Ncert Solutions Maths class 11th

20. $\underset{x\to 0}{lim}\frac{sinax+bx}{ax+sinbx}=\frac{\underset{x\to 0}{lim}\frac{sinax}{ax}*ax+bx}{\underset{x\to 0}{lim}ax+\frac{sinbx}{bx}*bx}.$

$\frac{\underset{x\to 0}{lim}\frac{sinax}{ax}*\underset{x\to 0}{lim}ax+\underset{x\to 0}{lim}bx}{\underset{x\to 0}{lim}ax+\underset{x\to 0}{lim}\frac{sinbx}{bx}*\underset{x\to 0}{lim}bx}$

$=\frac{\underset{x\to 0}{lim}ax+\underset{x\to 0}{lim}bx}{\underset{x\to 0}{lim}ax+\underset{x\to 0}{lim}bx}$

$=\underset{x\to 0}{lim}\frac{ax+bx}{ax+bx}$

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

= 1.

19. $\underset{x\to 0}{lim}xsecx=\underset{x\to 0}{lim}\frac{x}{cosx}$

$=\frac{0}{cos0}$

$=\frac{0}{1}$

= 0.

35. Since the 6-digit numbers to be formed from the digits 0, 1, 3, 5, 7 and 9 has to be divisible by 10 we have to fix the unit place as 0. Now, the remaining 5 places can be filled only by the digits 1, 3, 5, 7 and 9.

Therefore, the required number of ways

= 5!

= 5 * 4 * 3 * 2 * 1

= 120

16. $\underset{x\to 0}{lim}\frac{cosx}{\pi -x}=\frac{cos0}{\pi -0}=\frac{1}{\pi }$

15. $\underset{x\to \pi }{lim}\frac{sin(\pi -x)}{\pi (\pi -x)}=\frac{1}{\pi }*\underset{x\to \pi }{lim}\frac{sin(\pi -x)}{\pi -x}$

$=\frac{1}{\pi }.$

14. $\underset{x\to 0}{lim}\frac{sinax}{sinbx}=\underset{x\to 0}{lim}\frac{\frac{sinax}{ax}*ax}{\frac{sinbx}{bx}*bx}$

$=\frac{a}{b}$

34. In the eleven-letter word EXAMINATION there are 2A's, 2I's and 2N's and the rest are all different.

Since in a dictionary the words are listed according to the English alphabet we can only find words starting with A (as B, C, D are not a part of the letters forming the word EXAMINATION) listed before E.

Hence after fixing one A as first word we can rearrange the remaining 10 letters of which 2 are I, 2 are N and rest are all different.

Therefore, the required number of ways = $\frac{10!}{2! 2!}$

= 10 * 9 * 8 * 7 * 6 * 5 * 2 * 3

= 9,07,200

13. $\underset{x\to 0}{lim}\frac{sinax}{bx}=\frac{1}{b}\underset{x\to 0}{lim}\frac{sinax}{ax}*a$

$=\frac{a}{b}\underset{x\to 0}{lim}\frac{sinax}{ax}$ 

$=\frac{a}{b}$

Kindly go through the solution

$=\frac{1}{2*(-2)}=-\frac{1}{4}.$

11. $\underset{x\to 1}{lim}\frac{a{x}^2+bx+c}{c{x}^2+bx+a}=\frac{a+b+c}{c+b+a}=1$