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NCERT Exemplar · Class 11 Mathematics Binomial Theorem

40 questions · 40 still being checked

EXERCISE 8.3 21–30 (part 3 of 4)

  1. Choose the correct answer from the given options in each of the Exercises $\displaystyle 18$ to $\displaystyle 24$ (M.C.Q.).

    Exercise 21

    The coefficient of xn\displaystyle x^n in the expansion of (1+x)2n\displaystyle (1+x)^{2 n} and (1+x)2n−1\displaystyle (1+x)^{2 n-1} are in the ratio.
    (A)
    1\displaystyle 1 : 2\displaystyle 2
    (B)
    1:3\displaystyle 1: 3
    (C)
    3\displaystyle 3 : 1\displaystyle 1
    (D)
    2\displaystyle 2 : 1\displaystyle 1
    [Hint : 2nCn:2n−1Cn\displaystyle { }^{2 n} \mathrm{C}_n:{ }^{2 n-1} \mathrm{C}_n ]

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    (D)
    (D) \(\displaystyle 2:1\). The coefficients are \(\displaystyle {}^{2n}C_n\) and \(\displaystyle {}^{2n-1}C_n\).\[\frac{{}^{2n}C_n}{{}^{2n-1}C_n}=\frac{(2n)!}{n!\,n!}\cdot\frac{n!\,(n-1)!}{(2n-1)!}=\frac{2n}{n}=2 \]

  2. Exercise 22

    If the coefficients of 2nd ,3rd \displaystyle 2^{\text {nd }}, 3^{\text {rd }} and the 4th \displaystyle 4^{\text {th }} terms in the expansion of (1+x)n\displaystyle (1+x)^n are in A.P., then value of n\displaystyle n is
    (A)
    2\displaystyle 2 (B) 7\displaystyle 7 (c) 11\displaystyle 11
    (D)
    14\displaystyle 14
    [Hint: 2nC2=nC1+nC3⇒n2−9n+14=0⇒n=2\displaystyle 2{ }^n \mathrm{C}_2={ }^n \mathrm{C}_1+{ }^n \mathrm{C}_3 \Rightarrow n^2-9 n+14=0 \Rightarrow n=2 or 7\displaystyle 7]

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    (B)
    (B) \(\displaystyle 7\). The coefficients of the 2nd, 3rd, 4th terms are \(\displaystyle {}^nC_1,{}^nC_2,{}^nC_3\).\[2\,{}^nC_2={}^nC_1+{}^nC_3 \]\[n(n-1)=n+\frac{n(n-1)(n-2)}{6} \]\[6(n-1)=6+(n-1)(n-2)\Rightarrow n^2-9n+14=0\Rightarrow n=2\ \text{or}\ 7 \]\(\displaystyle n=2\) is rejected: the expansion has only $\displaystyle 3$ terms, so no 4th term.\[n=7:\quad 2\cdot21=7+35 \]

  3. Exercise 23

    If A and B are coefficient of xn\displaystyle x^n in the expansions of (1+x)2n\displaystyle (1+x)^{2 n} and (1+x)2n−1\displaystyle (1+x)^{2 n-1} respectively, then AB\displaystyle \frac{\mathrm{A}}{\mathrm{B}} equals
    (A)
    1\displaystyle 1 (B) 2\displaystyle 2 (C) 12\displaystyle \frac{1}{2}
    (D)
    1n\displaystyle \frac{1}{n}
    [Hint: AB=2nCn2n−1Cn=2\displaystyle \frac{\mathrm{A}}{\mathrm{B}}=\frac{{ }^{2 n} \mathrm{C}_n}{{ }^{2 n-1} \mathrm{C}_n}=2 ]

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    (B)
    (B) \(\displaystyle 2\). Here \(\displaystyle A={}^{2n}C_n\) and \(\displaystyle B={}^{2n-1}C_n\).\[\frac{A}{B}=\frac{(2n)!}{n!\,n!}\cdot\frac{n!\,(n-1)!}{(2n-1)!}=\frac{2n}{n}=2 \]

