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New answer posted•a year ago

Exemplar Solution Class 11th

If the seventh terms from the beginning and the end in the expansion of are equal, then $n$ equals _________________ .

[Hint: $T_{7} = T_{n - 7 + 2}$ , = ⁿC⁶ (2 /3)^n-6 (1/3 ^{1/3 6}) = ⁿC^{n – 6} (2 ^1/3 ⁶ (1/3 ^1/3) ⁿ^{- 6} solve for $n .$ ]

= (2 ^1/3)ⁿ^{– 12} = 1/3 ^{1/3 n -12} = only problem when n – 12 = 0 = n =12].

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} T h e g i v e n expansion i s {(+ \frac{1}{})}^{n} \\ ∴ T_{7} = T_{6 + 1} = {}_{n}{C_{6}} {(2^{1 / 3})}^{n - 6} . \frac{1}{{(3^{1 / 3})}^{6}} = {}_{n}{C_{6}} {(2)}^{\frac{n - 6}{3}} {(- 1)}^{r} . \frac{1}{{(3)}^{2}} \\ N o w t h e T_{7} f r o m t h e e n d = T_{7} f r o m t h e b e g i n n i n g i n {(\frac{1}{} +)}^{n} . \\ ∴ T_{7} = T_{6 + 1} = {}_{n}{C_{6}} {(\frac{1}{3^{1 / 3}})}^{n - 6} . {(2^{1 / 3})}^{6} \\ A s p e r t h e q u e s t i o n s, w e g e t \\ {}_{n}{C_{6}} {(2)}^{\frac{n - 6}{3}} . (\frac{1}{3^{2}}) = {}_{n}{C_{6}} . \frac{1}{3^{\frac{n - 6}{3}}} . {(2)}^{2} \\ \Rightarrow {(2)}^{\frac{n - 6}{3}} . {(3)}^{- 2} = {(3)}^{- (\frac{n - 6}{3})} . {(2)}^{2} \\ \Rightarrow {(2)}^{\frac{n - 6}{3} - 2} . {(3)}^{- 2 + \frac{n - 6}{3}} = 1 \\ \Rightarrow {(2)}^{\frac{n - 12}{3}} . {(3)}^{\frac{n - 12}{3}} = 1 \\ \Rightarrow {(6)}^{\frac{n - 12}{3}} = {(6)}^{0} \\ \Rightarrow \frac{n - 12}{3} = 0 \Rightarrow n = 12 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 12 . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

Find the $r$ -th term of an A.P. sum of whose first $n$ terms is $2 n + 3 n^{2}$ .
[Hint: $a_{n} = S_{n} - S_{n - 1}$ ]

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n t h a t S_{n} = 2 n + 3 n^{2} \\ \Rightarrow S_{1} = 2 \times 1 + 3 {(1)}^{2} = 5 \\ \Rightarrow S_{2} = 2 \times 2 + 3 \times 4 = 16 \\ \Rightarrow S_{3} = 2 \times 3 + 3 \times 9 = 33 \\ \dots \dots \dots \\ ∴ S_{1} = a_{1} = 5 \\ S_{2} - S_{1} = a_{2} = 16 - 5 = 11 \\ ∴ d = a_{2} - a_{1} = 11 - 5 = 6 \\ N o w T_{r} = a_{1} + (r - 1) d \\ = 5 + (r - 1) 6 = 5 + 6 r - 6 = 6 r - 1 \\ H e n c e, t h e r e q u i r e d r t h t e r m i s 6 r - 1 \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

The number of different four-digit numbers that can be formed with the digits 2, 3, 4, 7 and using each digit only once is:

(a) 120

(b) 96

(c) 24

(d) 100

10 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} F o u r - d i g i t n u m b e r s a r e t o b e f o r m e d f r o m t h e d i g i t s 2, 3, 4, 7 w i t h o u t r e p e t i t i o n \\ S o, t h e r e q u i r e d 4 - d i g i t n u m b e r s = {}_{4}{P_{4}} = 4! = 4 \times 3 \times 2 \times 1 = 24 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

Find the sum of the series $(3^{3} - 2^{3}) + (5^{3} - 4^{3}) + (7^{3} - 6^{3}) + . . .$ to (i) $n$ terms (ii) 10 terms.

