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New answer posted

a year ago

Maths Exemplar Solution

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

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

Maths Exemplar Solution

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

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

0 Follower 12 Views

Vishal Baghel

Contributor-Level 10

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

Maths Exemplar Solution

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

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

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

Maths Exemplar Solution

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]

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

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

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New answer posted

a year ago

Maths Exemplar Solution

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

(a) 36

(b) 64

(c) 12

(d) 32

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

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

Maths Exemplar Solution

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]

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

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

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New answer posted

a year ago

Maths Exemplar Solution

If ⁿC₁₂ = ⁿC₈, then n is equal to:

(a) 20

(b) 12

(c) 6

(d) 30

0 Follower 7 Views

Payal Gupta

Contributor-Level 10

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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New answer posted

a year ago

Maths Exemplar Solution

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 $\Rightarrow n^{2} - 9 n + 14 = 0 \Rightarrow n = 27 .$ ]

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

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

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New answer posted

a year ago

Maths Exemplar Solution

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

0 Follower 11 Views

Vishal Baghel

Contributor-Level 10

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

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

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New answer posted

a year ago

Maths Exemplar Solution

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

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

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

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

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