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Exemplar Solution Class 11th

If ⁿC₁₂ = ⁿC₈, then n is equal to:

(a) 20

(b) 12

(c) 6

(d) 30

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : {}_{n}{C_{12}} = {}_{n}{C_{8}} \\ \Rightarrow {}_{n}{C_{n - 12}} = {}_{n}{C_{8}} [? {}_{n}{C_{r}} = {}_{n}{C_{n - r}}] \\ \Rightarrow n - 12 = 8 \\ \Rightarrow n = 8 + 12 = 20 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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Exemplar Solution Class 11th

If the coefficients of the 2nd, 3rd, and 4th terms in the expansion of ${(1 + x)}^{n}$ are in A.P., then the value of $n$ is:

(a) 2

(b) 7

(c) 11

(d) 14

[Hint: 2 ⁿC₂ = ⁿC₁ + ⁿC₃ $\Rightarrow n^{2} - 9 n + 14 = 0 \Rightarrow n = 27 .$ ]

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n expression i s {(1 + x)}^{n} \\ {(1 + x)}^{n} = {}_{n}{C_{0}} + {}_{n}{C_{1}} x + {}_{n}{C_{2}} x^{2} + {}_{n}{C_{3}} x^{3} + \dots + {}_{n}{C_{n}} x^{n} \\ H e r e, c o e f f i c i e n t o f 2 n d t e r m = {}_{n}{C_{1}} \\ C o e f f i c i e n t o f 3 r d t e r m = {}_{n}{C_{2}} \\ a n d c o e f f i c i e n t o f 4 t h t e r m = {}_{n}{C_{3}} \\ G i v e n t h a t {}_{n}{C_{1}}, {}_{n}{C_{2}} a n d {}_{n}{C_{3}} a r e i n A . P \\ ∴ 2 . {}_{n}{C_{2}} = {}_{n}{C_{1}} + {}_{n}{C_{3}} \\ \Rightarrow 2 . \frac{n (n - 1)}{2} = n + \frac{n (n - 1) (n - 2)}{3.2.1} \\ \Rightarrow n (n - 1) = n + \frac{n (n - 1) (n - 2)}{6} \\ \Rightarrow n - 1 = 1 + \frac{(n - 1) (n - 2)}{6} \\ \Rightarrow 6 n - 6 = 6 + n^{2} - 3 n + 2 \\ \Rightarrow n^{2} - 3 n - 6 n + 14 = 0 \\ \Rightarrow n^{2} - 9 n + 14 = 0 \Rightarrow n^{2} - 7 n - 2 n + 14 = 0 \\ \Rightarrow n (n - 7) - 2 (n - 7) = 0 \Rightarrow (n - 2) (n - 7) = 0 \\ \Rightarrow n = 2, 7 \Rightarrow n = 7 w h e r e a s n = 2 i s n o t p o s s i b l e \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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Exemplar Solution Class 11th

The coefficient of $x^{n}$ in the expansion of ${(1 + x)}^{2 n}$ and ${(1 + x)}^{2 n - 1}$ are in the ratio:

(a) 1:2

(b) 1:3

(c) 3:1

(d) 2:1

11 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ I n t h e expansion o f {(1 + x)}^{2 n}, w e g e t T_{r + 1} = {}_{2 n}{C_{r}} . x^{r} \\ T o g e t t h e c o e f f i c i e n t s o f x^{n}, p u t r = n \\ ∴ C o e f f i c i e n t s o f x^{n} = {}_{2 n}{C_{n}} \\ I n t h e expansion o f {(1 + x)}^{2 n - 1}, w e g e t T_{r + 1} = {}_{2 n - 1}{C_{r}} . x^{r} \\ ∴ C o e f f i c i e n t s o f x^{n} = {}_{2 n - 1}{C_{n - 1}} \\ T h e r e q u i r e d r a t i o i s \frac{{}_{2 n}{C_{n}}}{{}_{2 n - 1}{C_{n - 1}}} \\ = \frac{\frac{2 n!}{n! (n)!}}{\frac{(2 n - 1)!}{(n - 1)! (2 n - 1 - n + 1)!}} = \frac{\frac{2 n!}{n! n!}}{\frac{(2 n - 1)!}{(n - 1)! (n)!}} \\ = \frac{2 n!}{n! n!} \times \frac{(n - 1)! n!}{(2 n - 1)!} = \frac{2 n (2 n - 1)!}{n! n (n - 1)!} \times \frac{(n - 1)! n!}{(2 n - 1)!} \\ = \frac{2}{1} = 2 : 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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Exemplar Solution Class 11th

