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NCERT Exemplar · Class 11 Mathematics Binomial Theorem

40 questions · 40 still being checked

EXERCISE 8.3 31–40 (part 4 of 4)

  1. Fill in the blanks in Exercises $\displaystyle 25$ to 33.

    Exercise 31

    The ratio of the coefficients of xp\displaystyle x^p and xq\displaystyle x^q in the expansion of (1+x)p+q\displaystyle (1+x)^{p+q} is ____\displaystyle \_\_\_\_[Hint: p+qCp=p+qCq\displaystyle { }^{p+q} \mathrm{C}_p={ }^{p+q} \mathrm{C}_q ]

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    NCERT’s answer
    $\displaystyle 1$
    \(\displaystyle 1\), that is \(\displaystyle 1:1\).\[\text{coefficient of } x^{p} = \binom{p+q}{p}, \qquad \text{coefficient of } x^{q} = \binom{p+q}{q} \] \[\binom{p+q}{q} = \binom{p+q}{(p+q)-q} = \binom{p+q}{p} \] \[\binom{p+q}{p} : \binom{p+q}{q} = 1 : 1 \] Answer: \(\displaystyle 1:1\)

  2. Exercise 32

    The position of the term independent of x\displaystyle x in the expansion of (x3+32x2)10\displaystyle \left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^2}\right)^{10} is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    Third term
    3rd term\[T_{r+1} = \binom{10}{r}\left(\frac{x}{3}\right)^{\frac{10-r}{2}}\left(\frac{3}{2x^{2}}\right)^{r} \] Power of \(\displaystyle x\): \[\frac{10-r}{2} - 2r = 0 \Rightarrow r = 2 \] \[T_3 = \binom{10}{2}\cdot\frac{1}{3^{4}}\cdot\frac{9}{4} = 45\cdot\frac{1}{81}\cdot\frac{9}{4} = \frac54 \] Answer: \(\displaystyle T_3\), the 3rd term

  3. Exercise 33

    If 2515\displaystyle 25^{15} is divided by 13\displaystyle 13, the reminder is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    $\displaystyle 12$
    $\displaystyle 12$\[25^{15} = (26-1)^{15} = \sum_{r=0}^{15} {}^{15}\mathrm{C}_r\, 26^{15-r}(-1)^r \]\[r \le 14: \quad 26^{15-r} \text{ is a multiple of } 13 \]\[r = 15: \quad (-1)^{15} = -1 \]\[25^{15} = 13m - 1 = 13(m-1) + 12 \]Answer: remainder \(\displaystyle 12\).
  4. State which of the statement in Exercises $\displaystyle 34$ to $\displaystyle 40$ is True or False.

    Exercise 34

    The sum of the series ∑r=01020Cr\displaystyle \sum_{r=0}^{10}{ }^{20} \mathrm{C}_r is 219+20C102\displaystyle 2^{19}+\frac{{ }^{20} \mathrm{C}_{10}}{2}

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    True\[\sum_{r=0}^{20} {}^{20}\mathrm{C}_r = 2^{20} \]\[{}^{20}\mathrm{C}_r = {}^{20}\mathrm{C}_{20-r} \;\Rightarrow\; \sum_{r=0}^{9} {}^{20}\mathrm{C}_r = \sum_{r=11}^{20} {}^{20}\mathrm{C}_r \]\[2\sum_{r=0}^{9} {}^{20}\mathrm{C}_r + {}^{20}\mathrm{C}_{10} = 2^{20} \]\[\sum_{r=0}^{10} {}^{20}\mathrm{C}_r = \frac{2^{20} - {}^{20}\mathrm{C}_{10}}{2} + {}^{20}\mathrm{C}_{10} = 2^{19} + \frac{{}^{20}\mathrm{C}_{10}}{2} \]

  5. Exercise 35

    The expression 79+97\displaystyle 7^9+9^7 is divisible by 64. Hint: 79+97=(1+8)7−(1−8)9\displaystyle 7^9+9^7=(1+8)^7-(1-8)^9

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    NCERT’s answer
    T
    True\[(1-8)^9 = (-7)^9 = -7^9 \;\Rightarrow\; 7^9 + 9^7 = (1+8)^7 - (1-8)^9 \]\[(1+8)^7 = 1 + 7\cdot 8 + 64a = 57 + 64a \]\[(1-8)^9 = 1 - 9\cdot 8 + 64b = -71 + 64b \]\[7^9 + 9^7 = 57 + 71 + 64(a-b) = 64(2 + a - b) \]Here \(\displaystyle a, b\) are integers, so the sum is divisible by \(\displaystyle 64\).

