# NCERT Solutions for Exercise 7.9 Class 12 Maths Chapter 7 - Integrals

In NCERT solutions for Class 12 Maths chapter 7 exercise 7.9 is one of the most important exercises from the exam perspective as it deals with the evaluation of definite integrals in a defined range. Such NCERT book questions are more often seen in the Board as well as competitive examination. Solutions to exercise 7.9 Class 12 Maths which are provided here are prepared in detail by the experienced subject matter experts. Considering the importance of NCERT solutions for Class 12 Maths chapter 7 exercise 7.9, it is highly recommended to students to practice at least a few questions before the examination. Also students can refer NCERT chapter Integrals for more practice.

• 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 7.10

• Integrals Exercise 7.11

• Integrals Miscellaneous Exercise

Integrals Class 12 Chapter 7 Exercise 7.9

### Question:1 Evaluate the definite integrals in Exercises 1 to 20.

$\int_{-1}^{1} (x+1)dx$

Given integral: $I = \int_{-1}^{1} (x+1)dx$

Consider the integral $\int (x+1)dx$

$\int (x+1)dx = \frac{x^2}{2}+x$

So, we have the function of $x$ , $f(x) = \frac{x^2}{2}+x$

Now, by Second fundamental theorem of calculus, we have

$I = f(1)-f(-1)$

$= \left ( \frac{1}{2}+1\right ) - \left (\frac{1}{2}-1 \right ) = \frac{1}{2}+1-\frac{1}{2}+1 = 2$

### Question:2 Evaluate the definite integrals in Exercises 1 to 20.

$\int_2^3\frac{1}{x}dx$

Given integral: $I = \int_2^3\frac{1}{x}dx$

Consider the integral $\int_2^3\frac{1}{x}dx$

$\int \frac{1}{x}dx = \log|x|$

So, we have the function of $x$ , $f(x) = \log|x|$

Now, by Second fundamental theorem of calculus, we have

$I = f(3)-f(2)$

$=\log|3|-\log|2| = \log \frac{3}{2}$

### Question:3 Evaluate the definite integrals in Exercises 1 to 20.

$\int_1^2(4x^3-5x^2 + 6x +9)dx$

Given integral: $I = \int_1^2(4x^3-5x^2 + 6x +9)dx$

Consider the integral $I = \int (4x^3-5x^2 + 6x +9)dx$

$\int (4x^3-5x^2 + 6x +9)dx = 4\frac{x^4}{4} -5\frac{x^3}{3}+6\frac{x^2}{2}+9x$

$= x^4 -\frac{5x^3}{3}+3x^2+9x$

So, we have the function of $x$ , $f(x) = x^4 -\frac{5x^3}{3}+3x^2+9x$

Now, by Second fundamental theorem of calculus, we have

$I = f(2)-f(1)$

$=\left \{ 2^4-\frac{5(2)^3}{3}+3(2)^2+9(2)\right \} - \left \{ 1^4-\frac{5(1)^3}{3}+3(1)^2+9(1) \right \}$

$=\left \{ 16-\frac{40}{3}+12+18\right \} - \left \{ 1-\frac{5}{3}+3+9 \right \}$

