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NCERT Solutions for Exercise 7.11 Class 12 Maths Chapter 7 - Integrals

NCERT Solutions for Exercise 7.11 Class 12 Maths Chapter 7 - Integrals

Edited By Ramraj Saini | Updated on Dec 03, 2023 08:58 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.11

NCERT Solutions for Exercise 7.11 Class 12 Maths Chapter 7 Integrals are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. In NCERT solutions for Class 12 Maths chapter 7 exercise 7.11 is last in the series of exercises apart from miscellaneous exercise. Solutions to exercise 7.11 Class 12 Maths mainly deals with some of the complex problems. Definite integral evaluation is done in this exercise. NCERT solutions for Class 12 Maths chapter 7 exercise 7.11 with 19 subjective questions and 2 Multiple choice questions provides a bulky source to practice questions. The level of these NCERT book questions is at par with that of JEE Main and NEET.

12th class Maths exercise 7.11 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Integrals Class 12 Chapter 7 Exercise: 7.11

Question:1 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^\frac{\pi}{2}\cos^2 x dx

Answer:

We have I\ =\ \int_0^\frac{\pi}{2}\cos^2 x dx ............................................................. (i)

By using

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get :-

I\ =\ \int_0^\frac{\pi}{2}\cos^2 x dx\ =\ \int_0^\frac{\pi}{2}\cos^2\ (\frac{\pi}{2}- x) dx

or

I\ =\ \int_0^\frac{\pi}{2}\sin^2 x dx ................................................................ (ii)

Adding both (i) and (ii), we get :-

\int_0^\frac{\pi}{2}\cos^2 x dx +\ \int_0^\frac{\pi}{2}\sin^2 x dx\ =\ 2I

or \int_0^\frac{\pi}{2}\ (cos^2 x\ +\ sin^2 x) dx\ =\ 2I

or \int_0^\frac{\pi}{2}1. dx\ =\ 2I

or 2I\ =\ \left [ x \right ] ^\frac{\Pi }{2}_0\ =\ \frac{\Pi }{2}

or I\ =\ \frac{\Pi }{4}


Question:2 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

. \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx

Answer:

We have I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx .......................................................................... (i)

By using ,

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin (\frac{\pi}{2}-x)}}{\sqrt{\sin (\frac{\pi}{2}-x)}+ \sqrt{\cos (\frac{\pi}{2}-x)}}dx


or I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+ \sqrt{\sin x}}dx .......................................................(ii)

Adding (i) and (ii), we get,

2I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}\ +\ \sqrt{\cos x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx

or 2I\ =\ \int_0^\frac{\pi}{2}1.dx


or 2I\ =\ \left [ x \right ]^\frac{\Pi }{2}_0\ =\ \frac{\Pi }{2}

Thus I\ =\ \frac{\Pi }{4}



Question:​​​​​​​3 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}

Answer:

We have I\ =\ \int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x} ..................................................................(i)

By using :

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}(\frac{\pi}{2}-x)dx}{\sin^\frac{3}{2}(\frac{\pi}{2}-x) + \cos^{\frac{3}{2}}(\frac{\pi}{2}-x)}


or I\ =\ \int^{\frac{\pi}{2}}_0\frac{\cos^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x} . ............................................................(ii)

Adding (i) and (ii), we get :

2I\ =\ \int^{\frac{\pi}{2}}_0\frac{\ (sin^{\frac{3}{2}}x+cos^{\frac{3}{2}}x)dx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}

or 2I\ = \int_{0}^{{\frac{\pi}{2}}}1.dx

or 2I\ = \left [ x \right ]^{\frac{\pi}{2}}_ 0\ =\ {\frac{\pi}{2}}

Thus I\ =\ {\frac{\pi}{4}}



Question:4 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

. \int_0^\frac{\pi}{2} \frac{\cos^5 xdx}{\sin^5x + \cos^5x}

Answer:

We have I\ =\ \int_0^\frac{\pi}{2} \frac{\cos^5 xdx}{\sin^5x + \cos^5x} ..................................................................(i)

By using :

