Maths

37. Since out of 8 total questions at least 3 questions has to be attempted from each of part I and II containing 5 and 7 questions respectively we can have the choices.

(a) 3 questions from I and 5 questions from II selected in ⁵C₃*⁷C₅ ways.

(b) 4 questions from I and 4 questions from II selected in ⁵C₄*⁷C₄ ways.

(c) 5 questions from I and 3 questions from II selected in ⁵C₅*⁷C₃ ways.

Therefore, the required number of ways.

= (⁵C₃*⁷C₅) + (⁵C₄*⁷C₄) + (⁵C₅*⁷C₃)

= $\frac{5!}{3!\left(5-3\right)!}$ * $\frac{7!}{5!\left(7-5\right)!}$ + $\frac{5!}{4!\left(5-4\right)!}$ * $\frac{7!}{4!\left(7-4\right)!}$ + $\frac{5!}{5!\left(5-5\right)!}$ * $\frac{7!}{3!\left(7-3\right)!}$

= (10 * 21) + (5 * 35) + 35

= 210 + 175 + 35

= 420

32. Given, f (x) = $\begin{array}{l}\{\begin{array}{ll}m{x}^2+n,\hfill & x<0\hfill \\ nx+m,\hfill & 0\le x\le 1\hfill \end{array}\\ n{x}^3+m,x>1.\end{array}$

For $\underset{x\to 0}{lim}f(x)\underset{x\to 0^{-}}{lim,}f(x)=\underset{x\to 0^{+}}{lim}f(x)$

$\Rightarrow \underset{x\to 0^{-}}{lim}\left(m{x}^2+n\right)=\underset{x\to 0^{+}}{lim}\left(m{x}^3+m\right)$

n = m

So, $\underset{x\to 0}{lim}f(x)$ exist for n = m.

Again, $\underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{-}}{lim}nx+m=n+m.$

$\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1^{+}}{lim}n{x}^3+m=n+m$

So, $\underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1^{-}}{lim}f(x)=n+m,$ Thus, $\underset{x\to 1}{lim}f(x)$ exist for any integral value of m and n.

36. In an English word there are 5 vowels and 21 consonants.

The number of ways of selecting 2 vowel out of 5 = ⁵C₂

= $\frac{5!}{2!\left(5?2\right)!}$

= 5 * 2 = 10

The number of ways of selecting 2 consonants out of 21 = ²¹C₂

= $\frac{21!}{2!\left(21?2\right)!}$

= 21 * 10

= 210

Therefore, the number of combinations of 2 vowels and 2 consonants is 10 * 210 = 2100

Each of these 2100 combinations has 4 letters which can be rearranged among themselves in 4! Ways.

Therefore, the required number of ways

= 4! * 2100

= 4 * 3 * 2 * 1 * 2100

= 50400

31. Given, $\underset{x\to 1}{lim}\frac{f(x)-2}{x^2-1}=\pi$

$\frac{\underset{x\to 1}{lim}f(x)-2}{\underset{x\to 1}{lim}{x}^2-1}=\pi$

$\Rightarrow \underset{x\to 1^{}}{lim}[f(x)-2]=\pi \underset{x\to 1}{lim}\left(x^2-1\right)$

$\Rightarrow \underset{x\to 1}{lim}f(x)-\underset{x\to 1}{lim}2=\pi \left(1^2-1\right)$

$\Rightarrow \underset{x\to 1}{lim}f(x)-2=0.$

$\Rightarrow \underset{x\to 1}{lim}f(x)=2$

30. Given, $f(x)=\{\begin{array}{ccc}\hfill \left|x\right|+1,\hfill & \hfill x<0\hfill & \hfill 0,\hfill \\ \hfill x=0\hfill & \hfill \left|x\right|-1,\hfill & \hfill x>0\hfill \end{array}\underset{x\to a}{lim}f(x)=?$

As | x | = $\{\begin{array}{l}-x,x<0\\ x,x>0\end{array}$

We can rewrite f (x) = $\begin{array}{l}\{\begin{array}{l}-x+1,x<0\\ 0,x=0\end{array}\\ x-1,x>0.\end{array}$

Case 1: when a<0,

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

So, $\underset{x\to a}{lim}f(x)$ = exist such that a< 0

Case II when a> 0,

$\underset{x\to a}{lim}f(x)=\underset{x\to a}{lim}(x-1)=a-1$

So, $\underset{x\to a}{lim}f(x)$ exist such that a>0.

Case III when a = 0.

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

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

Thus, $\underset{x\to a^{-}}{lim}f(x)\ne \underset{x\to a^{+}}{lim}f(x)$

So, $\underset{x\to a}{lim}f(x)$ does not exist at a = 0.

28. Given, f (x) = $\begin{array}{l}\{\begin{array}{l}a+bx,x<1\\ 4,x=1\end{array}\\ b-ax,x>1\end{array}$

Since we need $\underset{x\to 1}{lim}f(x)$ we need,

LHL =

$\underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{-}}{lim}(a+bx)$ = a + b * 1 = a + b

and RHL =

$\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1^{+}}{lim}(b-ax)$ = b - a * 1 = b - a

Given,  $\underset{x\to 1}{lim}f(x)=f(1):$ we have the following equations

a + b = 4 ____ (1)

b - a = 4 ____ (2)

Adding (1) and (2) we get,

2b = 8

b = 4

Putting b = 4 in (1) we get

a + 4 = 4

a = 0

27. $\underset{x\to 5}{lim}f(x)=\underset{x\to 5}{lim}\left|x\right|-5$

= | 5 |$-\mathrm{}$ 5

= 5 $-\mathrm{}$5

= 0

26. Given, f (x) $\{\begin{array}{ll}\frac{x}{\left|x\right|},\hfill & x\ne 0\hfill \\ 0,\hfill & x=0\hfill \end{array}$

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

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

Thus,  $\underset{x\to 0^{-}}{lim}f(x)\ne \underset{x\to 0^{+}}{lim}f(x)$

i e,  $\underset{x\to 0}{lim}f(x)$ does not exist.

25. Given f (x) = $\{\begin{array}{cc}\hfill \frac{\left|x\right|}{x},x\ne 0\hfill & \hfill 0,x=0\hfill \end{array},\underset{x\to 0}{lim}f(x)=?$

We know that, $\left|x\right|=\{\begin{array}{cc}\hfill x,\hfill & \hfill x\ge 0\hfill \\ \hfill -x,\hfill & \hfill x<0\hfill \end{array}$

Now,

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

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

Thus, $\underset{x\to 0^{-}}{lim}f(x)\ne \underset{x\to 0^{+}}{lim}f(x)$

i e, $\underset{x\to 0}{lim}f(x)$ does not exist.

24. Given $f(x)=\{\begin{array}{cc}\hfill x^2-1,x\le 1\hfill & \hfill -x^2-1,x>1\hfill \end{array},\underset{x\to 1}{lim}f(x)=?$

Now, L.H.L = $\underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{-}}{lim}\left(x^2-1\right)$

1²- 1 = 0

And R.H.L = $\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1^{+}}{lim}\left(-x^2-1\right)$  (1)2 1 = 1 = 2.

Thus,  $\underset{x\to 1^{-}}{lim}f(x)\ne \underset{x\to 1^{+}}{lim}f(x)$

So,  $\underset{x\to 1}{lim}f(x)$ does not exist.