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n s e r i e s \\ \Rightarrow (3^{3} - 2^{3}) + (5^{3} - 4^{3}) + (7^{3} - 6^{3}) + \dots = (3^{3} + 5^{3} + 7^{3} + \dots) - (2^{3} + 4^{3} + 6^{3} + \dots) \\ = [3^{3} + 5^{3} + 7^{3} + \dots {(2 n + 1)}^{3}] - [2^{3} + 4^{3} + 6^{3} + \dots {(2 n)}^{3}] \\ ∴ T_{n} = {(2 n + 1)}^{3} - {(2 n)}^{3} \\ = (2 n + 1 - 2 n) [{(2 n + 1)}^{2} + (2 n + 1) (2 n) + {(2 n)}^{2}] [? a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})] \\ = 1 . [4 n^{2} + 1 + 4 n + 4 n^{2} + 2 n + 4 n^{2}] = 12 n^{2} + 6 n + 1 \\ (i) S_{n} = \sum_{}^{} T_{n} = 12 \sum_{}^{} n^{2} + 6 \sum_{}^{} n + n \\ = 12 . \frac{n (n + 1) (2 n + 1)}{6} + \frac{6 n (n + 1)}{2} + n \\ = 2 n (n + 1) (2 n + 1) + 3 n (n + 1) + n \\ = n [2 (n + 1) (2 n + 1) + 3 (n + 1) + 1] \\ = n [2 (2 n^{2} + 3 n + 1) + 3 n + 3 + 1] \\ = n [4 n^{2} + 6 n + 2 + 3 n + 3 + 1] = n [4 n^{2} + 9 n + 6] \\ = 4 n^{3} + 9 n^{2} + 6 n \\ (i i) S_{10} = 4 {(10)}^{3} + 9 {(10)}^{2} + 6 (10) = 4 \times 1000 + 9 \times 100 + 60 \\ = 4000 + 960 = 4960 \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

In the expansion of x² – 1/x² ¹⁶ $,$ the value of the constant term is ________________ .

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t T_{r + 1} b e t h e c o n s t a n t t e r m i n t h e expansion o f {(x^{2} - \frac{1}{x^{2}})}^{16} \\ ∴ T_{r + 1} = {}_{16}{C_{r}} {(x^{2})}^{16 - r} {(\frac{- 1}{x^{2}})}^{r} = {}_{16}{C_{r}} {(x)}^{32 - 2 r} {(- 1)}^{r} . \frac{1}{x^{2 r}} \\ = {(- 1)}^{r} . {}_{16}{C_{r}} {(x)}^{32 - 2 r - 2 r} = {(- 1)}^{r} . {}_{16}{C_{r}} {(x)}^{32 - 4 r} \\ F o r g e t t i n g c o n s t a n t t e r m, 32 - 4 r = 0 \\ \Rightarrow r = 8 \\ ∴ T_{r + 1} = {(- 1)}^{8} . {}_{16}{C_{8}} = {}_{16}{C_{8}} \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s {}_{16}{C_{8}} . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

The number of terms in the expansion of ${(x + y + z)}^{n}$ is _________________

[Hint: ${(x + y + z)}^{n} = {[x + (y + z)]}^{n} .$ ]

13 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} T h e expression {(x + y + z)}^{n} c a n b e w r i t t e n a {[x + (y + z)]}^{n} \\ ∴ {[x + (y + z)]}^{n} = {}_{n}{C_{0}} x^{n} {(y + z)}^{0} + {}_{n}{C_{1}} {(x)}^{n - 1} (y + z) + {}_{n}{C_{2}} {(x)}^{n - 2} {(y + z)}^{2} + \dots + {}_{n}{C_{n}} {(y + z)}^{n} \\ ∴ N u m b e r o f t e r m s 1 + 2 + 3 + 4 + \dots + (n + 1) = \frac{(n + 1) (n + 2)}{2} \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s \frac{(n + 1) (n + 2)}{2} \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

The largest coefficient in the expansion of ${(1 + x)}^{30}$ is _________________ .