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int_0^\frac{\pi}{2} \frac{\cos^5 (\frac{\pi}{2}-x)dx}{\sin^5(\frac{\pi}{2}-x) + \cos^5(\frac{\pi}{2}-x)}

or I\ =\ \int_0^\frac{\pi}{2} \frac{\sin^5 xdx}{\sin^5x + \cos^5x} . ............................................................(ii)

Adding (i) and (ii), we get :

2I\ =\ \int_0^\frac{\pi}{2} \frac{\ ( sin^5x \ +\ cos^5 x)dx}{\sin^5x + \cos^5x}

or 2I\ = \int_{0}^{{\frac{\pi}{2}}}1.dx

or 2I\ = \left [ x \right ]^{\frac{\pi}{2}}_ 0\ =\ {\frac{\pi}{2}}

Thus I\ =\ {\frac{\pi}{4}}



Question:5 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_{-5}^5|x+2|dx

Answer:

We have, I\ =\ \int_{-5}^5|x+2|dx

For opening the modulas we need to define the bracket :

If (x + 2) < 0 then x belongs to (-5, -2). And if (x + 2) > 0 then x belongs to (-2, 5).

So the integral becomes :-

I\ =\ \int_{-5}^{-2} -(x+2)dx\ +\ \int_{-2}^{5} (x+2)dx

or I\ =\ -\left [ \frac{x^2}{2}\ +\ 2x \right ]^{-2} _{-5}\ +\ \left [ \frac{x^2}{2}\ +\ 2x \right ]^{5} _{-2}

This gives I\ =\ 29


Question:6​​​​​​​ By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_2^8|x-5|dx

Answer:

We have, I\ =\ \int_{2}^8|x-5|dx

For opening the modulas we need to define the bracket :

If (x - 5) < 0 then x belongs to (2, 5). And if (x - 5) > 0 then x belongs to (5, 8).

So the integral becomes:-

I\ =\ \int_{2}^{5} -(x-5)dx\ +\ \int_{5}^{8} (x-5)dx

or I\ =\ -\left [ \frac{x^2}{2}\ -\ 5x \right ]^{5} _{2}\ +\ \left [ \frac{x^2}{2}\ -\ 5x \right ]^{8} _{5}

This gives I\ =\ 9


Question:7 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int^1_0x(1-x)^ndx

Answer:

We have I\ =\ \int^1_0x(1-x)^ndx

U sing the property : -

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get : -

I\ =\ \int^1_0x(1-x)^ndx\ =\ \int^1_0(1-x)(1-(1-x))^ndx

or I\ =\ \int^1_0(1-x)x^n\ dx

or I\ =\ \int^1_0(x^n\ -\ x^{n+1}) \ dx

or =\ \left [ \frac{x^{n+1}}{n+1}\ -\ \frac{x^{n+2}}{n+2} \right ]^1_0

or =\ \left [ \frac{1}{n+1}\ -\ \frac{1}{n+2} \right ]

or I\ =\ \frac{1}{(n+1)(n+2)}


Question:8 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^\frac{\pi}{4}\log(1+\tan x)dx

Answer:

We have I\ =\ \int_0^\frac{\pi}{4}\log(1+\tan x)dx

By using the identity

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int_0^\frac{\pi}{4}\log(1+\tan x)dx\ =\ \int_0^\frac{\pi}{4}\log(1+\tan (\frac{\pi}{4}-x))dx

or I\ =\ \int_0^\frac{\pi}{4}\log(1+\frac{1-\tan x}{1+\tan x})dx

or I\ =\ \int_0^\frac{\pi}{4}\log(\frac{2}{1+\tan x})dx

or I\ =\ \int_0^\frac{\pi}{4}\log{2}dx\ -\ \int_0^\frac{\pi}{4}\log(1+ \tan x)dx

or I\ =\ \int_0^\frac{\pi}{4}\log{2}dx\ -\ I

or 2I\ =\ \left [ x\log2 \right ]^{\frac{\Pi }{4}}_0

or I\ =\ \frac{\Pi }{8}\log2



Question:9 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^2x\sqrt{2-x}dx

Answer:

We have I\ =\ \int_0^2x\sqrt{2-x}dx

By using the identity

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get :

I\ =\ \int_0^2x\sqrt{2-x}dx\ =\ \int_0^2(2-x)\sqrt{2-(2-x)}dx

or I\ =\ \int_0^2(2-x)\sqrt{x}dx

or I\ =\ \int_0^2(2\sqrt{x}\ -\ x^\frac{3}{2} dx

or =\ \left [ \frac{4}{3}x^\frac{3}{2}\ -\ \frac{2}{5}x^\frac{5}{2} \right ]^2_0

or =\ \frac{4}{3}(2)^\frac{3}{2}\ -\ \frac{2}{5}(2)^\frac{5}{2}

or I\ =\ \frac{16\sqrt{2}}{15}


Question:10 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^\frac{\pi}{2} (2\log\sin x- \log\sin 2x)dx

Answer:

We have I\ =\ \int_0^\frac{\pi}{2} (2\log\sin x- \log\sin 2x)dx

or I\ =\ \int_0^\frac{\pi}{2} (2\log\sin x- \log(2\sin x\cos x))dx

or I\ =\ \int_0^\frac{\pi}{2} (\log\sin x- \log\cos x\ -\ \log2)dx ..............................................................(i)

By using the identity :

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get :

I\ =\ \int_0^\frac{\pi}{2} (\log\sin (\frac{\pi}{2}-x)- \log\cos (\frac{\pi}{2}-x)\ -\ \log2)dx

or I\ =\ \int_0^\frac{\pi}{2} (\log\cos x- \log\sin x\ -\ \log2)dx ....................................................................(ii)


Adding (i) and (ii) we get :-

2I\ =\ \int_0^\frac{\pi}{2} (- \log 2 -\ \log 2)dx

or I\ =\ -\log 2\left [ \frac{\Pi }{2} \right ]

or I\ =\ \frac{\Pi }{2}\log\frac{1}{2}


Question:11 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^2 x dx

Answer:

We have I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^2 x dx

We know that sin 2 x is an even function. i.e., sin 2 (-x) = (-sinx) 2 = sin 2 x.

Also,

I\ =\ \int_{-a}^af(x) dx\ =\ 2\int_{0}^af(x) dx

So,

I\ =\ 2\int_0^\frac{\pi}{2}\sin^2 x dx\ =\ 2\int_0^\frac{\pi}{2}\frac{(1-\cos2x)}{2} dx

or =\ \left [ x\ -\ \frac{\sin2x}{2} \right ]^{\frac{\Pi }{2}}_0

or I\ =\ \frac{\Pi }{2}


Question:12 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^\pi\frac{xdx}{1+\sin x}

Answer:

We have I\ =\ \int_0^\pi\frac{xdx}{1+\sin x} ..........................................................................(i)

By using the identity :-

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int_0^\pi\frac{xdx}{1+\sin x}\ =\ \int_0^\pi\frac{(\Pi -x)dx}{1+\sin (\Pi -x)}

or I\ =\ \int_0^\pi\frac{(\Pi -x)dx}{1+\sin x} ............................................................................(ii)


Adding both (i) and (ii) we get,

2I\ =\ \int_0^\pi\frac{\Pi}{1+\sin x} dx

or 2I\ =\ \Pi \int_0^\pi\frac{1-\sin x}{(1+\sin x)(1-\sin x)} dx\ =\ \Pi \int_0^\pi\frac{1-\sin x}{\cos^2 x} dx

or 2I\ =\ \Pi \int_0^\pi (\sec^2\ -\ \tan x \sec x) x dx

or I\ =\ \Pi



Question:13 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^7xdx

Answer:

We have I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^7xdx

We know that \sin^7x is an odd function.

