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Exemplar Solution Class 11th

Show that the middle term in the expansion of (x – 1/x)²ⁿis 1 × 3 ×5 × … (2n -1)/n × (-2)ⁿ.

6 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n expression i s {(x - \frac{1}{x})}^{2 n} \\ N u m b e r o f t e r m s = 2 n + 1 = 9 (o d d) \\ ∴ M i d d l e t e r m {(\frac{n + 1}{2})}^{t h} t e r m = \frac{2 n + 1 + 1}{2} = (n + 1) t h t e r m . \\ G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ ∴ T_{n + 1} = {}_{2 n}{C_{n}} {(x)}^{2 n - n} {(- \frac{1}{x})}^{n} = {}_{2 n}{C_{n}} {(x)}^{n} {(- 1)}^{n} . \frac{1}{x^{n}} \\ = {(- 1)}^{n} . {}_{2 n}{C_{n}} = {(- 1)}^{n} . \frac{2 n!}{n! (2 n - n)!} \\ = {(- 1)}^{n} . \frac{2 n!}{n! n!} = {(- 1)}^{n} . \frac{2 n (2 n - 1) (2 n - 2) (2 n - 3) \dots 1}{n! n (n - 1) (n - 2) (n - 3) \dots 1} \\ = {(- 1)}^{n} . \frac{2 n (2 n - 1) . 2 (n - 1) (2 n - 3) \dots 1}{n! n (n - 1) (n - 2) (n - 3) \dots 1} \\ = \frac{{(- 1)}^{n} . 2^{n} . [n (n - 1) (n - 1) \dots] . [(2 n - 1) . (2 n - 3) \dots 5.3.1]}{n! . n (n - 1) (n - 2) \dots 1} \\ = \frac{{(- 2)}^{n} [(2 n - 1) . (2 n - 3) \dots 5.3.1]}{n!} \\ = \frac{1 \times 3 \times 5 \times \dots (2 n - 1)}{n!} \times {(- 2)}^{n} \\ H e n c e, t h e m i d d l e t e r m = \frac{1 \times 3 \times 5 \times \dots (2 n - 1)}{n!} \times {(- 2)}^{n} \end{array}$

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Exemplar Solution Class 11th

If $p$ is a real number and if the middle term in the expansion of ${(\frac{p}{2} + 2)}^{8}$ is 1120, find $p$ .

4 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n expression i s {(\frac{P}{2} + 2)}^{8} \\ N u m b e r o f t e r m s = 8 + 1 = 9 (o d d) \\ ∴ M i d d l e t e r m {(\frac{n + 1}{2})}^{t h} t e r m = \frac{9 + 1}{2} = \frac{10}{2} = 5^{t h} t e r m \\ T_{5} = T_{4 + 1} = {}_{8}{C_{4}} {(\frac{P}{2})}^{8 - 4} {(2)}^{4} \\ = {}_{8}{C_{4}} \frac{P^{4}}{2^{4}} \times 2^{4} = {}_{8}{C_{4}} P^{4} \\ N o w {}_{8}{C_{4}} P^{4} = 1120 \Rightarrow \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} . P^{4} = 1120 \\ \Rightarrow 70 P^{4} = 1120 \Rightarrow P^{4} = \frac{1120}{70} = 16 \\ \Rightarrow P^{4} = 2^{4} \Rightarrow P = \pm 2 \\ H e n c e, t h e r e q u i r e d v a l u e o f P = \pm 2 . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?

5 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:

$\begin{array}{l} H e r e, f i r s t t e r m a = 5 a n d t h e c o m m o n d i f f e r e n c e d = 2 l e t t h e c a r p e n t e r w i l l t a k e n d a y s \\ t o f i n i s h t h e j o b . \\ S_{n} = 192 \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ 192 = \frac{n}{2} [2 \times 5 + (n - 1) 2] \\ \Rightarrow 192 \times 2 = n [10 + 2 n - 2] \Rightarrow 384 = n (2 n + 8) \\ \Rightarrow 384 = 2 n^{2} + 8 n \Rightarrow 2 n^{2} + 8 n - 384 = 0 \\ \Rightarrow n^{2} + 4 n - 192 = 0 \Rightarrow n^{2} + 16 n - 12 n - 192 = 0 \\ \Rightarrow n (n + 16) - 12 (n + 16) = 0 \Rightarrow (n - 12) (n + 16) = 0 \\ \Rightarrow n = 12 [? n \neq - 16] \\ H e n c e, t h e r e q u i r e d n u m b e r o f d a y s = 12 . \end{array}$

