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New answer posted

a year ago

Class 11th Exemplar Solution

The total number of 9-digit numbers which have all different digits is:

(a) 10!

(b) 9!

(c) 9 * 9!

(d) 10 * 10!

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e t o f o r m 9 d i g i t n u m b e r s f r o m 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 a n d w e k n o w t h a t 0 c a n n o t b e p u t \\ o n e x t r e m e l y l e f t p l a c e . \\ S o, f i r s t p l a c e f r o m t h e l e f t s i d e c a n b e f i l l e d i n 9 w a y s . \\ N o w r e p e t i t i o n i s n o t a l l o w e d . S o, t h e r e m a i n i n g 8 p l a c e s c a n b e f i l l e d i n 9! \\ ∴ T h e r e q u i r e d n u m b e r o f w a y s = 9 * 9! \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

a year ago

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

If $p$ -th, $q$ -th, and $r$ -th terms of an A.P. and G.P. are both $a, b, c$ respectively, show that $\frac{a}{b - c} \cdot \frac{b}{c - a} \cdot \frac{c}{a - b} = 1 .$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t A b e t h e f i r s t t e r m a n d d b e t h e c o m m o n d i f f e r e n c e o f t h e g i v e n A . P . a n d x a n d R b e t h e \\ f i r s t t e r m a n d c o m m o n r a t i o r e s p e c t i v e l y o f t h e G . P . \\ ∴ A + (p - 1) d = a \dots (i) \\ A + (q - 1) d = b \dots (i i) \\ a n d A + (r - 1) d = c \dots (i i i) \\ F o r G . P ., w e h a v e \\ x R^{p - 1} = a \dots (i v) \\ x R^{q - 1} = b \dots (v) \\ a n d x R^{r - 1} = c \dots (v i) \\ S u b t r a c t i n g e q n . (i i) f r o m e q n . (i) w e g e t \\ (p - q) d = a - b \dots (v i i) \\ S i m i l a r l y, (q - r) d = b - c \dots (v i i i) \\ a n d (r - p) d = c - a \dots (i x) \\ N o w w e h a v e t o p r o v e t h a t \\ a^{b - c} . b^{c - a} . c^{a - b} = 1 \\ L . H . S . = a^{b - c} . b^{c - a} . c^{a - b} \\ = {[x R^{p - 1}]}^{(q - r) d} . {[x R^{q - 1}]}^{(r - p) d} . {[x R^{r - 1}]}^{(p - q) d} [\begin{array}{l} F r o m e q . (i), (i i), (i i i), (i v), (v), (v i), \\ (v i i), (v i i i), (i x) \end{array}] \\ = x^{(q - r) d} . R^{(p - 1) (q - r) d} . x^{(r - p) d} . R^{(q - 1) (r - p) d} . x^{(p - q) d} . R^{(r - 1) (p - q) d} \\ = x^{(q - r) d + (r - p) d + (p - q) d} . R^{(p - 1) (q - r) d + (q - 1) (r - p) d + (r - 1) (p - q) d} \\ = x^{(q - r + r - p + p - q) d} . R^{(p q - p r - q + r + q r - p q - r + p + p r - q r - p + q) d} \\ = x^{(0) d} . R^{(0) d} = x^{0} . R^{0} = 1 R . H . S . \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

The number of terms in the expansion of ${(a + b)}^{n}$ where $n \in ℕ$ is one less than the power $n$ .

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} Since, t h e n u m b e r o f t e r m s i n t h e g i v e n expression i s {(a + b)}^{n} i s 1 m o r e t h a n n i . e ., n + 1 \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

If the expansion of ${(x - \frac{1}{x^{2}})}^{2 n}$ contains a term independent of $x$ , then $n$ is a multiple of 2.

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} G i v e n expression i s {(x - \frac{1}{x^{2}})}^{2 n} \\ T_{r + 1} = {}_{2 n}{C_{r}} {(x)}^{2 n - r} {(- \frac{1}{x^{2}})}^{r} = {}_{2 n}{C_{r}} {(x)}^{2 n - r} {(- 1)}^{r} . \frac{1}{x^{2 r}} \\ = {}_{2 n}{C_{r}} {(x)}^{2 n - r - 2 r} {(- 1)}^{r} = {}_{2 n}{C_{r}} {(x)}^{2 n - 3 r} {(- 1)}^{r} \\ F o r t h e t e r m i n d e p e n d e n t o f x, 2 n - 3 r = 0 \\ ∴ r = \frac{2 n}{3} w h i c h i s n o t a n integer a n d t h e expression i s n o t p o s s i b l e t o b e t r u e . \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