$=\left \{ 46-\frac{40}{3}\right \} - \left \{ 13-\frac{5}{3}\right \}$

$=\left \{ 33-\frac{35}{3} \right \} = \left \{ \frac{99-35}{3} \right \}$

$= \frac{64}{3}$

### Question:4 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^\frac{\pi}{4}\sin 2x dx$

Given integral: $\int_0^\frac{\pi}{4}\sin 2x dx$

Consider the integral $\int \sin 2x dx$

$\int \sin 2x dx = \frac{-\cos 2x }{2}$

So, we have the function of $x$ , $f(x) = \frac{-\cos 2x }{2}$

Now, by Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{4})-f(0)$

$= \frac{-\cos 2(\frac{\pi}{4})}{2} + \frac{\cos 0}{2}$

$=\frac{1}{2} - 0$

$= \frac{1}{2}$

### Question:5 Evaluate the definite integrals in Exercises 1 to 20.

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

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

Consider the integral $\int \cos 2x dx$

$\int \cos 2x dx = \frac{\sin 2x }{2}$

So, we have the function of $x$ , $f(x) = \frac{\sin 2x }{2}$

Now, by Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{2})-f(0)$

$= \frac{1}{2}\left \{ \sin 2(\frac{\pi}{2}) - \sin 0 \right \}$

$= \frac{1}{2}\left \{ 0 - 0 \right \} = 0$

### Question:6 Evaluate the definite integrals in Exercises 1 to 20.

$\int_4^5 e^x dx$

Given integral: $\int_4^5 e^x dx$

Consider the integral $\int e^x dx$

$\int e^x dx = e^x$

So, we have the function of $x$ , $f(x) = e^x$

Now, by Second fundamental theorem of calculus, we have

$I = f(5) -f(4)$

$= e^5 -e^4$

$= e^4(e-1)$

### Question:7 Evaluate the definite integrals in Exercises 1 to 20.

$\int^\frac{\pi}{4}_0 \tan x dx$

Given integral: $\int^\frac{\pi}{4}_0 \tan x dx$

Consider the integral $\int \tan x dx$

$\int \tan x dx = -\log|\cos x |$

So, we have the function of $x$ , $f(x) = -\log|\cos x |$

Now, by Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{4}) -f(0)$

$= -\log\left | \cos \frac{\pi}{4} \right | +\log|\cos 0|$

$= -\log\left | \cos \frac{1}{\sqrt2} \right | +\log|1|$

$= -\log\left | \frac{1}{\sqrt2} \right | + 0 = -\log (2)^{-\frac{1}{2}}$

$= \frac{1}{2}\log (2)$

### Question:8 Evaluate the definite integrals in Exercises 1 to 20.

$\int_\frac{\pi}{6}^\frac{\pi}{4}\textup{cosec}xdx$

Given integral: $\int_\frac{\pi}{6}^\frac{\pi}{4}\textup{cosec}xdx$

Consider the integral $\int\textup{cosec}xdx$

$\int\textup{cosec}xdx = \log|cosec x -\cot x |$

So, we have the function of $x$ , $f(x) =\log|cosec x -\cot x |$

Now, by Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{4}) -f(\frac{\pi}{6})$

$= \log|cosec \frac{\pi}{4} -\cot \frac{\pi}{4} | - \log|cosec \frac{\pi}{6} -\cot \frac{\pi}{6} |$

$= \log|\sqrt2 -1 | - \log|2 -\sqrt3 |$

$= \log \left ( \frac{\sqrt2 -1}{2-\sqrt3} \right )$

### Question:9 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^1\frac{dx}{\sqrt{1-x^2}}$

Given integral: $\int_0^1\frac{dx}{\sqrt{1-x^2}}$

Consider the integral $\int \frac{dx}{\sqrt{1-x^2}}$

$\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}x$

So, we have the function of $x$ , $f(x) = \sin^{-1}x$

Now, by Second fundamental theorem of calculus, we have

$I = f(1) -f(0)$

$= \sin^{-1}(1) -\sin^{-1}(0)$

$= \frac{\pi}{2} - 0$

$= \frac{\pi}{2}$

### Question:10 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^1\frac{dx}{1 + x^2}$

Given integral: $\int_0^1\frac{dx}{1 + x^2}$

Consider the integral $\int\frac{dx}{1 + x^2}$

$\int\frac{dx}{1 + x^2} = \tan^{-1}x$

So, we have the function of $x$ , $f(x) =\tan^{-1}x$

Now, by Second fundamental theorem of calculus, we have

$I = f(1) -f(0)$

$= \tan^{-1}(1) -\tan^{-1}(0)$

$= \frac{\pi}{4} - 0$

$= \frac{\pi}{4}$

### Question:11 Evaluate the definite integrals in Exercises 1 to 20.

$\int_2^3 \frac{dx}{x^2 -1 }$

Given integral: $\int_2^3 \frac{dx}{x^2 -1 }$

Consider the integral $\int \frac{dx}{x^2 -1 }$

$\int \frac{dx}{x^2 -1 } = \frac{1}{2}\log\left | \frac{x-1}{x+1} \right |$

So, we have the function of $x$ , $f(x) =\frac{1}{2}\log\left | \frac{x-1}{x+1} \right |$