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Exemplar Solution Class 11th

Find the coefficient of $x^{4}$ in the expansion of ${(1 + x + x^{2} + x^{3})}^{11} .$

4 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:

$\begin{array}{l} G i v e n expression i s {(1 + x + x^{2} + x^{3})}^{11} \\ = {[(1 + x) + x^{2} (1 + x)]}^{11} = {[(1 + x) (1 + x^{2})]}^{11} \\ = {(1 + x)}^{11} . {(1 + x^{2})}^{11} \\ E x p a n d i n g t h e a b o v e expression, w e g e t \\ ({}_{11}{C_{0}} + {}_{11}{C_{1}} x + {}_{11}{C_{2}} x^{2} + {}_{11}{C_{3}} x^{3} + {}_{11}{C_{4}} x^{4} + \dots) . ({}_{11}{C_{0}} + {}_{11}{C_{1}} x^{2} + {}_{11}{C_{2}} x^{4} + \dots) \\ = (1 + 11 x + 55 x^{2} + 165 x^{3} + 330 x^{4} + \dots) . (1 + 11 x^{2} + 55 x^{4} + \dots) \\ C o l l e c t i n g t h e t e r m s c o n t a i n i n g x^{4}, w e g e t \\ (55 + 605 + 330) x^{4} = 990 x^{4} \\ H e n c e, t h e c o e f f i c i e n t o f x^{4} = 990 \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

If the $p$ -th and $q$ -th terms of a G.P. are $q$ and $p$ respectively, show that its $(p + q)$ -th term is ${(\frac{q}{p})}^{p - q}$

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:

$\begin{array}{l} L e t a b e t h e f i r s t t e r m a n d r b e t h e c o m m o n r a t i o o f a G . P . \\ G i v e n t h a t a_{p} = q \Rightarrow a r^{p - 1} = q \dots (i) \\ a n d a_{q} = p \Rightarrow a r^{q - 1} = p \dots (i i) \\ D i v i d i n g e q n . (i) b y e q n . (i i) w e g e t, \\ \frac{a r^{p - 1}}{a r^{q - 1}} = \frac{q}{p} \Rightarrow \frac{r^{p - 1}}{r^{q - 1}} = \frac{q}{p} \\ \Rightarrow r^{p - q} = \frac{q}{p} \Rightarrow r = {(\frac{q}{p})}^{\frac{1}{p - q}} \\ P u t t i n g t h e v a l u e o f r i n e q n . (i) w e g e t \\ a {(\frac{q}{p})}^{\frac{1}{p - q} \times p - 1} = q \\ a {(\frac{q}{p})}^{\frac{p - 1}{p - q}} = q \\ ∴ a = q . {(\frac{p}{q})}^{\frac{p - 1}{p - q}} \\ N o w T_{p + q} = a r^{p + q - 1} = q . {(\frac{p}{q})}^{\frac{p - 1}{p - q}} . {(\frac{q}{p})}^{\frac{1}{p - q} \times (p + q - 1)} \\ = q . {(\frac{p}{q})}^{\frac{p - 1}{p - q}} . {(\frac{q}{p})}^{\frac{p + q - 1}{p - q}} = q . {(\frac{p}{q})}^{\frac{p - 1}{p - q}} . {(\frac{p}{q})}^{\frac{- (p + q - 1)}{p - q}} \\ = q . {(\frac{p}{q})}^{\frac{p - 1}{p - q} - \frac{p + q - 1}{p - q}} = q . {(\frac{p}{q})}^{\frac{p - 1 - p - q + 1}{p - q}} \\ = q . {(\frac{p}{q})}^{\frac{- q}{p - q}} = q . {(\frac{q}{p})}^{\frac{q}{p - q}} = \frac{q^{\frac{q}{p - q} + 1}}{p^{\frac{q}{p - q}}} = \frac{q^{\frac{p}{p - q}}}{p^{\frac{q}{p - q}}} = {[\frac{q^{p}}{p^{q}}]}^{\frac{1}{p - q}} \\ H e n c e, t h e r e q u i r e d t e r m = {[\frac{q^{p}}{p^{q}}]}^{\frac{1}{p - q}} . \end{array}$

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Exemplar Solution Class 11th

A convex polygon has 44 diagonals. Find the number of its sides.

