RD Sharma Solutions Class 12 Mathematics Chapter 8 FBQ

# RD Sharma Solutions Class 12 Mathematics Chapter 8 FBQ

Edited By Kuldeep Maurya | Updated on Jan 27, 2022 01:56 PM IST

The RD Sharma Solutions Class 12 Chapter 8 – Continuity is planned by our specialists to support certainty among students in understanding the ideas canvassed in this chapter and strategies to take care of issues in a more limited period. At Career360 RD Sharma class 12th exercise FBQ helps students who seek to get a decent scholarly score in the exam.

Also Read - RD Sharma Solutions For Class 9 to 12 Maths

## Continuity Excercise: FBQ

### Continuity exercise Fill in the blanks question 1

Answer: $3a^{2}$
Hint:
Use the formula $(x^{3}-a^{3})$ so that $f(x)\neq \frac{0}{0}$ at $x\neq a$
Given:
$f(x)=\left\{\begin{array}{cl} \frac{x^{3}-a^{3}}{x-a} & , x \neq a \\ b & , x=a \end{array}\right.$ is continuous at $x=a$
Solution:
If $f(x)$ is continuous at $x=a$ , then RHL= LHL
LHL,
\begin{aligned} &\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a} \frac{x^{3}-a^{3}}{x-a} \\ &=\lim _{x \rightarrow a^{-}} \frac{(x-a)\left(x^{2}+a x+a^{2}\right)}{(x-a)} \\ &=\lim _{x \rightarrow a^{-}} x^{2}+a x+a^{2} \\ &=\lim _{\mathbb{\Xi} \rightarrow 0} a^{2}+a^{2}+a^{2} \\ &=3 a^{2} \end{aligned}$\left [ \because x=a \right ]$
RHL,
\begin{aligned} &\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{+}} b \\ &=b \end{aligned}
As$f(x)$ is continuous at $x=a$, RHL = LHL
$b=3a^{2}$

### Continuity exercise Fill in the blanks question 2

Answer: $\pm 1$
Hint:
Use the formula of $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$ to solve $\frac{\sin ^{2} a x}{x^{2}}$
Given:
$f(x)=\left\{\begin{array}{cl} \frac{\sin ^{2} a x}{x^{2}}, & x \neq 0 \\ 1, & x=0 \end{array}\right.$is continuous at $x=0$
Solution:
If $f(x)$ is continuous at $x=0$, then $f(x)=\frac{\sin ^{2} a x}{x^{2}}$ and $g(x)=1$ are equal.
LHL,
\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{\sin ^{2} a x}{x^{2}} \\ &=\lim _{x \rightarrow 0^{-}} \frac{\sin ^{2} a x}{a^{2} x^{2}}\left(a^{2}\right) \\ \end{aligned}
$=\lim_{x\rightarrow 0^{-}}\left ( \frac{\sin ax}{ax} \right )^{2}\left ( a^{2} \right )$
$=a^{2}$ $\left[\because \frac{\sin ^{2} a x}{a^{2} x^{2}}=1\right]$
RHL,
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} 1=1$
As $f(x)$ is continuous at $x=0$, LHL =RHL
\begin{aligned} &a^{2}=1 \\ &a=\pm 1 \end{aligned}
Hint: You must know about the concept of continuous function
Given:
$f(x)=\left\{\begin{array}{l} a x^{2}-b, 0 \leq x<1 \\ 2, x=1 \\ x+1,1
Solution:

If $f(x)$ is continuous at $x=1$, then
\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} f(x) \\ &\lim _{x \rightarrow 1^{-}} a x^{2}-b=\lim _{x \rightarrow 1^{+}} x+1=\lim _{x \rightarrow 1} 2 \end{aligned}
$\lim _{h \rightarrow 0} a(1-h)^{2}-b=\lim _{h \rightarrow 0}(1+h)+1=\lim _{x \rightarrow 1} 2$
\begin{aligned} &a-b=2=2 \\ &a-b=2 \end{aligned}
Hint: If $f(x)$is continuous, then all the functions are equal
Given:
$f(x)=\left\{\begin{array}{l} x+k, x<3 \\ 4, x=3 \quad \text { is continuous at } x=3 \\ 3 x-5, x>3 \end{array}\right.$
Solution
:
We know that as function is continuous at $x=3$ ,
\begin{aligned} &\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3} f(x) \\\\ &\lim _{x \rightarrow 3^{-}} x+k=\lim _{x \rightarrow 3^{+}} 3 x-5=\lim _{x \rightarrow 3} 4 \end{aligned}
$\lim _{h \rightarrow 0}(3-h)+k=\lim _{h \rightarrow 0} 3(3+h)-5=\lim _{x \rightarrow 3} 4$
\begin{aligned} &3+k=3(3)-5=4 \\ &3+k=4=4 \\ &k=1 \end{aligned}

Answer: $b=-1, a=1,a+b=0$

Hint: $\left | x \right |=x$, When $x< 0$ and $\left | x \right |=x$ , when $x> 0$
Given:
$f(x)=\left\{\begin{array}{l} \frac{x-4}{|x-4|}+a, x<4 \\ a+b, x=4 \quad \text { is continuous at } x=4 \\ \frac{x-4}{|x-4|}+b, x>4 \end{array}\right.$

Solution:
As the function is continuous at $x=4$,
\begin{aligned} &\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{+}} f(x)=\lim _{x \rightarrow 4} f(x) \\ &\lim _{x \rightarrow 4^{-}} \frac{x-4}{|x-4|}+a=\lim _{x \rightarrow 4^{+}} \frac{x-4}{|x-4|}+b=\lim _{x \rightarrow 4} a+b \end{aligned}
$\lim _{h \rightarrow 0} \frac{(4-h)-4}{|(4-h)-4|}+a=\lim _{h \rightarrow 0} \frac{(4+h)-4}{|(4+h)-4|}+b=\lim _{x \rightarrow 4} a+b$
\begin{aligned} &\lim _{h \rightarrow 0} \frac{-h}{|h|}+a=\lim _{x \rightarrow 4^{+}} \frac{h}{|h|}+b=a+b \\ &-1+a=1+b=a+b \end{aligned}
\begin{aligned} &a+b=-1+a \quad, \quad a+b=1+b \\ &b=-1 \quad, \quad a=1 \end{aligned}
Now, $a+b=1-1=0$

Hint: Use the formula of $\cos 3 x=4 \cos ^{3} x-3 \cos x$
Given:

$f(x)=\left\{\begin{array}{cl} \frac{\cos 3 x-\cos x}{x^{2}}, & x \neq 0 \\ \lambda & , x=0 \end{array}\right.$ is continuous at $x=0$
Solution:

We know that
$\lim _{x \rightarrow 0} f(x)=f(0),$ when function is continuous
$\therefore \lim _{x \rightarrow 0} f(x)=f(0)$
\begin{aligned} &\lim _{x \rightarrow 0} \frac{\cos 3 x-\cos x}{x^{2}}=\lambda \\ &\lim _{x \rightarrow 0} \frac{\left(4 \cos ^{3} x-3 \cos x\right)-\cos x}{x^{2}}=\lambda \\ &\lim _{x \rightarrow 0} \frac{4 \cos ^{3} x-4 \cos x}{x^{2}}=\lambda \\ &\lim _{x \rightarrow 0} \frac{4 \cos x\left(\cos ^{2} x-1\right)}{x^{2}}=\lambda \\ &\lim _{x \rightarrow 0} \frac{4 \cos x\left(-\sin ^{2} x\right)}{x^{2}}=\lambda \\ &\lim _{x \rightarrow 0}-4 \cos x=\lambda \\ &-4(1)=\lambda \end{aligned}
$\lambda=-4 \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \because \cos 3 x=4 \cos ^{3} x-3 \cos x \\ \because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \\ \because \cos 0^{\circ}=1 \end{array}\right]$