Now, by Second fundamental theorem of calculus, we have

$I = f(3) -f(2)$

$= \frac{1}{2}\left \{ \log\left | \frac{3-1}{3+1} \right | - \log\left | \frac{2-1}{2+1} \right | \right \}$

$= \frac{1}{2}\left \{ \log\left | \frac{2}{4} \right | -\log\left | \frac{1}{3} \right | \right \}$

$= \frac{1}{2}\left \{ \log \frac{1}{2} -\log \frac{1}{3} \right \} = \frac{1}{2}\log\frac{3}{2}$

### Question:12 Evaluate the definite integrals in Exercises 1 to 20.

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

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

Consider the integral $\int \cos^2 x dx$

$\int \cos^2 x dx = \int \frac{1+\cos 2x}{2} dx = \frac{x}{2}+\frac{\sin 2x }{4}$

$= \frac{1}{2}\left ( x+\frac{\sin 2x}{2} \right )$

So, we have the function of $x$ , $f(x) =\frac{1}{2}\left ( x+\frac{\sin 2x}{2} \right )$

Now, by Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{2}) -f(0)$

$= \frac{1}{2}\left \{ \left ( \frac{\pi}{2}-\frac{\sin \pi}{2} \right ) -\left ( 0+\frac{\sin 0}{2} \right ) \right \}$

$= \frac{1}{2}\left \{ \frac{\pi}{2}+0-0-0 \right \}$

$= \frac{\pi}{4}$

### Question:13 Evaluate the definite integrals in Exercises 1 to 20.

$\int_2^3\frac{xdx}{x^2+1}$

Given integral: $\int_2^3\frac{xdx}{x^2+1}$

Consider the integral $\int \frac{xdx}{x^2+1}$

$\int \frac{xdx}{x^2+1} = \frac{1}{2}\int \frac{2x}{x^2+1}dx =\frac{1}{2}\log(1+x^2)$

So, we have the function of $x$ , $f(x) =\frac{1}{2}\log(1+x^2)$

Now, by Second fundamental theorem of calculus, we have

$I = f(3) -f(2)$

$= \frac{1}{2}\left \{ \log(1+(3)^2)-\log(1+(2)^2) \right \}$

$= \frac{1}{2}\left \{ \log(10)-\log(5) \right \} = \frac{1}{2}\log\left ( \frac{10}{5} \right ) = \frac{1}{2}\log2$

### Question:14 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^1\frac{2x+3}{5x^2+1}dx$

Given integral: $\int_0^1\frac{2x+3}{5x^2+1}dx$

Consider the integral $\int \frac{2x+3}{5x^2+1}dx$

Multiplying by 5 both in numerator and denominator:

$\int \frac{2x+3}{5x^2+1}dx = \frac{1}{5}\int \frac{5(2x+3)}{5x^2+1}dx$

$=\frac{1}{5}\int \frac{10x+15}{5x^2+1}dx$

$= \frac{1}{5} \int \frac{10x}{5x^2+1} dx +3\int \frac{1}{5x^2+1} dx$

$= \frac{1}{5}\int \frac{10x}{5x^2+1}+3\int \frac{1}{5\left ( x^2+\frac{1}{5} \right )}dx$

$= \frac{1}{5}\log(5x^2+1) +\frac{3}{5}\times \frac{1}{\frac{1}{\sqrt5}} \tan^{-1}\frac{x}{\frac{1}{\sqrt5}}$

$= \frac{1}{5}\log(5x^2+1) +\frac{3}{\sqrt5}\tan^{-1}(\sqrt5 x )$

So, we have the function of $x$ , $f(x) = \frac{1}{5}\log(5x^2+1) +\frac{3}{\sqrt5}\tan^{-1}(\sqrt5 x )$