[Hint: Polygon of n sides has (ⁿC_{2 – n}) number of diagonals.]

6 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

$\begin{array}{l} G i v e n t h a t q u e s t i o n n u m b e r 1 a n d 2 a r e c o m p u l s o r y . \\ ∴ T h e r e m a i n i n g q u e s t i o n s a r e 5 - 2 = 3 \\ L e t n b e t h e n u m b e r o f s i d e s i n a p o l y g o n . \\ Since, P o l y g o n o f n s i d e s h a s ({}_{n}{C_{2}} - n) n u m b e r o f d i a g o n a l s \\ ∴ {}_{n}{C_{2}} - n = 44 = \frac{n!}{2! (n - 2)!} - n = 44 \\ \Rightarrow \frac{n (n - 1) (n - 2)!}{2 . (n - 2)!} - n = 44 \Rightarrow \frac{n (n - 1)}{2} - n = 44 \\ \Rightarrow \frac{n^{2} - n - 2 n}{2} = 44 \Rightarrow n^{2} - 3 n = 88 \Rightarrow n^{2} - 3 n - 88 = 0 \\ \Rightarrow n^{2} - 11 n + 8 n - 88 = 0 \Rightarrow n (n - 11) + 8 (n - 11) = 0 \\ \Rightarrow (n - 11) (n + 8) = 0 ∴ n = 11 a n d n = - 8 [? n \neq - 8] \\ S o, n = 11 \\ H e n c e, t h e r e q u i r e d n u m b e r o f s i d e s = 11 . \end{array}$

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Exemplar Solution Class 11th

If the coefficient of second, third, and fourth terms in the expansion of ${(1 + x)}^{2 n}$ are in A.P. Show that $2 n^{2} - 9 n + 7 = 0 .$

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:

$\begin{array}{l} G i v e n expression i s {(1 + x)}^{2 n} \\ H e r e, c o e f f i c i e n t o f 2 n d t e r m = {}_{2 n}{C_{1}} \\ C o e f f i c i e n t o f 3 r d t e r m = {}_{2 n}{C_{2}} \\ a n d c o e f f i c i e n t o f 4 t h t e r m = {}_{2 n}{C_{3}} \\ G i v e n t h a t {}_{2 n}{C_{1}}, {}_{2 n}{C_{2}} a n d {}_{2 n}{C_{3}} a r e i n A . P \\ ∴ {}_{2 n}{C_{2}} - {}_{2 n}{C_{1}} = {}_{2 n}{C_{3}} - {}_{2 n}{C_{2}} \\ \Rightarrow 2 . {}_{2 n}{C_{2}} = {}_{2 n}{C_{1}} + {}_{2 n}{C_{3}} \\ \Rightarrow 2 . \frac{2 n!}{2! (2 n - 2)!} = \frac{2 n!}{(2 n - 1)!} + \frac{2 n!}{3! (2 n - 3)!} \\ \Rightarrow 2 [\frac{2 n (2 n - 1) (2 n - 2)!}{2 \times 1 \times (2 n - 2)!}] = \frac{2 n (2 n - 1)!}{(2 n - 1)!} + \frac{2 n (2 n - 1) (2 n - 2) (2 n - 3)!}{3 \times 2 \times 1 \times (2 n - 3)!} \\ \Rightarrow n (2 n - 1) = n + \frac{n (2 n - 1) (2 n - 2)}{6} \\ \Rightarrow 2 n - 1 = 1 + \frac{(2 n - 1) (2 n - 2)}{6} \\ \Rightarrow 12 n - 6 = 6 + 4 n^{2} - 4 n - 2 n + 2 \\ \Rightarrow 12 n - 12 = 4 n^{2} - 6 n + 2 \\ \Rightarrow 4 n^{2} - 6 n - 12 n + 2 + 12 = 0 \\ \Rightarrow 4 n^{2} - 18 n + 14 = 0 \\ \Rightarrow 2 n^{2} - 9 n + 7 = 0 \\ H e n c e p r o v e d . \end{array}$

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Exemplar Solution Class 11th

In an examination, a student has to answer 4 questions out of 5 questions; questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice.