The two successive terms in the expansion of ${(1 + x)}^{24}$ whose coefficients are in the ratio 1:4 are:

(a) 3^rd and 4^th

(b) 4^th and 5^th

(c) 5^th and 6^th

(d) 6^th and 7^th

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t r^{t h} a n d {(r + 1)}^{t h} b e t w o s u c c e s s i v e t e r m s i n t h e expansion {(1 + x)}^{24} \\ ∴ T_{r + 1} = {}_{24}{C_{r}} . x^{r} \\ a n d T_{r + 2} = T_{r + 1 + 1} = {}_{24}{C_{r + 1}} . x^{r + 1} \\ A s p e r q u e s t i o n, w e h a v e \frac{{}_{24}{C_{r}}}{{}_{24}{C_{r + 1}}} = \frac{1}{4} \\ \Rightarrow \frac{\frac{24!}{r! (24 - r)!}}{\frac{24!}{(r + 1)! (24 - r - 1)!}} = \frac{1}{4} \\ \Rightarrow \frac{24!}{r! (24 - r)!} \times \frac{(r + 1)! (24 - r - 1)!}{24!} = \frac{1}{4} \\ \Rightarrow \frac{(r + 1) . r! (24 - r - 1)!}{r! (24 - r) (24 - r - 1)!} = \frac{1}{4} \\ \Rightarrow \frac{r + 1}{24 - r} = \frac{1}{4} \Rightarrow 4 r + 4 = 24 - r \\ \Rightarrow 5 r = 20 \Rightarrow r = 4 \\ ∴ T_{4 + 1} = T_{5} a n d T_{4 + 2} = T_{6} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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Exemplar Solution Class 11th

Given the integers $r > 1, n > 2$ , and the coefficients of the $(3 r)$ th and $(r + 2)$ nd terms in the binomial expansion of ${(1 + x)}^{2 n}$ are equal, then:

(a) $n = 2 r$

(b) $n = 3 r$

(c) $n = 2 r + 1$

(d) None of these

4 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t r > 1 a n d n > 2 \\ t h e n T_{3 r} = T_{3 r - 1 + 1} = {}_{2 n}{C_{3 r - 1}} . x^{3 r - 1} \\ a n d T_{r + 2} = T_{r + 1 + 1} = {}_{2 n}{C_{r + 1}} . x^{r + 1} \\ A s p e r q u e s t i o n, w e h a v e \\ {}_{2 n}{C_{3 r - 1}} = {}_{2 n}{C_{r + 1}} \\ \Rightarrow 3 r - 1 + r + 1 = 2 n [? {}_{n}{C_{p}} = {}_{n}{C_{q}} \Rightarrow n = p + q] \\ \Rightarrow 4 r = 2 n \\ \Rightarrow n = 2 r \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

If $a_{1}, a_{2}, a_{3}, . . ., a_{n}$ are in A.P., where $a_{i} > 0$ for all $i$ , show that $\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + . . . + \frac{1}{a_{n - 1} a_{n}} = \frac{1}{a_{1}} - \frac{1}{a_{n}} .$