  4. Exercise 24

    If the middle term of (1x+xsin⁡x)10\displaystyle \left(\frac{1}{x}+x \sin x\right)^{10} is equal to 778\displaystyle 7 \frac{7}{8}, then value of x\displaystyle x is
    (A)
    2nπ+π6\displaystyle 2 n \pi+\frac{\pi}{6}
    (B)
    nπ+π6\displaystyle n \pi+\frac{\pi}{6}
    (C)
    nπ+(−1)nπ6\displaystyle n \pi+(-1)^n \frac{\pi}{6}
    (D)
    nπ+(−1)nπ3\displaystyle n \pi+(-1)^n \frac{\pi}{3}
    [Hint: T6=10C51x5⋅x5sin⁡5x=638⇒sin⁡5x=125⇒sin⁡x=12\displaystyle \mathrm{T}_6={ }^{10} \mathrm{C}_5 \frac{1}{x^5} \cdot x^5 \sin ^5 x=\frac{63}{8} \Rightarrow \sin ^5 x=\frac{1}{2^5} \Rightarrow \sin x=\frac{1}{2} ⇒x=nπ+(−1)nπ6\displaystyle \Rightarrow x=n \pi+(-1)^n \frac{\pi}{6}]

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    NCERT’s answer
    (C)
    (C) \(\displaystyle n\pi+(-1)^n\dfrac{\pi}{6}\). There are $\displaystyle 11$ terms, so the middle term is \(\displaystyle T_6\).\[T_6={}^{10}C_5\left(\frac1x\right)^5(x\sin x)^5=252\sin^5x \]\[252\sin^5x=7\tfrac78=\tfrac{63}{8}\Rightarrow \sin^5x=\tfrac{1}{32}\Rightarrow \sin x=\tfrac12 \]\[x=n\pi+(-1)^n\frac{\pi}{6} \]
  5. Fill in the blanks in Exercises $\displaystyle 25$ to 33.

    Exercise 25

    The largest coefficient in the expansion of (1+x)30\displaystyle (1+x)^{30} is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle { }^{30} \mathrm{C}_{15}\)
    \(\displaystyle 155117520\), that is \(\displaystyle \binom{30}{15}\).The coefficients are \(\displaystyle \binom{30}{r}\); they rise while the ratio below is at least $\displaystyle 1$, then fall. \[\frac{\binom{30}{r+1}}{\binom{30}{r}} = \frac{30-r}{r+1} \ge 1 \iff r \le 14 \] \[\text{largest coefficient} = \binom{30}{15} = \frac{30!}{15!\,15!} = 155117520 \] Answer: \(\displaystyle 155117520\)

  6. Exercise 26

    The number of terms in the expansion of (x+y+z)n\displaystyle (x+y+z)^n ____\displaystyle \_\_\_\_. [Hint: (x+y+z)n=[x+(y+z)]n\displaystyle (x+y+z)^n=[x+(y+z)]^n ]

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    NCERT’s answer
    \(\displaystyle \frac{(n+1)(n+2)}{2}\)
    \(\displaystyle \dfrac{(n+1)(n+2)}{2}\)Write \(\displaystyle x+y+z = x+(y+z)\). \[(x+y+z)^n = \sum_{r=0}^{n}\binom{n}{r}\,x^{n-r}(y+z)^r \] \[(y+z)^r \text{ has } r+1 \text{ terms} \] The powers of \(\displaystyle x,y,z\) differ from term to term, so nothing combines. \[\sum_{r=0}^{n}(r+1) = \frac{(n+1)(n+2)}{2} \] Answer: \(\displaystyle \dfrac{(n+1)(n+2)}{2}\)

  7. Exercise 27

    In the expansion of (x2−1x2)16\displaystyle \left(x^2-\frac{1}{x^2}\right)^{16}, the value of constant term is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle { }^{16} C_8\)
    \(\displaystyle 12870\)\[T_{r+1} = \binom{16}{r}(x^2)^{16-r}\left(-\frac{1}{x^2}\right)^{r} = (-1)^r\binom{16}{r}\,x^{32-4r} \] Constant term: \[32-4r = 0 \Rightarrow r = 8 \] \[T_9 = (-1)^8\binom{16}{8} = 12870 \] Answer: \(\displaystyle 12870\)