So the following property holds here:-

\int_{-a}^{a}f(x)dx\ =\ 0

Hence

I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^7xdx\ =\ 0



Question:14 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^{2\pi}\cos^5xdx

Answer:

We have I\ =\ \int_0^{2\pi}\cos^5xdx

I t is known that :-

\int_0^{2a}f(x)dx\ =\ 2\int_0^{a}f(x)dx If f (2a - x) = f(x)

=\ 0 If f (2a - x) = - f(x)

Now, using the above property

\cos^5(\Pi - x)\ =\ - \cos^5x

Therefore, I\ =\ 0



Question:15 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int^\frac{\pi}{2} _0\frac{\sin x - \cos x }{1+\sin x\cos x}dx

Answer:

We have I\ =\ \int^\frac{\pi}{2} _0\frac{\sin x - \cos x }{1+\sin x\cos x}dx ................................................................(i)


By using the property :-

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get ,

I\ =\ \int^\frac{\pi}{2} _0\frac{\sin (\frac{\pi}{2}-x) - \cos (\frac{\pi}{2}-x) }{1+\sin (\frac{\pi}{2}-x)\cos (\frac{\pi}{2}-x)}dx

or I\ =\ \int^\frac{\pi}{2} _0\frac{\cos x - \sin x }{1+\sin x\cos x}dx ......................................................................(ii)


Adding both (i) and (ii), we get


2I\ =\ \int^\frac{\pi}{2} _0\frac{0 }{1+\sin x\cos x}dx

Thus I = 0


Question:16 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^\pi\log(1 +\cos x)dx


Answer:

We have I\ =\ \int_0^\pi\log(1 +\tan x)dx .....................................................................................(i)

By using the property:-

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,


or

I\ =\ \int_0^\pi\log(1 +\cos (\Pi -x))dx

I\ =\ \int_0^\pi\log(1 -\cos x)dx ....................................................................(ii)


Adding both (i) and (ii) we get,

2I\ =\ \int_0^\pi\log(1 +\cos x)dx\ +\ \int_0^\pi\log(1 -\cos x)dx

or 2I\ =\ \int_0^\pi\log(1 -\cos^2 x)dx\ =\ \int_0^\pi\log \sin^2 xdx

or 2I\ =\ 2\int_0^\pi\log \sin xdx

or I\ =\ \int_0^\pi\log \sin xdx ........................................................................(iii)

or I\ =\ 2\int_0^ \frac{\pi}{2} \log \sin xdx ........................................................................(iv)

or I\ =\ 2\int_0^ \frac{\pi}{2} \log \cos xdx .....................................................................(v)

Adding (iv) and (v) we get,

I\ =\ -\pi \log2



Question:17 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^a \frac{\sqrt x}{\sqrt x + \sqrt{a-x}}dx

Answer:

We have I\ =\ \int_0^a \frac{\sqrt x}{\sqrt x + \sqrt{a-x}}dx ................................................................................(i)

By using, we get


\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int_0^a \frac{\sqrt x}{\sqrt x + \sqrt{a-x}}dx\ =\ \int_0^a \frac{\sqrt {(a-x)}}{\sqrt {(a-x)} + \sqrt{x}}dx .................................................................(ii)


Adding (i) and (ii) we get :

2I\ =\ \int_0^a \frac{\sqrt x\ +\ \sqrt{a-x}}{\sqrt x + \sqrt{a-x}}dx

or 2I\ =\ \left [ x \right ]^a_0 = a

or I\ =\ \frac{a}{2}


Question:18 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^4 |x-1|dx

Answer:

We have, I\ =\ \int_{0}^4|x-1|dx

For opening the modulas we need to define the bracket :

If (x - 1) < 0 then x belongs to (0, 1). And if (x - 1) > 0 then x belongs to (1, 4).

So the integral becomes:-

I\ =\ \int_{0}^{1} -(x-1)dx\ +\ \int_{1}^{4} (x-1)dx

or I\ =\ \left [ x\ -\ \frac{x^2}{2}\ \right ]^{1} _{0}\ +\ \left [ \frac{x^2}{2}\ -\ x \right ]^{4} _{1}

This gives I\ =\ 5


Question:19 Show that \int_0^a f(x)g(x)dx = 2\int_0^af(x)dx if f and g are defined as f(x) = f(a-x) and g(x) + g(a-x) = 4


Answer:

Let I\ =\ \int_0^a f(x)g(x)dx ........................................................(i)

This can also be written as :

I\ =\ \int_0^a f(a-x)g(a-x)dx

or I\ =\ \int_0^a f(x)g(a-x)dx ................................................................(ii)

Adding (i) and (ii), we get,

2I\ =\ \int_0^a f(x)g(a-x)dx +\ \int_0^a f(x)g(x)dx

2I\ =\ \int_0^a f(x)4dx

or I\ =\ 2\int_0^a f(x)dx



Question:​​​​​​​20 Choose the correct answer in Exercises 20 and 21.