3 Views|Posted a year ago

Asked by Alok Kumar Singh

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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_{2}, a_{3}, a_{4}, \dots, a_{n} a r e i n A . P . \\ ∴ C o m m o n d i f f e r e n c e d = a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = \dots = a_{n} - a_{n - 1} \\ I f a_{2} - a_{1} = d t h e n \sqrt[]{a_{2}^{2}} - \sqrt[]{a_{1}^{2}} = d \\ \Rightarrow (\sqrt[]{a_{2}} - \sqrt[]{a_{1}}) (\sqrt[]{a_{2}} + \sqrt[]{a_{1}}) = d [? a^{2} - b^{2} = (a + b) (a - b)] \\ \Rightarrow \frac{1}{\sqrt[]{a_{1}} + \sqrt[]{a_{2}}} = \frac{\sqrt[]{a_{2}} - \sqrt[]{a_{1}}}{d} \\ S i m i l a r l y, \frac{1}{\sqrt[]{a_{2}} + \sqrt[]{a_{3}}} = \frac{\sqrt[]{a_{3}} - \sqrt[]{a_{2}}}{d} \\ \frac{1}{\sqrt[]{a_{3}} + \sqrt[]{a_{4}}} = \frac{\sqrt[]{a_{4}} - \sqrt[]{a_{3}}}{d} \\ \dots \dots \dots \\ \frac{1}{\sqrt[]{a_{n - 1}} + \sqrt[]{a_{n}}} = \frac{\sqrt[]{a_{n}} - \sqrt[]{a_{n - 1}}}{d} \\ A d d i n g t h e a b o v e t e r m s, w e g e t \\ \frac{1}{\sqrt[]{a_{1}} + \sqrt[]{a_{2}}} + \frac{1}{\sqrt[]{a_{2}} + \sqrt[]{a_{3}}} + \frac{1}{\sqrt[]{a_{3}} + \sqrt[]{a_{4}}} + \dots + \frac{1}{\sqrt[]{a_{n - 1}} + \sqrt[]{a_{n}}} = \frac{1}{d} [\begin{array}{l} \sqrt[]{a_{2}} - \sqrt[]{a_{1}} + \sqrt[]{a_{3}} - \sqrt[]{a_{2}} + \\ \sqrt[]{a_{4}} - \sqrt[]{a_{3}} + \dots + \sqrt[]{a_{n}} - \sqrt[]{a_{n - 1}} \end{array}] \\ \frac{1}{\sqrt[]{a_{1}} + \sqrt[]{a_{2}}} + \frac{1}{\sqrt[]{a_{2}} + \sqrt[]{a_{3}}} + \frac{1}{\sqrt[]{a_{3}} + \sqrt[]{a_{4}}} + \dots + \frac{1}{\sqrt[]{a_{n - 1}} + \sqrt[]{a_{n}}} = \frac{1}{d} [\sqrt[]{a_{n}} - \sqrt[]{a_{1}}] \dots (i) \\ N o w a_{n} = a_{1} + (n - 1) d \\ \Rightarrow a_{n} - a_{1} = (n - 1) d \\ \Rightarrow \sqrt[]{a_{n}^{2}} - \sqrt[]{a_{1}^{2}} = (n - 1) d \\ \Rightarrow (\sqrt[]{a_{n}} + \sqrt[]{a_{1}}) (\sqrt[]{a_{n}} - \sqrt[]{a_{1}}) = (n - 1) d \\ \Rightarrow \sqrt[]{a_{n}} - \sqrt[]{a_{1}} = \frac{(n - 1) d}{\sqrt[]{a_{n}} + \sqrt[]{a_{1}}} \\ \Rightarrow \frac{\sqrt[]{a_{n}} - \sqrt[]{a_{1}}}{d} = \frac{(n - 1)}{\sqrt[]{a_{n}} + \sqrt[]{a_{1}}} \dots (i i) \\ F r o m e q n . (i) a n d e q n . (i i) w e g e t \\ \frac{1}{\sqrt[]{a_{1}} + \sqrt[]{a_{2}}} + \frac{1}{\sqrt[]{a_{2}} + \sqrt[]{a_{3}}} + \frac{1}{\sqrt[]{a_{3}} + \sqrt[]{a_{4}}} + \dots + \frac{1}{\sqrt[]{a_{n - 1}} + \sqrt[]{a_{n}}} = \frac{(n - 1)}{\sqrt[]{a_{n}} + \sqrt[]{a_{1}}} \\ H e n c e p r o v e d . \end{array}$

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Exemplar Solution Class 11th

The total number of terms in the expansion of ${(x + a)}^{100} + {(x - a)}^{100}$ after simplification is:

(a) 50

(b) 202

(c) 51

(d) None of these

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} N u m b e r o f t e r m s i n t h e expansion o f {(x + a)}^{100} = 101 \\ N u m b e r o f t e r m s i n t h e expansion o f {(x - a)}^{100} = 101 \\ N o w 50 t e r m s o f expansion w i l l c a n c e l o u t w i t h n e g a t i v e 50 t e r m s o f {(x - a)}^{100} \\ S o, t h e r e m a i n i n g 51 t e r m s o f f i r s t expansion w i l l b e a d d e d t o 51 t e r m s o f o t h e r . \\ T h e r e f o r e, t h e n u m b e r o f t e r m s = 51 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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Exemplar Solution Class 11th

A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has:

(i) No girls.