  8. Exercise 28

    If the seventh terms from the beginning and the end in the expansion of (23+133)n\displaystyle \left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n are equal, then n\displaystyle n equals ____\displaystyle \_\_\_\_. [Hint : T7=Tn−7+2⇒nC6(213)n−6(1313)6=nCn−6(213)6(1313)n−6\displaystyle \mathrm{T}_7=\mathrm{T}_{n-7+2} \Rightarrow{ }^n \mathrm{C}_6\left(2^{\frac{1}{3}}\right)^{n-6}\left(\frac{1}{3^{\frac{1}{3}}}\right)^6={ }^n \mathrm{C}_{n-6}\left(2^{\frac{1}{3}}\right)^6\left(\frac{1}{3^{\frac{1}{3}}}\right)^{n-6} ⇒(213)n−12=(1313)n−12⇒\displaystyle \Rightarrow\left(2^{\frac{1}{3}}\right)^{n-12}=\left(\frac{1}{3^{\frac{1}{3}}}\right)^{n-12} \Rightarrow only problem when n−12=0⇒n=12\displaystyle n-12=0 \Rightarrow n=12].

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    NCERT’s answer
    \(\displaystyle n=12\)
    \(\displaystyle 12\)The 7th term from the end is \(\displaystyle T_{n-5}\). \[T_7 = \binom{n}{6}\left(2^{\frac13}\right)^{n-6}\left(3^{-\frac13}\right)^{6} \] \[T_{n-5} = \binom{n}{n-6}\left(2^{\frac13}\right)^{6}\left(3^{-\frac13}\right)^{n-6} \] \[\binom{n}{6}=\binom{n}{n-6} \text{ cancels, and } T_7 = T_{n-5} \text{ gives} \] \[2^{\frac{n-6}{3}}\cdot 3^{-2} = 2^{2}\cdot 3^{-\frac{n-6}{3}} \] \[2^{\frac{n-12}{3}}\cdot 3^{\frac{n-12}{3}} = 1 \] \[6^{\frac{n-12}{3}} = 1 \Rightarrow n = 12 \] Answer: \(\displaystyle n = 12\)

  9. Exercise 29

    The coefficient of a−6b4\displaystyle a^{-6} b^4 in the expansion of (1a−2b3)10\displaystyle \left(\frac{1}{a}-\frac{2 b}{3}\right)^{10} is ____\displaystyle \_\_\_\_. [Hint : T5=10C4(1a)b(−2b3)4=112027a−6b4\displaystyle \mathrm{T}_5={ }^{10} \mathrm{C}_4\left(\frac{1}{a}\right)^b\left(\frac{-2 b}{3}\right)^4=\frac{1120}{27} a^{-6} b^4 ]

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    NCERT’s answer
    \(\displaystyle \frac{1120}{27} a^{-6} a^4\)
    \(\displaystyle \dfrac{1120}{27}\)\[T_{r+1} = \binom{10}{r}\left(\frac{1}{a}\right)^{10-r}\left(-\frac{2b}{3}\right)^{r} \] \[b^{4}:\ r = 4, \qquad a^{-(10-4)} = a^{-6} \] \[\binom{10}{4}\left(-\frac{2}{3}\right)^{4} = 210\cdot\frac{16}{81} = \frac{1120}{27} \] Answer: \(\displaystyle \dfrac{1120}{27}\)

  10. Exercise 30

    Middle term in the expansion of (a3+ba)28\displaystyle \left(a^3+b a\right)^{28} is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle { }^{28} \mathrm{C}_{14} a^{56} b^{14}\)
    \(\displaystyle \binom{28}{14}\,a^{56}b^{14}\)The exponent $\displaystyle 28$ is even, so there is one middle term, \(\displaystyle T_{15}\). \[T_{15} = \binom{28}{14}(a^{3})^{14}(ba)^{14} = \binom{28}{14}\,a^{42}\cdot a^{14}b^{14} \] \[T_{15} = \binom{28}{14}\,a^{56}b^{14} = 40116600\,a^{56}b^{14} \] Answer: \(\displaystyle \binom{28}{14}\,a^{56}b^{14}\)