Page 58 - Exemplar Solution - Updates, Questions & Discussions on Exemplar Solution at Shiksha

Exemplar Solution Class 11th

Find the value of $r$ , if the coefficients of the (2r + 4)^th and ^th terms in the expansion of ${(1 + x)}^{18}$ are equal.

6 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ F o r c o e f f i c i e n t s o f {(2 r + 4)}^{t h} t e r m, w e h a v e \\ T_{2 r + 4} = T_{2 r + 3 + 1} = {}_{18}{C_{2 r + 3}} {(1)}^{18 - 2 r - 3} . x^{2 r + 3} \\ ∴ C o e f f i c i e n t s o f {(2 r + 4)}^{t h} t e r m = {}_{18}{C_{2 r + 3}} \\ S i m i l a r l y, T_{r - 2} = T_{r - 3 + 1} = {}_{18}{C_{r - 3}} {(1)}^{18 - r + 3} . x^{r - 3} \\ ∴ C o e f f i c i e n t s o f {(r - 2)}^{t h} t e r m = {}_{18}{C_{r - 3}} \\ A s p e r t h e c o n d i t i o n o f t h e q u e s t i o n s, w e h a v e {}_{18}{C_{2 r + 3}} = {}_{18}{C_{r - 3}} \\ \Rightarrow 2 r + 3 + r - 3 = 18 \Rightarrow 3 r = 18 \Rightarrow r = 6 \end{array}$

Upvote

Exemplar Solution Class 11th

If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, in how many points will they intersect each other?

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t o u t o f 20 l i n e s, n o t w o l i n e s a r e p a r a l l e l a n d n o t h r e e l i n e s a r e c o n c u r r e n t . \\ T h e r e f o r e, n u m b e r o f point o f intersection \\ = {}_{20}{C_{2}} [? f o r a n y point o f intersection, w e n e e d t w o l i n e s] \\ = \frac{20.19}{2.1} = 190 \\ H e n c e, t h e r e q u i r e d n u m b e r o f points = 190 . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

A man saved Rs 66000 in 20 years. In each succeeding year after the first year, he saved Rs 200 more than what he saved in the previous year. How much did he save in the first year?

3 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t x b e s a v e d i n f i r s t y e a r . \\ A n n u a l i n c r e m e n t = 200 \\ w h i c h f o r m s a n A . P . \\ f i r s t t e r m = a a n d c o m m o n d i f f e r e n c e d = 200 \\ n = 20 y e a r s \\ ∴ S_{n} = \frac{n}{2} [2 a + (n - 1) d] = S_{20} = \frac{20}{2} [2 a + (20 - 1) 200] \\ \Rightarrow 66000 = 10 [2 a + 3800] \Rightarrow 6600 = 2 a + 3800 \\ \Rightarrow 2 a = 6600 - 3800 \Rightarrow 2 a = 2800 \Rightarrow a = 1400 \\ H e n c e, t h e m a n s a v e d 1400 i n t h e f i r s t y e a r . \end{array}$

Upvote

Exemplar Solution Class 11th

Find the number of integers greater than 7000 that can be formed with the digits 3, 5, 7, 8, and 9 where no digits are repeated.

[Hint: Besides 4-digit integers greater than 7000, five-digit integers are always greater than 7000.]

4 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t a l l t h e 5 d i g i t n u m b e r s a r e g r e a t e r t h a n 7000 . \\ S o, t h e w a y s o f f o r m a t i n g 5 - d i g i t n u m b e r s = 5 \times 4 \times 3 \times 2 \times 1 = 120 \\ N o w a l l t h e f o u r d i g i t n u m b e r g r e a t e r t h a n 7000 c a n b e f o r m e d a s f o l l o w s . \\ T h o u s a n d p l a c e c a n b e f i l l e d w i t h 3 w a y s \\ H u n d r e d p l a c e c a n b e f i l l e d w i t h 4 w a y s \\ T e n t h s p l a c e c a n b e f i l l e d w i t h 3 w a y s \\ U n i t s p l a c e c a n b e f i l l e d w i t h 2 w a y s \\ S o, t h e t o t a l n u m b e r o f 4 - d i g i t n u m b e r s = 3 \times 4 \times 3 \times 2 = 72 \\ ∴ T o t a l n u m b e r o f integers = 120 + 72 = 192 \\ H e n c e, t h e r e q u i r e d n u m b e r o f integers = 192 \end{array}$