Now, by Second fundamental theorem of calculus, we have

$I = f(1) -f(0)$

$= \left \{ \frac{1}{5}\log(1+5)+\frac{3}{\sqrt5}\tan^{-1}(\sqrt5) \right \} - \left \{ \frac{1}{5}\log(1)+\frac{3}{\sqrt5}\tan^{-1}(0) \right \}$

$= \frac{1}{5}\log 6 +\frac{3}{\sqrt 5}\tan^{-1}{\sqrt5}$

### Question:15 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^1xe^{x^2}dx$

Given integral: $\int_0^1xe^{x^2}dx$

Consider the integral $\int xe^{x^2}dx$

Putting $x^2 = t$ which gives, $2xdx =dt$

As, $x\rightarrow0 ,t \rightarrow0$ and as $x\rightarrow1 ,t \rightarrow1$ .

So, we have now:

$\therefore I = \frac{1}{2}\int_0^1 e^t dt$

$= \frac{1}{2}\int e^t dt = \frac{1}{2} e^t$

So, we have the function of $x$ , $f(x) = \frac{1}{2} e^t$

Now, by Second fundamental theorem of calculus, we have

$I = f(1) -f(0)$

$= \frac{1}{2}e^1 -\frac{1}{2}e^0 = \frac{1}{2}(e-1)$

### Question:16 Evaluate the definite integrals in Exercises 1 to 20.

$\int_1^2\frac{5x^2}{x^2 + 4x +3}$

Given integral: $I = \int_1^2\frac{5x^2}{x^2 + 4x +3}$

So, we can rewrite the integral as;

$I = \int_1^2 \frac{5x^2}{x^2 + 4x +3}= \int_1^2 \left ( 5 - \frac{20x+15}{x^2 + 4x +3} \right ) dx$

$= \int_1^2 5 dx - \int_1^2 \frac{20x+15}{x^2+4x+3}dx$

$= [5x]_1^2 - \int_1^2 \frac{20x+15}{x^2+4x+3}dx$

$I = 5-I_1$ where $I = \int_1^2 \frac{20x+15}{x^2+4x+3}dx$ . ................(1)

Now, consider $I = \int_1^2 \frac{20x+15}{x^2+4x+3}dx$

Take numerator $20x+15 = A \frac{d}{dx}\left ( x^2+4x+3 \right )+B$

$= 2A x+(4A+B)$

We now equate the coefficients of x and constant term, we get

A=10 and B=-25

$\Rightarrow I_1 = 10\int_1^2 \frac{2x+4}{x^2+4x+3}dx -25\int_1^2 \frac{dx}{x^2+4x+3}$

Now take denominator $x^2+4x+3 = t$

Then we have $(2x+4)dx =dt$

$\Rightarrow I_{1} =10\int \frac{dt}{t} -25\int \frac{dx}{(x+2)^2-1^2}$

$= 10\log t -25\left [ \frac{1}{2}\log\left ( \frac{x+2-1}{x+2+1} \right ) \right ]$

$=[10\log(x^2+4x+3)]_1^2 -25 \left [ \frac{1}{2}\log\left ( \frac{x+1}{x+3} \right ) \right ]_1^2$

$= \left [ 10\log15 -10\log 8 \right ] -25 \left [ \frac{1}{2}\log\frac{3}{5} -\frac{1}{2}\log\frac{2}{4} \right ]$

$= \left [ 10\log5 +10\log3 -10\log4-10\log2 \right ] -\frac{25}{2}\left [ \log3 -\log5-\log2+\log4 \right ]$ $= \left ( 10+\frac{25}{2} \right )\log5 + \left ( -10-\frac{25}{2} \right )\log 4 + \left ( 10-\frac{25}{2} \right )\log 3 + \left ( -10+\frac{25}{2} \right )\log 2$ $= \frac{45}{2}\log5 -\frac{45}{2}\log4 - \frac{5}{2}\log3 +\frac{5}{2}\log2$