(ii) At least one boy and one girl.

(iii) At least three girls.

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} W e h a v e 4 g i r l s a n d 7 b o y s a n d a t e a m o f 5 m e m b e r s i s t o b e s e l e c t e d . \\ (i) I f n o g i r l i s s e l e c t e d, t h e n a l l t h e 5 m e m b e r s a r e t o b e s e l e c t e d \\ o u t o f 7 b o y s i . e . {}_{7}{C_{5}} = \frac{7!}{5! 2!} = \frac{7 \times 6.5!}{5! \times 2} = 21 w a y s \\ (i i) W h e n a t l e a s t o n e b o y a n d o n e g i r l a r e t o b e s e l e c t e d, t h e n \\ N u m b e r o f w a y s = {}_{4}{C_{1}} \times {}_{7}{C_{4}} + {}_{4}{C_{2}} \times {}_{7}{C_{3}} + {}_{4}{C_{3}} \times {}_{7}{C_{2}} + {}_{4}{C_{4}} \times {}_{7}{C_{1}} \\ = 4 \times \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} + \frac{4 \times 3}{2 \times 1} \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1} + 4 \times \frac{7 \times 6}{2 \times 1} + 1 \times 7 \\ = 4 \times 35 + 6 \times 35 + 4 \times 21 + 7 = 140 + 210 + 84 + 7 = 441 w a y s \\ (i i i) W h e n a t l e a s t 3 g i r l s a r e i n c l u d e d, t h e n \\ N u m b e r o f w a y s = {}_{4}{C_{3}} \times {}_{7}{C_{2}} + {}_{4}{C_{4}} \times {}_{7}{C_{1}} \\ = 4 \times \frac{7 \times 6}{2 \times 1} + 1 \times 7 = 84 + 7 = 91 w a y s \\ H e n c e, t h e r e q u i r e d n u m b e r o f w a y s a r e \\ (i) 21 w a y s (i i) 441 w a y s (i i i) 91 w a y s \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

In a cricket tournament, 16 school teams participated. A sum of Rs 8000 is to be awarded among themselves as prize money. If the last placed team is awarded Rs 275 in prize money and the award increases by the same amount for successive finishing places, how much amount will the first place team receive?

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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} L e t t h e p r i z e a m o u n t g o t b y f i r s t p l a c e t e a m b e a \\ Since, t h e p r i z e m o n e y i n c r e a s e s b y t h e s a m e a m o u n t f o r s u c c e s s i v e f i n i s h i n g p l a c e s, \\ T h e r e f o r e, t h e s e r i e s w i l l b e A . P . \\ ∴ a_{n} = 275, n = 16 a n d S_{16} = 8000 \\ a_{n} = a + (n - 1) d \\ 275 = a + (16 - 1) (- d) [? C o m m o n d i f f e r e n c e d i s (-) a s t h e s e r i e s i s decreasing] \\ \Rightarrow 275 = a - 15 d \dots (i) \\ N o w S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ \Rightarrow S_{16} = \frac{16}{2} [2 a + 15 (- d)] \\ \Rightarrow 8000 = 8 [2 a - 15 d] = 2 a - 15 d = 1000 \dots (i i) \\ S o l v i n g e q . (i) a n d e q . (i i) w e g e t \\ a = 725 a n d d = 30 \\ H e n c e, t h e r e q u i r e d a w a r d r e c e i v e d b y f i r s t p l a c e t e r m = 725 . \end{array}$

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Exemplar Solution Class 11th

Find the term independent of $x$ in the expansion of ${(1 + x + 2 x^{3})}^{}$ 3/2 x² – 1/3x ⁹.