Upvote

Exemplar Solution Class 11th

Find the sixth term of the expansion ${(y^{\frac{1}{2}} + x^{\frac{1}{3}})}^{n}$ , if the binomial coefficient of the third term from the end is 45.

[Hint: Binomial coefficient of third term from the end = Binomial coefficient of third term from the beginning = ⁿC₂.]

5 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} T h e g i v e n expression i s {(y^{\frac{1}{2}} + x^{\frac{1}{3}})}^{n}, since t h e b i n o m i a l c o e f f i c i e n t o f t h i r d \\ t e r m f r o m t h e e n d = b i n o m i a l c o e f f i c i e n t o f t h i r d t e r m f r o m t h e b e g i n n i n g = {}_{n}{C_{2}} \\ ∴ {}_{n}{C_{2}} = 45 \\ \Rightarrow \frac{n (n - 1)}{2} = 45 \Rightarrow n^{2} - n = 90 \\ \Rightarrow n^{2} - n - 90 = 0 \Rightarrow n^{2} - 10 n + 9 n - 90 = 0 \\ \Rightarrow n (n - 10) + 9 (n - 10) = 0 \Rightarrow (n - 10) (n + 9) = 0 \\ \Rightarrow n = 10, n = - 9 \Rightarrow n = 10, n \neq - 9 \\ S o, t h e g i v e n expression b e c o m e s {(y^{\frac{1}{2}} + x^{\frac{1}{3}})}^{10} \\ S i x t h t e r m i n t h i s expression \\ T_{6} = T_{5 + 1} = {}_{10}{C_{5}} {(y^{\frac{1}{2}})}^{10 - 5} {(x^{\frac{1}{3}})}^{5} = {}_{10}{C_{5}} y^{\frac{5}{2}} . x^{\frac{5}{3}} \\ = 252 y^{\frac{5}{2}} . x^{\frac{5}{3}} \\ H e n c e, t h e r e q u i r e d t e r m = 252 y^{\frac{5}{2}} . x^{\frac{5}{3}} \end{array}$

Upvote

Exemplar Solution Class 11th

If ⁿC_{r – 1} = 36, ⁿC_r = 84, and ⁿC_{r + 1}= 126, then find ^rC₂.

[Hint: Form equation using C(n, r-1), C(n, r), and C(n, r+1) to find the value of r.]

15 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t {}_{n}{C_{r - 1}} = 36 \dots (i) \\ {}_{n}{C_{r}} = 84 \dots (i i) \\ {}_{n}{C_{r + 1}} = 126 \dots (i i i) \\ D i v i d i n g e q n . (i) b y e q n . (i i) w e g e t \\ \frac{{}_{n}{C_{r - 1}}}{{}_{n}{C_{r}}} = \frac{36}{84} \\ \Rightarrow \frac{\frac{n!}{(r - 1)! (n - r + 1)!}}{\frac{n!}{r! (n - r)!}} = \frac{3}{7} [? {}_{n}{C_{r}} = \frac{n!}{r! (n - r)!}] \\ \Rightarrow \frac{n!}{(r - 1)! (n - r + 1)!} \times \frac{r! (n - r)!}{n!} = \frac{3}{7} \\ \Rightarrow \frac{r . (r - 1)! (n - r)!}{(r - 1)! (n - r + 1)! + (n - r)!} = \frac{3}{7} \\ \Rightarrow \frac{r}{n - r + 1} = \frac{3}{7} \Rightarrow 3 n - 3 r + 3 = 7 r \\ \Rightarrow 3 n - 10 r = - 3 \dots (i v) \\ D i v i d i n g e q n . (i i) b y e q n . (i i i) w e g e t \\ \frac{{}_{n}{C_{r}}}{{}_{n}{C_{r + 1}}} = \frac{84}{126} \Rightarrow \frac{\frac{n!}{r! (n - r)!}}{\frac{n!}{(r + 1)! (n - r - 1)!}} = \frac{2}{3} \\ \Rightarrow &th \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