  6. Exercise 36

    The number of terms in the expansion of [(2x+y3)4]7\displaystyle \left[\left(2 x+y^3\right)^4\right]^7 is 8\displaystyle 8

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    NCERT’s answer
    F
    False\[\left[(2x+y^3)^4\right]^7 = (2x+y^3)^{28} \]\[\text{number of terms} = 28 + 1 = 29 \neq 8 \]

  7. Exercise 37

    The sum of coefficients of the two middle terms in the expansion of (1+x)2n−1\displaystyle (1+x)^{2 n-1} is equal to 2n−1Cn\displaystyle { }^{2 n-1} \mathrm{C}_n.

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    NCERT’s answer
    F
    FalseThe middle terms of \(\displaystyle (1+x)^{2n-1}\) are \(\displaystyle T_n\) and \(\displaystyle T_{n+1}\).\[{}^{2n-1}\mathrm{C}_{n-1} + {}^{2n-1}\mathrm{C}_{n} = {}^{2n}\mathrm{C}_{n} \]\[{}^{2n-1}\mathrm{C}_{n-1} = {}^{2n-1}\mathrm{C}_{n} \;\Rightarrow\; {}^{2n}\mathrm{C}_{n} = 2\,{}^{2n-1}\mathrm{C}_{n} \neq {}^{2n-1}\mathrm{C}_{n} \]\[n = 2: \quad 3 + 3 = 6 \neq {}^{3}\mathrm{C}_{2} = 3 \]

  8. Exercise 38

    The last two digits of the numbers 3400\displaystyle 3^{400} are 01.

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    NCERT’s answer
    T
    True\[3^{400} = 9^{200} = (10-1)^{200} \]\[= \sum_{r=0}^{198} {}^{200}\mathrm{C}_r\, 10^{200-r}(-1)^r + {}^{200}\mathrm{C}_{199}\cdot 10\cdot(-1)^{199} + 1 \]\[r \le 198: \quad 10^{200-r} \text{ is a multiple of } 100 \]\[3^{400} = 100k - 2000 + 1 = 100(k-20) + 1 \]The last two digits are \(\displaystyle 01\).

  9. Exercise 39

    If the expansion of (x−1x2)2n\displaystyle \left(x-\frac{1}{x^2}\right)^{2 n} contains a term independent of x\displaystyle x, then n\displaystyle n is a multiple of 2.

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    NCERT’s answer
    F
    False\[T_{r+1} = {}^{2n}\mathrm{C}_r\, x^{2n-r}\left(-\frac{1}{x^2}\right)^r = (-1)^r\,{}^{2n}\mathrm{C}_r\, x^{2n-3r} \]\[2n - 3r = 0 \;\Rightarrow\; r = \frac{2n}{3} \]A term free of \(\displaystyle x\) needs \(\displaystyle n\) to be a multiple of \(\displaystyle 3\), not of \(\displaystyle 2\).\[n = 3: \quad T_3 = {}^{6}\mathrm{C}_2 = 15 \]Here \(\displaystyle n = 3\) is not a multiple of \(\displaystyle 2\).

  10. Exercise 40

    Number of terms in the expansion of (a+b)n\displaystyle (a+b)^n where n∈N\displaystyle n \in \mathbf{N} is one less than the power n\displaystyle n.

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    NCERT’s answer
    F
    False\[(a+b)^n \text{ has } n + 1 \text{ terms} \]This is one more than \(\displaystyle n\), not one less.