Solution :
The area bounded by the curves, {(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9}, is represented as
The points of intersection of both the curves are (1/2,√2) and (1/2, -√2).
The required area is given by OABCO.
It can be observed that area OABCO is symmetrical about
x
-axis.
∴ Area OABCO = 2 × Area OBC
Area OBCO = Area OMC + Area MBC
Therefore, the required area is
units.
Question 16.Choose the correct answer:
Area bounded by the curve y = x 3 , the x -axis and the ordinates x = –2 and x = 1 is
(A) -9 (B) -15/4 (C) 15/4 (D) 17/4 Solution :
Therefore, option (D) is correct.
Question 17.Choose the correct answer:
The area bounded by the curve y = x|x|, axis and the ordinates
x
= –1 and
x
= 1 is given by:
[Hint: y = x 2 if x > 0 and y = – x 2 if x < 0]
(A) 0 (B) 1/3 (C) 2/3 (D) 4/3 Solution :
Therefore, option (C) is correct.
Question 18.Choose the correct answer:
The area of the circle x 2 + y 2 = 16 exterior to the parabola y 2 = 6 x is
Solution :
The given equations are
x
2
+
y
2
= 16 … (1)
y
2
= 6
x
… (2)
Area bounded by the circle and parabola
Thus, the correct answer is C.
Question 19.Choose the correct answer:
The area bounded by the y -axis, y = cos x and y = sin x when 0 ≤ x ≤ π/2.
(A) 2(√2 - 1) (B) √2 - 1 (C) √2 + 1 (D) √2 Solution : The given equations are y = cos x … (1) And, y = sin x … (2)
Required area = Area (ABLA) + area (OBLO)
Required area = Area (AB
Therefore, option (B) is correct.
NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise FAQs
What is the Miscellaneous Exercise in Class 12 Maths Chapter 8 about?
The Miscellaneous Exercise in Class 12 Maths Chapter 8 includes additional problems related to the concepts covered in the chapter. It serves as an opportunity for students to practice and reinforce their understanding of the material.
Why is it important to practice the Miscellaneous Exercise in Class 12 Maths Chapter 8?
The Miscellaneous Exercise provides additional practice problems, helping students strengthen their grasp of the chapter's concepts. It allows them to apply the learned principles to solve diverse mathematical scenarios, enhancing their problem-solving skills.
How can NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise aid in exam preparation?
NCERT Solutions serve as comprehensive guides to solve problems effectively. By using these solutions for the Miscellaneous Exercise, students can understand the step-by-step process of solving each problem, reinforcing their understanding and preparing them for exams.
Can NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise be used for self-study?
Absolutely, NCERT Solutions are designed for self-study. Students can utilize these solutions independently to practice and understand the solutions to problems in the Miscellaneous Exercise, reinforcing their mathematical skills.
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