Continuity exercise Fill in the blanks question 7

Hint: Use L-Hospital’s rule
Given:$f(x)=\left\{\begin{array}{cc} \frac{1-\sin x}{\pi-2 x}, & x \neq \frac{\pi}{2} \\ k & , x=\frac{\pi}{2} \end{array}\right.$ is continuous at $x=\frac{\pi }{2}$

Solution:
If $f(x)$ is continuous at $x=\frac{\pi }{2}$ , then
\begin{aligned} &\lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right) \\ &\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\pi-2 x}=k \end{aligned}
Applying L-Hospital’s rule on LHS that is differentiating both numerator and denominator,
\begin{aligned} &\lim _{x \rightarrow \frac{\pi}{2}} \frac{0-(\cos x)}{0-2(1)}=k \\ &\lim _{x \rightarrow \frac{\pi}{2}} \frac{-\cos x}{-2}=k \\ &\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{2}=k \\ &\frac{\cos \frac{\pi}{2}}{2}=k \end{aligned}
$k=0 \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \cos \frac{\pi}{2}=0\right]$

### Continuity exercise Fill in the blanks question 8

Answer:$f(0)=\frac{-1}{8}$

Hint: Use L-Hospital’s rule

Given:$f(x)=\frac{2-\sqrt{x+4}}{\sin 2 x}, x \neq 0$and $f(x)$ is continuous at $x=0$

Solution:
As $f(x)$ is continuous at $x=0$,
\begin{aligned} &f(0)=\lim _{x \rightarrow 0} f(x) \\ &f(0)=\lim _{x \rightarrow 0} \frac{2-\sqrt{x+4}}{\sin 2 x} \end{aligned}
Using L-Hospital’s rule,
$f(0)=\lim _{x \rightarrow 0} \frac{0-\frac{1}{2 \sqrt{x+4}}}{2 \cos 2 x}$
\begin{aligned} &f(0)=\lim _{x \rightarrow 0} \frac{-1}{(2 \sqrt{x+4})(2 \cos 2 x)} \\ &f(0)=\lim _{x \rightarrow 0} \frac{-1}{4(\sqrt{x+4})(\cos 2 x)} \end{aligned}
\begin{aligned} &f(0)=\frac{-1}{4(1)(\sqrt{4})} \\ &f(0)=\frac{-1}{4(2)} \\ &f(0)=\frac{-1}{8} \end{aligned}

Continuity exercise Fill in the blanks question 9

Answer: $k=0$

Hint:Use the formula of $(a^{2}-b^{2})$

Given:
$f(x)=\left\{\begin{array}{l} \frac{x^{2}-9}{x-3}, x \neq 3 \\ 2 x+k, x=3 \end{array}\right.$ is continuous at $x=3$
Solution:
If $f(x)$ is continuous at $x=3$ , then
\begin{aligned} &\lim _{x \rightarrow 3} f(x)=f(3) \\ &\lim _{x \rightarrow 3} \frac{x^{2}-9}{x-3}=2(3)+k \end{aligned}
\begin{aligned} &\lim _{x \rightarrow 3} \frac{(x-3)(x+3)}{x-3}=2(3)+k \\ &\lim _{x \rightarrow 3} x+3=6+k \end{aligned}
\begin{aligned} &3+3=6+k \\ &6=6+k \\ &k=0 \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because\left(a^{2}-b^{2}\right)=(a+b)(a-b)\right] \end{aligned}

