Class 12th

24. Given, f(x) = $\{\begin{array}{cc}\hfill sinx-cosx,\hfill & \hfill if x\ne 0\hfill \\ \hfill -1,\hfill & \hfill if x=0.\hfill \end{array}$

For x = c $\cancel{=}$ 0,

f(c) = sin c cos c.

$\underset{x\to c}{lim}$ f (x) = $\underset{x\to c}{lim}$ (sin x cos x) = sin c cos c = f(c)

So, f is continuous at $x\ne 0$

For x = 0,

f(0) = 1

$\underset{x\to 0}{lim}$ f (x) = $\underset{x\to 0}{lim}$ (sin x cos x) = sin 0 cos 0 = 0 1 = 1

$\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{+}}{lim}(sinx-corx)=sin0-cos0=-1.$

∴ $\underset{x\to 0-}{lim}$ f(x) = $\underset{x\to 0+}{lim}$ f (x) = f (0)

So, f is continuous at x = 0.

Find the values of $k$ so that the function $f$ is continuous at the indicated point in Exercises 26 to 29.

23. Given f (x) = $\{\begin{array}{ll}{x}^{2}sin\frac{1}{x}, if x\ne 0.\hfill & 0 if x=0.\hfill \end{array}$

For x = c $\cancel{=}$ 0,

f (c) = $c^{2}sin\frac{1}{c}$

$\underset{x\to c}{lim}f(x)=\underset{x\to c}{lim}{x}^{2}sin\frac{1}{x}=c^{2}sin\frac{1}{c}.$

So, f is continuous for $x\ne 0.$

For x = 0,

f (0) = 0

$\underset{x\to 0}{lim}f(x)=\underset{x\to 0}{lim}\left(x^{2}sin\frac{1}{x}\right)$

As we have sin $\frac{1}{x}\in [-1,1]$

$\underset{x\to 0}{lim}$ f (x) = 02 a where $a\in [-1,1]$

= 0 = f (0).

∴ f is also continuous at x = 0.

22. Given f(x) = $\{\begin{array}{ll}\frac{sinx}{x},\hfill & if x<0.\hfill \\ x+1,\hfill & if x\ge 0.\hfill \end{array}$

For x = c < 0,

f(c) = $\frac{sinc}{c}$

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ $\frac{sinx}{x}=\frac{sinc}{c}=f(c)$

So, f is continuous for x < 0

For x = c > 0

f(c) = c + 1

$\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ x + 1 = c + 1 = f(c)

So, f is continuous for x > 0.

For x = 0.

L.H.L. = $\underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{-}}{lim}\frac{sinx}{x}=1.$

R.H.S. = $\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{+}}{lim}x+1=0+1=1$

And f(0) = 0 + 1 = 1

L.H.L = R.H.L. = f(0)

So, f is continuous at x = 1.

Hence, discontinuous point of x does not exit.

Students who complete Class 12 Humanities stream have a number of UG courses to join. They can go for BA, LLB, BDes in Design, BSc in Hospitality & Travel, BJMC, BDes in Animation, BFA, BMM, Psychology, etc. They can also join for mass communication, journalism, social work, anthropology, hotel management, etc. Check courses after 12th for Arts/Humanities students.

Those who completed Class 12th in Commerce stream can go for Bcom, Company BBA, LLB, Economics, Economics, etc. There are so many degree courses for commerce students. Check few courses for commerce students below.

1. B.Com Accounting and Taxation
2. B.Com Applied Economics
3. B.Com (Honours)
4. Bachelor of Business Administration
5. Chartered Accountancy
6. Company Secretary
7. Company3 Accountancy
8. Company3 of Accounting and Finance
9. Bachelor of Management Studies
10. Bachelor of Foreign Trade

The Class 12 grade plays an important role in college admissions. Many universities or colleges have a minimum percentage requirement to apply for different courses. For engineering, 50% marks in Class 12 with Physics, Chemistry, and Mathematics (PCM) are generally required for admission to many colleges. For some degree courses, admissions are based completely on your 12th marks. For entrance exam-based admissions (like JEE, NEET), the 12th percentage will also be considered as a criteria. Hence, your 12th grade marks are important for your higher education admission.

19. Given f (x) = x2 sin x + 5.

At x = .

$f(\pi )={\pi }^2-sin\pi +5={\pi }^2-0+5={\pi }^2+5$

$\underset{x\to \pi }{lim}$ f (x) = $\underset{x\to \pi }{lim}$  [x2 sin x + 5]

If x = $\pi +h$ then as x, h 0, so,

$\underset{x\to \pi }{lim}$ f (x) = $\underset{x\to 0}{lim}$  [ ( + h)2 sin ( + h) + 5]

= ( + 0)2 $\underset{h\to 0}{lim}$ $\begin{array}{rr}\hfill [sin\pi cosh+cos\pi \cdot sina]& \hfill +5.\end{array}$

= 2 $\underset{h\to 0}{lim}$ sin cos h $\underset{h\to 0}{lim}$ cos sin h + 5

= x2 0 * (1) ( 1) 0 + 5.

= 2 + 5 = f (x)

So, f is continuous at x = .

18. Given, g (x) = x [x].

For $n\in z,$

g (n) = n [n] = nn = 0

$\underset{x\to n^{-}}{lim}$ f (x) = $\underset{x\to n^{-}}{lim}$  (x [x]) = n [n 1] = n + 1 = 1

$\underset{x\to n^{+}}{lim}$ g (x) = $\underset{x\to n^{+}}{lim}$ x [x] = n [n] = 0

So,  $\underset{x\to n^{-}}{lim}$ g (x) $\cancel{=}$ $\underset{x\to n^{+}}{lim}$ g (x).

g (x) is d is continuous at all x $\in z.$