Maths

35. Given, f (x)= 99x, f   (100)=?

So, f (100)= ${}_{h\to 0}\frac{f(100+h)-f(100)}{h}$

$=\underset{h\to 0}{lim}\frac{99(100+h)-99(100)}{h}$

$=\underset{h\to 0}{lim}\frac{99*100+99*h-99*100}{h}$

$=\underset{h\to 0}{lim}\frac{99h}{h}$

$=\underset{h\to 0}{lim}99$

=99

39. As out of the total 9 seats 4 women are to be at even places we can have the following arrangement.

Seat places

Also from this arrangement the women and men can rearrange among themselves.

Therefore, the required number of ways = 4! * 5!

= (4 * 3 * 2 * 1) * (5 * 4 * 3 * 2 * 1)

= 24 * 120

= 2880

34. Given, f (x)=x, f   (1)=?

We have,

$f^{\prime }(1)=\underset{h\to 0}{lim}\frac{f(1+h)-f(1)}{h}$

$$ $=\underset{h\to 0}{lim}\frac{1+h-1}{h}$

$=\underset{h\to 0}{lim}\frac{h}{h}$

$=\underset{h\to 0}{lim}1$

=1

38. In a deck of 52 card there are four kings.

So, number of ways of selecting exactly one king is ⁴C_1.

Now, after fixing one king card, we need to have the remaining 4 out of 5 cards to be a non-king i.e., only from the other 48 cards. So, number of ways of selecting is ⁴⁸C₄

Therefore, the required number of ways

= ⁴C₁*⁴⁸C₄

33. Given, f (x)=x²- 2 ., $f^{\prime }(10)=?$

We have,  

$f^{\prime }(10)=\underset{h\to 0}{lim}\frac{f(10+h)-f(10)}{h}$

$=\underset{h\to 0}{lim}\frac{\left[{(10+h)}^2-2\right]-\left[10^2-2\right]}{h}$

= $\underset{h\to 0}{lim}\frac{10^2+h^2+20h-2-10^2+2}{h}$

= $\underset{h\to 0}{lim}\frac{h(h+20)}{h}$

$\Rightarrow \underset{h\to 0}{lim}h+20$

=20

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.