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

$\begin{array}{l} G i v e n t h a t q u e s t i o n n u m b e r 1 a n d 2 a r e c o m p u l s o r y . \\ ∴ T h e r e m a i n i n g q u e s t i o n s a r e 5 - 2 = 3 \\ T o t a l n u m b e r o f q u e s t i o n s t o b e a t t e m p t e d = 4 q u e s t i o n s 1 a n d 2 a r e c o m p u l s o r y . \\ S o o n l y 2 q u e s t i o n s a r e t o b e d o n e o u t o f 3 q u e s t i o n s \\ T h e r e f o r e n u m b e r o f w a y s = {}_{3}{C_{2}} = {}_{3}{C_{3 - 2}} = {}_{3}{C_{1}} = 3 [? {}_{n}{C_{r}} = {}_{n}{C_{n - r}}] \\ H e n c e, t h e r e q u i r e d n u m b e r o f w a y s = 3 . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

A man accepts a position with an initial salary of Rs 5200 per month. It is understood that he will receive an automatic increase of Rs 320 in the very next month and each month thereafter.

(a) Find his salary for the tenth month.

(b) What is his total earnings during the first year?

15 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:

$\begin{array}{l} G i v e n t h a t f i x e d i n c r e m e n t i n t h e s a l a r y o f a m a n = 320 e a c h m o n t h \\ I n i t i a l s a l a r y = 5200 w h i c h m a k e s a n A . P . \\ w h o s e f i r s t t e r m (a) = 5200 a n d c o m m o n d i f f e r e n c e (d) = 320 \\ (i) S a l a r y f o r t h e t e n t h m o n t h \\ a_{10} = a + (n - 1) d \\ = 5200 + (10 - 1) \times 320 = 5200 + 2880 = 8080 \\ (i i) T o t a l e a r n i n g d u r i n g t h e f i r s t y e a r (12 m o n t h s) \\ S_{12} = \frac{12}{2} [2 \times 5200 + (12 - 1) \times 320] [? S_{n} = \frac{n}{2} [2 a + (n - 1) d]] \\ = 6 [10400 + 3520] = 6 \times 13920 = 83520 \\ H e n c e, t h e r e q u i r e d a m o u n t i s (i) 8080 a n d (i i) 83520 \end{array}$

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Exemplar Solution Class 11th

In a certain city, all telephone numbers have six digits, the first two digits always being 41 or 42 or 46 or 62 or 64. How many telephone numbers have all six digits distinct?

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

$\begin{array}{l} I f f i r s t t w o d i g i t s i s 41, t h e n t h e r e m a i n i n g 4 d i g i t s c a n b e a r r a n g e d i n {}_{8}{P_{4}} w a y s \\ = \frac{8!}{(8 - 4)!} = \frac{8!}{4!} = \frac{8 \times 7 \times 6 \times 5 \times 4!}{4!} = 1680 \\ S i m i l a r l y f i r s t t w o d i g i t s c a n b e 42 o r 46 o r 62 o r 64 . \\ ∴ T o t a l n u m b e r o f t e l e p h o n e n u m b e r s h a v e a l l d i g i t s d i s t i n c t = 5 \times 1680 = 8400 \\ H e n c e, t h e r e q u i r e d t e l e p h o n e n u m b e r s = 8400 . \end{array}$

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Show that the middle term in the expansion of (x – 1/x)²ⁿis 1 × 3 ×5 × … (2n -1)/n × (-2)ⁿ.

If $p$ is a real number and if the middle term in the expansion of ${(\frac{p}{2} + 2)}^{8}$ is 1120, find $p$ .

A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?

Find the coefficient of $x^{4}$ in the expansion of ${(1 + x + x^{2} + x^{3})}^{11} .$

If the $p$ -th and $q$ -th terms of a G.P. are $q$ and $p$ respectively, show that its $(p + q)$ -th term is ${(\frac{q}{p})}^{p - q}$

A convex polygon has 44 diagonals. Find the number of its sides.

[Hint: Polygon of n sides has (ⁿC_{2 – n}) number of diagonals.]

If the coefficient of second, third, and fourth terms in the expansion of ${(1 + x)}^{2 n}$ are in A.P. Show that $2 n^{2} - 9 n + 7 = 0 .$

In an examination, a student has to answer 4 questions out of 5 questions; questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice.

A man accepts a position with an initial salary of Rs 5200 per month. It is understood that he will receive an automatic increase of Rs 320 in the very next month and each month thereafter.

(a) Find his salary for the tenth month.

(b) What is his total earnings during the first year?

In a certain city, all telephone numbers have six digits, the first two digits always being 41 or 42 or 46 or 62 or 64. How many telephone numbers have all six digits distinct?