Continuity exercise Fill in the blanks question 10

Hint: Use the formula of $(a^{2}-b^{2})$
Given:$f(x)=\left\{\begin{array}{cl} \frac{x^{2}-1}{x-1}, & x \neq 1 \\ k & , x=1 \end{array} \text { is continuous at } x=1\right.$
Solution:
If $f(x)$ is continuous at $x=1$, then
\begin{aligned} &\lim _{x \rightarrow 1} f(x)=f(1) \\ &\lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1}=k \end{aligned}
$\lim _{x \rightarrow 1} \frac{(x-1)(x+1)}{x-1}=k \; \; \; \; \; \; \; \; \quad\left[\because\left(a^{2}-b^{2}\right)=(a+b)(a-b)\right]$
\begin{aligned} &\lim _{x \rightarrow 1} x+1=k \\ &1+1=k \\ &k=2 \end{aligned}

Continuity exercise Fill in the blanks question 12

Answer: $f\left(\frac{\pi}{4}\right)=\frac{-1}{2}$
Hint: You must know about L-hospital’s rule
Given:$f(x)=\frac{1-\tan x}{4 x-\pi}, x \neq \frac{\pi}{4}, x \in\left[0, \frac{\pi}{2}\right] \text { and } f(x) \text { is continuous in }\left[0, \frac{\pi}{2}\right]$
Solution:
$f(x)$ is continuous in $\left [ 0,\frac{\pi }{2} \right ]$
\begin{aligned} &\lim _{x \rightarrow \frac{\pi}{4}} f(x)=f\left(\frac{\pi}{4}\right) \\ &\lim _{x \rightarrow \frac{\pi}{4}} \frac{1-\tan x}{4 x-\pi}=f\left(\frac{\pi}{4}\right) \end{aligned}
Applying L-Hospital’s rule
\begin{aligned} &\lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sec ^{2} x}{4}=f\left(\frac{\pi}{4}\right) \\ &\frac{-2}{4}=f\left(\frac{\pi}{4}\right) \end{aligned} $\quad\left [\because \sec ^{2} \frac{\pi}{4}=2\right]$

Continuity exercise Fill in the blanks question 13

Answer: $\pi$
Hint: Use the identity $\frac{\sin x}{x}=1$
Given:
$f(x)=x\sin \left ( \frac{\pi }{x} \right )$ is continuous everywhere
Solution:
$f(x)=x\sin \left ( \frac{\pi }{3} \right )$ is continuous
\begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\ &\lim _{x \rightarrow 0} x \sin \left(\frac{\pi}{x}\right)=f(0) \\ &\lim _{x \rightarrow 0} \frac{x \sin \left(\frac{\pi}{x}\right)}{\frac{\pi}{x}} \times \frac{\pi}{x}=f(0) \end{aligned}
\begin{aligned} &\lim _{x \rightarrow 0} x \times \frac{\pi}{x}=f(0) \\ &\lim _{x \rightarrow 0} \pi=f(0) \\ &f(0)=\pi \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \end{aligned}\begin{aligned} &\lim _{x \rightarrow 0} x \times \frac{\pi}{x}=f(0) \\ &\lim _{x \rightarrow 0} \pi=f(0) \\ &f(0)=\pi \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \end{aligned}\begin{aligned} &\lim _{x \rightarrow 0} x \times \frac{\pi}{x}=f(0) \\ &\lim _{x \rightarrow 0} \pi=f(0) \\ &f(0)=\pi \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \end{aligned}

Continuity exercise Fill in the blanks question 14

Hint: Function is discontinuous where denominator is zero
Given: $f(x)=\frac{1}{\log |x|}$
Solution:
$\log x$ is not defined at $x=0$.
So,$x=0$, $f(x)$ is discontinuous.
$\log \left | x \right |=0 , when \left | x \right |=1$
i.e. $x=1,-1$
When $x=1,-1$ , value of $\frac{1}{\log \left | x \right |}=\infty$
$x=1,-1$ act as vertical asymptotes
Therefore,$f(x)$ is discontinuous at $x=-1,0,1$