The value of is \int_\frac{-\pi}{2}^\frac{\pi}{2}(x^3 + x\cos x + \tan^5 x + 1)dx is

(A) 0

(B) 2

(C) \pi

(D) 1

Answer:

We have

I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}(x^3 + x\cos x + \tan^5 x + 1)dx

This can be written as :

I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}x^3dx +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} x\cos x +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} \tan^5 x +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} 1dx

Also if a function is even function then \int_{-a}^{a}f(x)\ dx\ =\ 2\int_{0}^{a}f(x)\ dx

And if the function is an odd function then : \int_{-a}^{a}f(x)\ dx\ =\ 0

Using the above property I become:-

I\ =\ 0+0+0+ 2\int_{0}^{\frac{\Pi }{2}}1.dx

or I\ =\ 2\left [ x \right ]^{\frac{\Pi }{2}}_0

or I\ =\ \Pi



Question:​​​​​​​21 Choose the correct answer in Exercises 20 and 21.

The value of \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx is

Answer:

We have

I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx .................................................................................(i)


By using :

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin (\frac{\pi}{2}-x)}{4+3\cos (\frac{\pi}{2}-x)} \right )dx

or I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\cos x}{4+3\sin x} \right )dx .............................................................................(ii)

Adding (i) and (ii), we get:

2I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx\ +\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\cos x}{4+3\sin x} \right )dx

or 2I\ =\ \int_0^\frac{\pi}{2}\log1.dx

Thus I\ =\ 0

More About NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.11

The NCERT syllabus Class 12 Maths chapter Integrals is an exercise which can take a lot of time and brain churning. Exercise 7.11 Class 12 Maths deals with advanced level problems. NCERT solutions for Class 12 Maths chapter 7 exercise 7.11 should be done with high concentration as questions can demand presence of mind and application of a lot of concepts together.

Also Read| Integrals Class 12 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.11

  • The Class 12th Maths chapter 7 exercise has a substantial number of questions to practice.
  • Exercise 7.11 Class 12 Maths can take a lot of time without knowledge of basic concepts. Hence one should go for previous exercises before doing this.
  • These Class 12 Maths chapter 7 exercise 7.11 solutions have similar questions to that of exercise 7.10.
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Key Features Of NCERT Solutions for Exercise 7.11 Class 12 Maths Chapter 7

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 7.11 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 7.11, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 7.11 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 7.11 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 7.11 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 7.11 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus
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Frequently Asked Questions (FAQs)

1. Two types of integration are …………. And ………….. ?

Two types of integration are Definite and indefinite Integrals. 

2. What can be the limit of an integral ?

It can be anything in which the given function is real. 

3. Integrals are used in …………... ?

Integrals are used in finding area, volume, displacement etc. 

4. Can one skip exercise 7.11 Class 12 Maths ?

No, one should do this exercise as 5 marks questions can be asked in the Board examination. 

5. Which topics are dealt in Exercise 7.11 Class 12 Maths?

Some definite integrals of advance level are discussed in this exercise. 

6. Can we take some assumptions in proof related questions ?

Yes, provided it fulfills the demand of the question.

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It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.

  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Read Complete Answer
Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer
Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer
kumardurgesh1802 10 July,2024
I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
  • You have to appear for the 2025 12th board exams.
  • Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
  • Aim to register before late October to avoid extra fees.
  • Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

Read Complete Answer
kumardurgesh1802 10 July,2024

Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read More

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
Read More

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
Read More

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
Read More

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read More

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
Read More

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
Read More

With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
Read More

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
Read More

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
Read More

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