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Definite integrals are all about mastering the art of accumulation, it's like how the little changes in areas, distances, and volumes add up over an interval. In exercise 7.10 of the chapter Integrals, we will learn about some properties of definite integrals. These properties will help the students simplify and evaluate problems related to definite integrals with ease. This article on the NCERT Solutions for Exercise 7.10 Class 12 Maths Chapter 7 - Integrals, provides detailed and step-by-step solutions to the problems given in the exercise, so that students will be able to clear any doubts they have and understand the applications of the properties of definite integrals. For syllabus, notes, and PDF, refer to this link: NCERT.
Question 1: By using the properties of definite integrals, evaluate the integral
$\int_0^\frac{\pi}{2}\cos^2 x dx$
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 :-
$\int_0^{\frac{\pi}{2}} \text{cos}^2 x\ dx + \int_0^{\frac{\pi}{2}} \text{sin}^2 x\ dx = 2I$
$\int_0^{\frac{\pi}{2}} (\text{cos}^2 x + \text{sin}^2 x)\ dx = 2I$
$\int_0^\frac{\pi}{2}1. dx\ =\ 2I$
or $2I = \left[ x \right]_0^{\frac{\pi}{2}} = \frac{\pi}{2}$
or $I = \frac{\pi}{4}$
Question 2: By using the properties of definite integrals, evaluate the integral
. $\int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx$
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,
$2I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}\ +\ \sqrt{\cos x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx$
$2I\ =\ \int_0^\frac{\pi}{2}1.dx$
or $2I = \left[ x \right]_0^{\frac{\pi}{2}} = \frac{\pi}{2}$
or $I = \frac{\pi}{4}$
Question 3: By using the properties of definite integrals, evaluate the integral
$\int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}$
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 :
$2I = \int_0^{\frac{\pi}{2}} \frac{\ (\text{sin}^{\frac{3}{2}}x + \text{cos}^{\frac{3}{2}}x)\,dx}{\text{sin}^{\frac{3}{2}}x + \text{cos}^{\frac{3}{2}}x}$
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 integral
. $\int_0^\frac{\pi}{2} \frac{\cos^5 xdx}{\sin^5x + \cos^5x}$
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 :
$2I = \int_0^{\frac{\pi}{2}} \frac{(\text{sin}^5 x + \text{cos}^5 x)\,dx}{\text{sin}^5 x + \text{cos}^5 x}$
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 5: By using the properties of definite integrals, evaluate the integral
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 integral
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 integral
Answer:
We have $I\ =\ \int^1_0x(1-x)^ndx$
Using the property : -
$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$
We get : -
$I\ =\ \int^1_0x(1-x)^ndx\ =\ \int^1_0(1-x)(1-(1-x))^ndx$
$I\ =\ \int^1_0(1-x)x^n\ dx$
or $I\ =\ \int^1_0(x^n\ -\ x^{n+1}) \ dx$
$=\ \left [ \frac{x^{n+1}}{n+1}\ -\ \frac{x^{n+2}}{n+2} \right ]^1_0$
$=\ \left [ \frac{1}{n+1}\ -\ \frac{1}{n+2} \right ]$
or $I\ =\ \frac{1}{(n+1)(n+2)}$
Question 8: By using the properties of definite integrals, evaluate the integral
$\int_0^\frac{\pi}{4}\log(1+\tan x)dx$
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,
$I\ =\ \int_0^\frac{\pi}{4}\log(1+\tan x)dx\ =\ \int_0^\frac{\pi}{4}\log(1+\tan (\frac{\pi}{4}-x))dx$
$I\ =\ \int_0^\frac{\pi}{4}\log(1+\frac{1-\tan x}{1+\tan x})dx$
$I\ =\ \int_0^\frac{\pi}{4}\log(\frac{2}{1+\tan x})dx$
$I\ =\ \int_0^\frac{\pi}{4}\log{2}dx\ -\ \int_0^\frac{\pi}{4}\log(1+ \tan x)dx$
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 integral
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 integral
$\int_0^\frac{\pi}{2} (2\log\sin x- \log\sin 2x)dx$
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 integral.
$\int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^2 x dx$
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 integral.
$\int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^7xdx$
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 integral.
Answer:
We have $I\ =\ \int_0^{2\pi}\cos^5xdx$
It 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 integral.
$\int_0^{\frac{\pi}{2}} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx$
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 integral.
$\int_0^\pi\log(1 +\cos x)dx$
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 integral.
$\int_0^a \frac{\sqrt x}{\sqrt x + \sqrt{a-x}}dx$
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 integral.
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 $I\ =\ 5$
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
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:-
$I = 0 + 0 + 0 + 2\int_0^{\frac{\pi}{2}} 1\,dx$
$I = 2\left[ x \right]_0^{\frac{\pi}{2}}$
$I = \pi$
Thus, correct answer is $\pi$.
Question 21: Choose the correct answer
The value of $\int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx$ is
(A) 2
(B) 3/4
(C) 0
(D) -2
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, $I\ =\ 0$
Also Read,
The main topic covered in class 12 maths chapter 7 of Integrals, exercise 7.10 is:
Some properties of definite integrals: Some useful properties of definite integrals are given below that will help evaluate definite integrals easily.
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Also Read,
Below are some useful links for subject-wise NCERT solutions for class 12.
Here are some links to subject-wise solutions for the NCERT exemplar class 12.
Frequently Asked Questions (FAQs)
Two types of integration are Definite and indefinite Integrals.
It can be anything in which the given function is real.
Integrals are used in finding area, volume, displacement etc.
No, one should do this exercise as 5 marks questions can be asked in the Board examination.
Some definite integrals of advance level are discussed in this exercise.
Yes, provided it fulfills the demand of the question.
On Question asked by student community
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