Hint: $f(x)$ is continuous when $LHL=RHL$
Given: f(x)=\left\{\begin{aligned} a x+1, & \text { if } x \geq 1 \\ x+2, & \text { if } x<1 \end{aligned}\right.
Solution:
f(x)=\left\{\begin{aligned} a x+1, & \text { if } x \geq 1 \\ x+2, & \text { if } x<1 \end{aligned}\right.
$f(x)$ is continuous when $LHL=RHL$
$\lim _{x \rightarrow 1^{+}} a x+1=\lim _{x \rightarrow 1^{-}} x+2 \text { at } x=1$
$\lim _{h \rightarrow 0} a(1+h)+1=\lim _{h \rightarrow 0}(1-h)+2$
\begin{aligned} &a+1=3 \\ &a=2 \end{aligned}

Continuity exercise Fill in the blanks question 16

Answer: $f(a)$
Hint: $f(x)$ is continuous when $LHL=RHL$
Given: $f(x)$ is continuous at $x=a$
Solution:
$f(x)$ is continuous at $x=a$
\begin{aligned} &\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)=k \\ &L H L=R H L=f(a) \\ &k=f(a) \end{aligned}

Continuity exercise Fill in the blanks question 17

Hint: $f(x)$ is continuous when $LHL=RHL$
Given: $f(x)=\left\{\begin{array}{c} \frac{\sin 3 x}{x}, \text { if } x \neq 0 \\ \frac{k}{2}, \text { if } x=0 \end{array}\right.$
Solution:
$f(x)=\left\{\begin{array}{c} \frac{\sin 3 x}{x}, \text { if } x \neq 0 \\ \frac{k}{2}, \text { if } x=0 \end{array}\right.$is continuous at $x=0$
At $x=0$
$LHL=RHL$ $=f(a)$
\begin{aligned} &\lim _{x \rightarrow 0} \frac{\sin 3 x}{x}=\frac{k}{2} \\ &\lim _{x \rightarrow 0} \frac{3}{3} \frac{\sin 3 x}{x}=\frac{k}{2} \\ &3 \lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x}=\frac{k}{2} \end{aligned} $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
\begin{aligned} &3(1)=\frac{k}{2} \\ &k=6 \end{aligned}

Continuity exercise Fill in the blanks question 18

Answer: $(2 n+1) \frac{\pi}{2}, n \in I$
Hint: $f(x)$is discontinuous when $f\rightarrow \infty$
Given: $f(x)=\tan x$
Solution:
$f(x)=\tan x$
The points of discontinuity of $f(x)$ are there at which $\tan x$ is infinite
\begin{aligned} &\tan x \rightarrow \infty \\ &\tan x=\tan \frac{\pi}{2} \end{aligned} \; \; \; \; \; \; \; \; \; \; \; \; \; \left[\begin{array}{l} \because \tan x=\tan y \\ x=(2 n+1) y \end{array}\right]
$x=(2 n+1) \frac{\pi}{2}, n \in I$

Continuity exercise Fill in the blanks question 19

Hint: $f(x)$ is discontinuous when $f(x)$ is not defined
Given: $f(x)=[x]$
Solution:
$f(x)=[x]$
Continuity will be measured at
$\bullet$$x$ is an integer
$\bullet$ $x$ is not an integer
Case I: $x$ is not an integer
\begin{aligned} &\lim _{x \rightarrow d} f(x)=f(d) \\ &{[d]=[d]} \end{aligned}
$f(x)$ is continuous at all non-integer
Case II: $x$ is an integer
$f(x)=[x]$ is continuous if $LHL=RHL=f(c)$
\begin{aligned} &\lim _{x \rightarrow c^{-}} f(x)=\lim _{k \rightarrow 0} f(c-h)=c-h=c-1 \\ &\lim _{x \rightarrow c^{+}} f(x)=\lim _{k \rightarrow 0} f(c+h)=c+h=c \\ &c \neq c-1 \\ &L H L \neq R H L \end{aligned}
$f(x)$ is discontinuous at all integer