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} T h e g i v e n expression i s (1 + x + 2 x^{3}) {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9} \\ L e t u s c o n s i d e r {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9} \\ G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ T_{r + 1} = {}_{9}{C_{r}} {(\frac{3}{2} x^{2})}^{9 - r} {(- \frac{1}{3 x})}^{r} = {}_{9}{C_{r}} {(\frac{3}{2})}^{9 - r} {(x)}^{18 - 2 r} {(- \frac{1}{3})}^{r} . \frac{1}{x^{r}} \\ = {}_{9}{C_{r}} {(\frac{3}{2})}^{9 - r} {(x)}^{18 - 2 r - r} {(- \frac{1}{3})}^{r} = {}_{9}{C_{r}} {(\frac{3}{2})}^{9 - r} {(- \frac{1}{3})}^{r} . {(x)}^{18 - 3 r} \\ S o, t h e g e n e r a l t e r m i n t h e expansion o f \\ (1 + x + 2 x^{3}) {(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{9} \\ = {}_{9}{C_{r}} {(\frac{3}{2})}^{9 - r} {(- \frac{1}{3})}^{r} . {(x)}^{18 - 3 r} + {}_{9}{C_{r}} {(\frac{3}{2})}^{9 - r} {(- \frac{1}{3})}^{r} . {(x)}^{19 - 3 r} + 2 . {}_{9}{C_{r}} {(\frac{3}{2})}^{9 - r} {(- \frac{1}{3})}^{r} . {(x)}^{21 - 3 r} \\ F o r g e t t i n g t h e t e r m i n d e p e n d e n t o f x, \\ P u t 18 - 3 r = 0, 19 - 3 r = 0 a n d 21 - 3 r = 0 w e g e t \\ r = 6, r = \frac{19}{3} a n d r = 7 \\ T h e p o s s i b l e v a l u e o f r a r e 6 a n d 7 (? r \neq \frac{19}{3}) \\ ∴ T h e t e r m i n d e p e n d e n t o f x i s \\ = {}_{9}{C_{6}} {(\frac{3}{2})}^{9 - 6} {(- \frac{1}{3})}^{6} + . {}_{9}{C_{7}} {(\frac{3}{2})}^{9 - 7} {(- \frac{1}{3})}^{7} \\ = \frac{9 \times 8 \times 7 \times 6!}{3 \times 2 \times 1 \times 6!} . \frac{3^{3}}{2^{3}} . \frac{1}{3^{6}} - 2 . \frac{9 \times 8 \times 7!}{2 \times 1 \times 7!} . \frac{3^{2}}{2^{2}} . \frac{1}{3^{7}} \\ = \frac{84}{8} . \frac{1}{3^{3}} - \frac{36}{4} . \frac{2}{3^{5}} = \frac{7}{18} - \frac{2}{27} = \frac{21 - 4}{54} = \frac{17}{54} \\ H e n c e, t h e r e q u i r e d t e r m i s \frac{17}{54} \end{array}$

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If ⁿC₁₂ = ⁿC₈, then n is equal to:

(a) 20

(b) 12

(c) 6

(d) 30

If the coefficients of the 2nd, 3rd, and 4th terms in the expansion of ${(1 + x)}^{n}$ are in A.P., then the value of $n$ is:

(a) 2

(b) 7

(c) 11

(d) 14

[Hint: 2 ⁿC₂ = ⁿC₁ + ⁿC₃ $\Rightarrow n^{2} - 9 n + 14 = 0 \Rightarrow n = 27 .$ ]

The coefficient of $x^{n}$ in the expansion of ${(1 + x)}^{2 n}$ and ${(1 + x)}^{2 n - 1}$ are in the ratio:

(a) 1:2

(b) 1:3

(c) 3:1

(d) 2:1

The two successive terms in the expansion of ${(1 + x)}^{24}$ whose coefficients are in the ratio 1:4 are:

(a) 3^rd and 4^th

(b) 4^th and 5^th

(c) 5^th and 6^th

(d) 6^th and 7^th

Given the integers $r > 1, n > 2$ , and the coefficients of the $(3 r)$ th and $(r + 2)$ nd terms in the binomial expansion of ${(1 + x)}^{2 n}$ are equal, then:

(a) $n = 2 r$

(b) $n = 3 r$

(c) $n = 2 r + 1$

(d) None of these

The total number of terms in the expansion of ${(x + a)}^{100} + {(x - a)}^{100}$ after simplification is:

(a) 50

(b) 202

(c) 51

(d) None of these

A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has:

(i) No girls.

(ii) At least one boy and one girl.

(iii) At least three girls.

Find the term independent of $x$ in the expansion of ${(1 + x + 2 x^{3})}^{}$ 3/2 x² – 1/3x ⁹.