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New answer posted

a year ago

Class 11th Exemplar Solution

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

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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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New answer posted

a year ago

Class 11th 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 $\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + . . . + \frac{1}{a_{n - 1} a_{n}} = \frac{1}{a_{1}} - \frac{1}{a_{n}} .$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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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New answer posted

a year ago

Class 11th Exemplar Solution

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

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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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New answer posted

a year ago

Class 11th Exemplar Solution

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.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

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

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

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New answer posted

a year ago

Class 11th 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?

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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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New answer posted

a year ago

Class 11th Exemplar Solution

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

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

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

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New answer posted

a year ago

Class 11th Exemplar Solution

A sports team of 11 students is to be constituted, choosing at least 5 from Class XI and at least 5 from Class XII. If there are 20 students in each of these classes, in how many ways can the team be constituted?

0 Follower 29 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} T o t a l n u m b e r o f s t u d e n t s i n e a c h c l a s s = 20 \\ W e h a v e t o s e l e c t a t l e a s t 5 s t u d e n t s f r o m e a c h c l a s s . \\ W e h a v e t h e f o l l o w i n g c a s e s . \\ (i) 5 s t u d e n t s f r o m X I c l a s s a n d 6 s t u d e n t s f r o m X I I c l a s s \\ (i i) 6 s t u d e n t s f r o m X I c l a s s a n d 5 s t u d e n t s f r o m X I I c l a s s \\ S o, n u m b e r o f w a y s o f s e l e c t i o n o f a t e a m o f 11 p l a y e r s \\ = {}_{20}{C_{5}} * {}_{20}{C_{6}} + {}_{20}{C_{6}} * {}_{20}{C_{5}} = 2 [{}_{20}{C_{5}} * {}_{20}{C_{6}}] \\ H e n c e, t h e r e q u i r e d w a y s o f s e l e c t i o n = 2 [{}_{20}{C_{5}} * {}_{20}{C_{6}}] \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Nine

In a potato race, 20 potatoes are placed in a line at intervals of 4 meters with the first potato 24 meters from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} A s p e r t h e g i v e n information w e h a v e t h e f o l l o w i n g d i a g r a m \\ S t a r t i n g point = S \\ Distance t r a v e l l e d t o b r i n g t h e f i r s t p o t a t o = 24 + 24 = 48 m \\ Distance t r a v e l l e d t o b r i n g t h e second p o t a t o = 2 (24 + 4) = 56 m \\ Distance t r a v e l l e d t o b r i n g t h e t h i r d p o t a t o = 2 (24 + 4 + 4) = 64 m \\ T h e r f o r e, t h e s e r i e s w i l l b e = 48, 56, 64, \dots \\ w h i c h a n ? ? A . P i n w h i c h a = 48, d = 56 - 48 = 8 \\ W e h a v e t o f i n d t h e t o t a l distance t o b r i n g a l l t h e p o t a t o e s b a c k, s o, n = 20 \\ ∴ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ \Rightarrow S_{20} = \frac{20}{2} [2 * 48 + (20 - 1) 8] = 10 [96 + 152] \\ = 10 * 248 = 2480 m \\ H e n c e, t h e r e q u i r e d distance = 2480 m \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

If $x^{p}$ occurs in the expansion of ${(\frac{x^{2} + 1}{x})}^{2 n},$ prove that its coefficient is $\frac{4 n - p}{3} \frac{2 n + p}{3} .$

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} T h e g i v e n expression i s {(x^{2} + \frac{1}{x})}^{2 n} \\ G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ = {}_{2 n}{C_{r}} {(x^{2})}^{2 n - r} {(\frac{1}{x})}^{r} = {}_{2 n}{C_{r}} {(x)}^{4 n - 2 r} . \frac{1}{x^{r}} \\ = {}_{2 n}{C_{r}} {(x)}^{4 n - 2 r - r} = {}_{2 n}{C_{r}} {(x)}^{4 n - 3 r} \\ I f x^{p} o c c u r s i n {(x^{2} + \frac{1}{x})}^{2 n} \\ t h e n 4 n - 3 r = p \Rightarrow 3 r = 4 n - p \\ \Rightarrow r = \frac{4 n - p}{3} \\ ∴ C o e f f i c i e n t o f x^{p} = {}_{2 n}{C_{r}} = {}_{2 n}{C_{\frac{4 n - p}{3}}} \\ = \frac{(2 n)!}{(\frac{4 n - p}{3})! (2 n - \frac{4 n - p}{3})!} = \frac{(2 n)!}{(\frac{4 n - p}{3})! (\frac{6 n - 4 n + p}{3})!} \\ = \frac{(2 n)!}{(\frac{4 n - p}{3})! (\frac{2 n + p}{3})!} \\ H e n c e, c o e f f i c i e n t o f x^{p} = \frac{(2 n)!}{(\frac{4 n - p}{3})! (\frac{2 n + p}{3})!} \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

In how many ways can a football team of 11 players be selected from 16 players? How many of them will:

(i) Include 2 particular players?

(ii) Exclude 2 particular players?

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t t h e t o t a l n u m b e r o f p l a y e r s = 16 \\ W e h a v e t o s e l e c t 11 p l a y e r s o u t o f 16 p l a y e r s . \\ (i) I f 2 p l a y e r s a r e i n c l u d e d, t h e n n u m b e r o f w a y s o f s e l e c t i o n = {}_{16 - 2}{C_{11 - 2}} = {}_{14}{C_{9}} \\ (i i) I f 2 p l a y e r s a r e e x c l u d e d, t h e n n u m b e r o f w a y s o f s e l e c t i o n = {}_{16 - 2}{C_{11}} = {}_{14}{C_{11}} \\ H e n c e, t h e r e q u i r e d n u m b e r o f w a y s o f s e l e c t i o n a r e \\ (i) {}_{14}{C_{9}} (i i) {}_{14}{C_{11}} \end{array}$

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