$= \frac{45}{2}\log\frac{5}{4}-\frac{5}{2}\log \frac{3}{2}$

Then substituting the value of $I_{1}$ in equation (1), we get

$I= 5 -\left ( \frac{45}{2}\log\frac{5}{4}-\frac{5}{2}\log\frac{3}{2} \right )$

$= 5 -\frac{5}{2}\left ( 9\log\frac{5}{4}-\log\frac{3}{2} \right )$

### Question:17 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^\frac{\pi}{4}(2\sec^2x + x^3 + 2)dx$

Given integral: $\int_0^\frac{\pi}{4}(2\sec^2x + x^3 + 2)dx$

Consider the integral $\int (2\sec^2x + x^3 + 2)dx$

$\int (2\sec^2x + x^3 + 2)dx = 2\tan x +\frac{x^4}{4}+2x$

So, we have the function of $x$ , $f(x) = 2\tan x +\frac{x^4}{4}+2x$

Now, by Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{4}) -f(0)$

$= \left \{ \left ( 2\tan\frac{\pi}{4}+\frac{1}{4}\left ( \frac{\pi}{4} \right )^4+2\frac{\pi}{4} \right ) - \left ( 2\tan 0 +0 +0 \right ) \right \}$

$=2\tan\frac{\pi}{4} +\frac{\pi^4}{4^5} +\frac{\pi}{2}$

$2+\frac{\pi}{2}+\frac{\pi^4}{1024}$

### Question:18 Evaluate the definite integrals in Exercises 1 to 20.

$\int^\pi_0(\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx$

Given integral: $\int^\pi_0(\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx$

Consider the integral $\int (\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx$

can be rewritten as: $-\int (\cos^2\frac{x}{2} - \sin^2\frac{x}{2})dx = -\int_0^{\pi} \cos x dx$

$= \sin x$

So, we have the function of $x$ , $f(x) =\sin x$

Now, by Second fundamental theorem of calculus, we have

$I = f(\pi) - f(0)$

$\Rightarrow \sin \pi - \sin 0 = 0-0 =0$

### Question:19 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^2\frac{6x+3}{x^2+ 4}$

Given integral: $\int_0^2\frac{6x+3}{x^2+ 4}$

Consider the integral $\int \frac{6x+3}{x^2+ 4}$

can be rewritten as: $\int \frac{6x+3}{x^2+ 4} = 3\int \frac{2x+1}{x^2+4}dx$

$= 3\int \frac{2x}{x^2+4}dx +3\int \frac{1}{x^2+4}dx$

$= 3\log (x^2+4) +\frac{3}{2}\tan^{-1}\frac{x}{2}$

So, we have the function of $x$ , $f(x) =3\log (x^2+4) +\frac{3}{2}\tan^{-1}\frac{x}{2}$

Now, by Second fundamental theorem of calculus, we have

$I = f(2) - f(0)$

$= \left \{ 3\log(2^2+4)+\frac{3}{2}\tan^{-1}\left ( \frac{2}{2} \right ) \right \}- \left \{ 3\log(0+4)+\frac{3}{2}\tan^{-1}\left ( \frac{0}{2} \right ) \right \}$ $=3\log 8 +\frac{3}{2}\tan^{-1}1 -3\log 4 -\frac{3}{2}\tan^{-1} 0$

$=3\log 8 +\frac{3}{2}\times\frac{\pi}{4} -3\log 4 -0$

$=3\log \frac{8}{4} +\frac{3\pi}{8}$

or we have $=3\log 2 +\frac{3\pi}{8}$

### Question:20 Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^1(xe^x + sin\frac{\pi x}{4})dx$

Given integral: $\int_0^1(xe^x + sin\frac{\pi x}{4})dx$

Consider the integral $\int (xe^x + sin\frac{\pi x}{4})dx$

can be rewritten as: $x\int e^x dx - \int \left \{ \left ( \frac{d}{dx}x \right )\int e^x dx \right \}dx +\left \{ \frac{-\cos \frac{\pi x}{4}}{\frac{\pi}{4}} \right \}$