Continuity exercise Fill in the blanks question 20

Hint:$f(x)$ is discontinuous when $f(x)$ is not defined
Given: $f(x)=\frac{1}{x-[x]}$
Solution:
$f(x)=\frac{1}{x-[x]}$

$\left [ x \right ]$ is discontinuous at all integer points
$x-\left [ x \right ]$ is discontinuous at all integer points
$\frac{1}{x-\left [ x \right ]}$ is discontinuous at each integer value of $x$
All integral points, $f(x)$ is discontinuous

Continuity exercise Fill in the blanks question 3

Hint: You must know about the concept of continuous function
Given:
$f(x)=\left\{\begin{array}{l} a x^{2}-b, 0 \leq x<1 \\ 2, x=1 \\ x+1,1
Solution:

If $f(x)$ is continuous at $x=1$, then
\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} f(x) \\ \\&\lim _{x \rightarrow 1^{-}} a x^{2}-b=\lim _{x \rightarrow 1^{+}} x+1=\lim _{x \rightarrow 1} 2 \end{aligned}
$\lim _{h \rightarrow 0} a(1-h)^{2}-b=\lim _{h \rightarrow 0}(1+h)+1=\lim _{x \rightarrow 1} 2$
\begin{aligned} &a-b=2=2 \\ &a-b=2 \end{aligned}

This chapter of RD Sharma class 12th exercise FBQ principally centers around the idea of coherence. To find out about this theme students can download the RD Sharma class 12 solutions FBQ Chapter 8 Continuity. This chapter clarifies Continuity of a point, Continuity of an interval (open or closed) and its applications with tackled examples. RD Sharma class 12 solutions FBQ, the specialists give answer keys as well as some remarkable tips in the book that the students probably won't discover elsewhere. RD Sharma class 12th exercise FBQ has around 20 questions.

The class 12 RD Sharma chapter 8 exercise FBQ arrangement is exceptionally trusted and suggested by students and instructors across the whole country. The appropriate responses given in the RD Sharma class 12th exercise FBQ are handpicked and made by specialists, which makes them precise and reasonable enough for students.

## RD Sharma Chapter-wise Solutions

1. . Are the RD Sharma class 12 solutions chapter 8 FBQ considered the best review material for exam preparations?

The RD Sharma class 12 solution of Continuity exercise FBQ is made by the most recent CBSE rules and prospectus. Every one of the solutions is outlined by individual subject matter specialists to assist students with acing the exam. Further, the RD Sharma class 12th exercise FBQ are arranged as PD

2. Is the RD Sharma class 12 Solutions chapter 8 FBQ of the latest syllabus?

Yes, RD Sharma class 12th exercise FBQ  is as per the latest version. The solutions are yearly revised according to the syllabus of NCERT. This makes it more efficient for practice.

3. How could students plan for the yearly exam utilizing the RD Sharma Class 12 Continuity Solution exercise FBQ?

It's significant that students get familiar with the chapter and practice all the exercise questions completely to accomplish a solid grasp of the key ideas. The  RD Sharma class 12 solutions FBQ Chapter 8 are created by master subject mentors at CAREER360 and are an adept asset for exam preparations.

4. How might students utilize the RD Sharma class 12 exercise FBQ Chapter 8 to get ready for the yearly exam?

Class 12 RD Sharma chapter 8 exercise FBQ, to acquire an unmistakable comprehension of the key ideas, students should completely concentrate on the chapter and practice the entirety of the exercise questions.

5. What number of questions are there in RD Sharma Solutions Class 12 Maths Chapter 8 FBQ?

There is an aggregate of 20 questions in RD Sharma Solutions Class 12 RD Sharma chapter 8 exercise FBQ.

## Upcoming School Exams

#### National Means Cum-Merit Scholarship

Application Date:01 August,2024 - 16 September,2024

Exam Date:19 September,2024 - 19 September,2024