The number of ways in which we can choose a committee from four men and six women so that the committee includes at least two men and exactly twice as many women as men is:

(a) 94

(b) 126

(c) 128

(d) None

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} N u m b e r o f m e n = 4 \\ N u m b e r o f w o m e n = 6 \\ W e a r e g i v e n t h a t t h e c o m m i t t e e i n c l u d e s 2 m e n a n d e x a c t l y t w i c e a s m a n y w o m e n a s m e n . \\ T h u s, t h e p o s s i b l e s e l e c t i o n c a n b e \\ 2 m e n a n d 4 w o m e n a n d 3 m e n a n d 6 w o m e n . \\ S o, t h e n u m b e r o f c o m m i t t e e = {}_{4}{C_{2}} * {}_{6}{C_{4}} + {}_{4}{C_{3}} * {}_{6}{C_{6}} \\ = 6 * 15 + 4 * 1 = 90 + 4 = 94 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

The last two digits of the number $3^{400}$ are 01.

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} G i v e n t h a t 3^{400} = {(9)}^{200} = {(10 - 1)}^{200} \\ ∴ {(10 - 1)}^{200} = {}_{200}{C_{0}} {(10)}^{200} + {}_{200}{C_{1}} {(10)}^{199} + \dots - {}_{200}{C_{199}} {(10)}^{1} + {}_{200}{C_{200}} {(1)}^{200} \\ = 1 0^{200} - 200 * 1 0^{199} + \dots - 10 * 200 + 1 \\ S o, i t i s c l e a r t h a t l a s t t w o d i g i t s a r e 01 . \\ H e n c e, t h e g i v e n s t a t e m e n t i s' T r u e' . \end{array}$

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New answer posted

a year ago

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

If the sum of $p$ terms of an A.P. is $q$ and the sum of $q$ terms is $p$ , show that the sum of $p + q$ terms is $- (p + q)$ . Also, find the sum of the first $p - q$ terms ( $p > q$ ).

0 Follower 7 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

a year ago

Class 11th Exemplar Solution

The sum of coefficients of the two middle terms in the expansion of ${(1 + x)}^{2 n - 1}$ is equal to ${}_{2 n - 1}{C_{n}}$ .

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} G i v e n expression i s {(1 + x)}^{2 n - 1} \\ N u m b e r o f t e r m s = 2 n - 1 + 1 = 2 n (e v e n) \\ ∴ M i d d l e t e r m s a r e \frac{2 n}{2} t h t e r m a n d {(\frac{2 n}{2} + 1)}^{t h} t e r m s = n t h t e r m s a n d (n + 1) t h t e r m s \\ C o e f f i c i e n t o f n t h t e r m = {}_{2 n - 1}{C_{n - 1}} \\ a n d t h e c o e f f i c i e n t o f (n + 1) t h t e r m = {}_{2 n - 1}{C_{n}} \\ S u m o f t h e c o e f f i c i e n t s = {}_{2 n - 1}{C_{n - 1}} + {}_{2 n - 1}{C_{n}} \\ = {}_{2 n - 1 + 1}{C_{n}} = {}_{2 n}{C_{n}} [? {}_{n}{C_{r}} + {}_{n}{C_{r - 1}} = {}_{n + 1}{C_{r}}] \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

The number of 5-digit telephone numbers having at least one of their digits repeated is:

(a) 90,000

(b) 10,000

(c) 30,240

(d) 69,760

0 Follower 6 Views

Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f 5 - d i g i t t e l e p h o n e n u m b e r i f a l l t h e d i g i t s a r e r e p e a t e d = {(10)}^{5} \\ [? d i g i t s a r e f r o m 0 t o 9] \\ I f d i g i t s a r e n o t r e p e a t e d, t h e n 5 - d i g i t t e l e p h o n e s, c a n b e f o r m e d i n {}_{10}{P_{5}} w a y s \\ ∴ Required n u m b e r o f w a y s = {(10)}^{5} - {}_{10}{P_{5}} \\ = 100000 - \frac{10!}{(10 - 5)!} = 100000 - \frac{10 * 9 * 8 * 7 * 6 * 5!}{5!} \\ = 100000 - 30240 = 69760 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

0 0

New answer posted

a year ago

Class 11th Exemplar Solution

The number of terms in the expansion of ${[{(2 x + y^{3})}^{4}]}^{7}$ is 8.

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

This is a True or False Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} G i v e n expression i s {[{(2 x + y^{3})}^{4}]}^{7} = {(2 x + y^{3})}^{28} \\ H e r e, n = 28 \\ ∴ T h e n u m b e r o f t e r m s i n g i v e n expression i s (n + 1) \\ i . e ., 28 + 1 = 29 \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' . \end{array}$

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