The first term of an A.P. is $a$ , and the sum of the first $p$ terms is zero, show that the

sum of its next $q$ terms is $\frac{- a (p + q) q}{p - 1}$ .

[Hint: Required sum = $S_{p + q} - S_{p}$ ]

2 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t a_{1} = a a n d S_{p} \\ S u m o f n e x t q t e r m s o f t h e g i v e n A . P . = S_{p + q} - S_{p} \\ ∴ S_{p + q} = \frac{p + q}{2} [2 a + (p + q - 1) d] \\ a n d S_{p} = \frac{p}{2} [2 a + (p - 1) d] = 0 \\ \Rightarrow 2 a + (p - 1) d = 0 \Rightarrow (p - 1) d = - 2 a \\ \Rightarrow d = \frac{- 2 a}{p - 1} \\ S u m o f n e x t q t e r m s = S_{p + q} - S_{p} \\ = \frac{p + q}{2} [2 a + (p + q - 1) d] - 0 \\ = \frac{p + q}{2} [2 a + (p + q - 1) (\frac{- 2 a}{p - 1})] \\ = \frac{p + q}{2} [2 a + \frac{(p - 1) (- 2 a)}{p - 1} - \frac{2 a q}{p - 1}] \\ = \frac{p + q}{2} [2 a - 2 a - \frac{2 a q}{p - 1}] = \frac{p + q}{2} (- \frac{2 a q}{p - 1}) \\ = \frac{- a (p + q) q}{p - 1} \\ H e n c e, t h e r e q u i r e d s u m = \frac{- a (p + q) q}{p - 1} \end{array}$

Upvote

Exemplar Solution Class 11th

Find the coefficient of $\frac{1}{x^{17}}$ in the expansion of x⁴– 1/x^{3 15}

6 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} T h e g i v e n expression i s {(x^{4} - \frac{1}{x^{3}})}^{15} \\ G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ = {}_{15}{C_{r}} {(x^{4})}^{15 - r} {(- \frac{1}{x^{3}})}^{r} = {}_{15}{C_{r}} {(x)}^{60 - 4 r} {(- 1)}^{r} . \frac{1}{{x^{3}}^{r}} \\ = {}_{15}{C_{r}} {(- 1)}^{r} . \frac{1}{{x^{3}}^{r - 60 + 4 r}} = {}_{15}{C_{r}} {(- 1)}^{r} . \frac{1}{{x^{7}}^{r - 60}} \\ T o f i n d t h e c o e f f i c i e n t o f \frac{1}{x^{17}}, P u t 7 r - 60 = 17 \Rightarrow 7 r = 60 + 11 \Rightarrow r = \frac{77}{7} = 11 \\ P u t t i n g t h e v a l u e o f r i n t h e a b o v e expression, w e g e t \\ = {}_{15}{C_{11}} {(- 1)}^{11} . \frac{1}{x^{17}} = - {}_{15}{C_{4}} . \frac{1}{x^{17}} = - \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} . \frac{1}{x^{17}} \\ = - 1365 . \frac{1}{x^{17}} \\ H e n c e, t h e r e q u i r e d c o e f f i c i e n t o f \frac{1}{x^{17}} = - 1365 \end{array}$

Upvote

Exemplar Solution Class 11th

A box contains two white, three black and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?

[Hint: Required number of ways = ³C₁ × ⁶C₂ + ³C₂ × ⁶C₂+ ³C₃.]

