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

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. To find other exercises one can refer, the NCERT exercise given below.

• Integrals Exercise 7.1

• Integrals Exercise 7.2

• Integrals Exercise 7.3

• Integrals Exercise 7.4

• Integrals Exercise 7.5

• Integrals Exercise 7.6

• Integrals Exercise 7.7

• Integrals Exercise 7.8

• Integrals Exercise 7.9

• Integrals Exercise 7.10

• Integrals Miscellaneous Exercise

## 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.

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

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.

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

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:â€‹â€‹â€‹â€‹â€‹â€‹â€‹3 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

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

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.

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

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.

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

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:6â€‹â€‹â€‹â€‹â€‹â€‹â€‹ By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_2^8|x-5|dx$

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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.

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

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.

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

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.

$\int_0^4 |x-1|dx$

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 $g(x) + g(a-x) = 4$

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:â€‹â€‹â€‹â€‹â€‹â€‹â€‹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

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:â€‹â€‹â€‹â€‹â€‹â€‹â€‹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

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.

## 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.

## Also see-

• NCERT Exemplar Solutions Class 12 Maths Chapter 7

• NCERT Solutions for Class 12 Maths Chapter 7

## NCERT Solutions Subject Wise

• NCERT Solutions Class 12 Chemistry

• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Biology

• NCERT Solutions for Class 12 Mathematics

## Subject Wise NCERT Exemplar Solutions

• NCERT Exemplar Class 12 Maths

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• NCERT Exemplar Class 12 Chemistry

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