$= xe^x -\int e^x dx -\frac{4\pi}{\pi} \cos \frac{x}{4}$

$= xe^x -e^x -\frac{4\pi}{\pi} \cos \frac{x}{4}$

So, we have the function of $x$ , $f(x) = xe^x -e^x -\frac{4\pi}{\pi} \cos \frac{x}{4}$

Now, by Second fundamental theorem of calculus, we have

$I = f(1) - f(0)$

$= \left (1.e^t-e^t - \frac{4}{\pi}\cos \frac{\pi}{4} \right ) - \left ( 0.e^0 -e^0 -\frac{4}{\pi}\cos 0 \right )$

$= e-e -\frac{4}{\pi}\left ( \frac{1}{\sqrt2} \right )+1+\frac{4}{\pi}$

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

$\int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2}$

(A) $\frac{\pi}{3}$

(B) $\frac{2\pi}{3}$

(C) $\frac{\pi}{6}$

(D) $\frac{\pi}{12}$

Given definite integral $\int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2}$

Consider $\int \frac{dx}{1 +x^2} = \tan^{-1}x$

we have then the function of x, as $f(x) = \tan^{-1}x$

By applying the second fundamental theorem of calculus, we will get

$\int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2} = f(\sqrt3) - f(1)$

$= \tan^{-1}\sqrt{3} - \tan^{-1}1$

$=\frac{\pi}{3} - \frac{\pi}{4}$

$= \frac{\pi}{12}$

Therefore the correct answer is D.

### Question:22 Choose the correct answer in Exercises 21 and 22.

$\int_0^\frac{2}{3}\frac{dx}{4+ 9x^2}$ equals

(A) $\frac{\pi}{6}$

(B) $\frac{\pi}{12}$

(C) $\frac{\pi}{24}$

(D) $\frac{\pi}{4}$

Given definite integral $\int_0^\frac{2}{3}\frac{dx}{4+ 9x^2}$

Consider $\int \frac{dx}{4+ 9x^2} = \int \frac{dx}{2^2+(3x)^2}$

Now, putting $3x = t$

we get, $3dx=dt$

Therefore we have, $\int \frac{dx}{2^2+(3x)^2} = \frac{1}{3}\int \frac{dt}{2^2+t^2}$

$= \frac{1}{3}\left ( \frac{1}{2}\tan^{-1}\frac{t}{2} \right ) = \frac{1}{6}\tan^{-1}\left ( \frac{3x}{2} \right )$

we have the function of x , as $f(x) =\frac{1}{6}\tan^{-1}\left ( \frac{3x}{2} \right )$

So, by applying the second fundamental theorem of calculus, we get

$\int_0^\frac{2}{3}\frac{dx}{4+ 9x^2} = f(\frac{2}{3}) - f(0)$

$= \frac{1}{6}\tan^{-1}\left ( \frac{3}{2}.\frac{2}{3} \right ) -\frac{1}{6}\tan^{-1}0$

$= \frac{1}{6}\tan^{-1}1 - 0$

$= \frac{1}{6}\times \frac{\pi}{4} = \frac{\pi}{24}$

Therefore the correct answer is C.

## More About NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.9

The NCERT Class 12 Maths chapter Integrals is one of the most important chapters of Class 12 Maths NCERT syllabus. If we talk about exercise 7.9 Class 12 Maths, it holds good weightage in the Board examination. NCERT Solutions for Class 12 Maths chapter 7 exercise 7.9 can be learnt easily if practice of some questions is done on a regular basis.

Also Read| Integrals Class 12 Notes

## Benefits of NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.9

• The Class 12th Maths chapter 7 exercise is a one stop solution for clearing concepts as well as practicing questions.

• Practicing exercise 7.9 Class 12 Maths can be of great help to the students aspiring to score well in the examination.

• These Class 12 Maths chapter 7 exercise 7.9 solutions can be asked directly in Board examination which can be verified from looking Previous year questions.

• Along with NCERT solutions for Class 12 Maths chapter 7 exercise 7.9 one must also go through the previous year questions of the same topics.

## Also see-

• NCERT Exemplar Solutions Class 12 Maths Chapter 7

• NCERT Solutions for Class 12 Maths Chapter 7

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