3 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e 2 w h i t e, 3 b l a c k a n d 4 r e d b a l l s i n a b o x . 3 b a l l s a r e t o b e d r a w n o u t o f 9 b a l l s \\ a t l e a s t o n e b l a c k b a l l i s t o b e i n c l u d e d . \\ S o, t h e p o s s i b l e s e l e c t i o n i s \\ (1 b l a c k a n d 2 o t h e r b a l l s) o r (2 b l a c k a n d 1 o t h e r b a l l) o r (3 b l a c k a n d n o o t h e r b a l l) \\ S o, t h e n u m b e r o f p o s s i b l e s e l e c t i o n i s \\ = {}_{3}{C_{1}} \times {}_{6}{C_{2}} + {}_{3}{C_{2}} \times {}_{6}{C_{1}} + {}_{3}{C_{3}} \times {}_{6}{C_{0}} \\ = 3 \times 15 + 3 \times 6 + 1 \times 1 = 45 + 18 + 1 = 64 \\ H e n c e, t h e r e q u i r e d s e l e c t i o n = 64 . \end{array}$

Upvote

Exemplar Solution Class 11th

Find the coefficient of $x^{15}$ in the expansion of ${(x - x^{2})}^{10} .$

5 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} T h e g i v e n expression i s {(x - x^{2})}^{10} \\ G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ = {}_{10}{C_{r}} {(x)}^{10 - r} {(- x^{2})}^{r} = {}_{10}{C_{r}} {(x)}^{10 - r} {(- 1)}^{r} {(x^{2})}^{r} \\ = {(- 1)}^{r} . {}_{10}{C_{r}} {(x)}^{10 - r + 2 r} = {(- 1)}^{r} . {}_{10}{C_{r}} {(x)}^{10 + r} \\ T o f i n d t h e c o e f f i c i e n t o f x^{15}, P u t 10 + r = 15 \Rightarrow r = 5 \\ ∴ C o e f f i c i e n t o f x^{15} = {(- 1)}^{5} . {}_{10}{C_{5}} = - {}_{10}{C_{5}} = - 252 \\ H e n c e, t h e r e q u i r e d c o e f f i c i e n t = - 252 \end{array}$

Upvote

Exemplar Solution

Questions

Discussions

Active Users

Followers

Find the value of $r$ , if the coefficients of the (2r + 4)^th and ^th terms in the expansion of ${(1 + x)}^{18}$ are equal.

If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, in how many points will they intersect each other?

A man saved Rs 66000 in 20 years. In each succeeding year after the first year, he saved Rs 200 more than what he saved in the previous year. How much did he save in the first year?

Find the number of integers greater than 7000 that can be formed with the digits 3, 5, 7, 8, and 9 where no digits are repeated.

[Hint: Besides 4-digit integers greater than 7000, five-digit integers are always greater than 7000.]

Find the sixth term of the expansion ${(y^{\frac{1}{2}} + x^{\frac{1}{3}})}^{n}$ , if the binomial coefficient of the third term from the end is 45.

[Hint: Binomial coefficient of third term from the end = Binomial coefficient of third term from the beginning = ⁿC₂.]

If ⁿC_{r – 1} = 36, ⁿC_r = 84, and ⁿC_{r + 1}= 126, then find ^rC₂.

[Hint: Form equation using C(n, r-1), C(n, r), and C(n, r+1) to find the value of r.]

The first term of an A.P. is $a$ , and the sum of the first $p$ terms is zero, show that the

sum of its next $q$ terms is $\frac{- a (p + q) q}{p - 1}$ .

[Hint: Required sum = $S_{p + q} - S_{p}$ ]

Find the coefficient of $\frac{1}{x^{17}}$ in the expansion of x⁴– 1/x^{3 15}

A box contains two white, three black and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?

[Hint: Required number of ways = ³C₁ × ⁶C₂ + ³C₂ × ⁶C₂+ ³C₃.]

Find the coefficient of $x^{15}$ in the expansion of ${(x - x^{2})}^{10} .$