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} H e r e n = 30 w h i c h i s e v e n \\ ∴ t h e the largest c o e f f i c i e n t i n {(1 + x)}^{n} = {}_{n}{C_{n / 2}} \\ S o, t h e the largest c o e f f i c i e n t i n {(1 + x)}^{30} = {}_{30}{C_{15}} \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s {}_{30}{C_{15}} . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If the middle term of ${(1 + \frac{x sin x}{x})}^{10}$ is equal to $7 \frac{7}{8}$ , then the value of $x$ is:

(a) $2 n π + \frac{π}{6}$

(b) $n π + \frac{π}{6}$

(c) $n π + {(- 1)}^{n} \frac{π}{6}$

(d) $n π + {(- 1)}^{n} \frac{π}{3}$

[Hint: T₆ = ¹⁰C₅ 1/x₅ . x⁵ sin⁵ x = 63/3 = sin⁵ x = 1/25 sin ½= x = nπ + (-1)ⁿπ/6]

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n expression i s {(\frac{1}{x} + x s i n x)}^{10} \\ N u m b e r o f t e r m s = 10 + 1 = 11 (o d d) \\ ∴ M i d d l e t e r m {(\frac{n + 1}{2})}^{t h} t e r m = \frac{11 + 1}{2} = \frac{12}{2} = 6^{t h} t e r m \\ T_{6} = T_{5 + 1} = {}_{10}{C_{5}} {(\frac{1}{x})}^{10 - 5} {(x s i n x)}^{5} \\ ∴ {}_{10}{C_{5}} {(\frac{1}{x})}^{5} . x^{5} . s i n^{5} x = 7 \frac{7}{8} \Rightarrow {}_{10}{C_{5}} . s i n^{5} x = \frac{63}{8} \\ \Rightarrow \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} . s i n^{5} x = \frac{63}{8} \Rightarrow 252 . s i n^{5} x = \frac{63}{8} \\ \Rightarrow s i n^{5} x = \frac{63}{8 \times 252} \Rightarrow s i n^{5} x = \frac{1}{32} \\ \Rightarrow s i n^{5} x = {(\frac{1}{2})}^{5} \Rightarrow s i n x = \frac{1}{2} \\ \Rightarrow s i n x = s i n \frac{π}{6} \\ ∴ x = n π + {(- 1)}^{n} . \frac{π}{6} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

The number of possible outcomes when a coin is tossed 6 times is:

(a) 36

(b) 64

(c) 12

(d) 32

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e k n o w t h a t a c o i n h a s H e a d a n d T a i l (H, T) \\ ∴ W h e n a c o i n i s t o s s e d 6 t i m e s, t h e n t h e \\ P o s s i b l e o u t c o m e = 2^{6} = 64 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If $A$ and $B$ are coefficients of $x^{n}$ in the expansions of ${(1 + x)}^{2 n}$ and ${(1 + x)}^{2 n - 1}$ , respectively, then $\frac{A}{B}$

(a) 1

(b) 2

(c) $\frac{1}{2}$

(d) $\frac{1}{n}$

[Hint: $\frac{A}{B} =$ ²ⁿC_n/^{2n – 1}C_n = 2]

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n expression i s {(1 + x)}^{2 n} \\ T_{r + 1} = {}_{2 n}{C_{r}} x^{r} \\ ∴ C o e f f i c i e n t o f x^{n} = {}_{2 n}{C_{n}} = A (G i v e n) \\ I n t h e expression i s {(1 + x)}^{2 n - 1} \\ T_{r + 1} = {}_{2 n - 1}{C_{r}} x^{r} \\ ∴ C o e f f i c i e n t o f x^{n} = {}_{2 n - 1}{C_{n - 1}} = B (G i v e n) \\ S o, \frac{A}{B} = \frac{{}_{2 n}{C_{n}}}{{}_{2 n - 1}{C_{n - 1}}} \\ = \frac{\frac{2 n!}{n! (n)!}}{\frac{(2 n - 1)!}{(n - 1)! (2 n - 1 - n + 1)!}} = \frac{\frac{2 n!}{n! n!}}{\frac{(2 n - 1)!}{(n - 1)! (n)!}} \\ = \frac{2 n!}{n! n!} \times \frac{(n - 1)! n!}{(2 n - 1)!} = \frac{2 n (2 n - 1)!}{n! n (n - 1)!} \times \frac{(n - 1)! n!}{(2 n - 1)!} \\ = \frac{2}{1} = 2 : 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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