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NCERT Exemplar · Class 11 Mathematics Binomial Theorem

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EXERCISE 8.3 1–10 (part 1 of 4)

  1. Exercise 1

    Find the term independent of x,x≠0\displaystyle x, x \neq 0, in the expansion of (3x22−13x)15\displaystyle \left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^{15}.

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    NCERT’s answer
    \(\displaystyle { }^{15} \mathrm{C}_{10}\left(\frac{1}{6}\right)^5\)
    \[T_{r+1} = \binom{15}{r}\left(\frac{3x^2}{2}\right)^{15-r}\left(-\frac{1}{3x}\right)^{r} \] Power of \(\displaystyle x\) is zero: \[2(15-r) - r = 0 \Rightarrow r = 10 \] \[T_{11} = \binom{15}{10}\left(\frac{3}{2}\right)^{5}\left(\frac{1}{3}\right)^{10} = \frac{3003}{2^5\cdot 3^5} = \frac{1001}{2592} \] Answer: \(\displaystyle \dfrac{1001}{2592}\)

  2. Exercise 2

    If the term free from x\displaystyle x in the expansion of (x−kx2)10\displaystyle \left(\sqrt{x}-\frac{k}{x^2}\right)^{10} is 405\displaystyle 405 , find the value of k\displaystyle k.

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    NCERT’s answer
    \(\displaystyle k= \pm 3\)
    \[T_{r+1} = \binom{10}{r}(\sqrt{x})^{10-r}\left(-\frac{k}{x^2}\right)^{r} \] Term free from \(\displaystyle x\): \[\frac{10-r}{2} - 2r = 0 \Rightarrow r = 2 \] \[T_3 = \binom{10}{2}k^2 = 45k^2 = 405 \] \[k^2 = 9 \Rightarrow k = \pm 3 \] Answer: \(\displaystyle k = \pm 3\)

  3. Exercise 3

    Find the coefficient of x\displaystyle x in the expansion of (1−3x+7x2)(1−x)16\displaystyle \left(1-3 x+7 x^2\right)(1-x)^{16}.

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    NCERT’s answer
    -$\displaystyle 19$
    \[(1-x)^{16} = 1 - \binom{16}{1}x + \binom{16}{2}x^2 - \cdots = 1 - 16x + \cdots \] Terms giving \(\displaystyle x\) in \(\displaystyle (1-3x+7x^2)(1-x)^{16}\): \[1\cdot(-16x) + (-3x)\cdot 1 = -19x \] Answer: \(\displaystyle -19\)

  4. Exercise 4

    Find the term independent of x\displaystyle x in the expansion of, (3x−2x2)15\displaystyle \left(3 x-\frac{2}{x^2}\right)^{15}.

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    NCERT’s answer
    -$\displaystyle 3003$ (\(\displaystyle 3^{10}\)) (\(\displaystyle 2^{5}\))
    \[T_{r+1} = \binom{15}{r}(3x)^{15-r}\left(-\frac{2}{x^2}\right)^{r} \] Power of \(\displaystyle x\) is zero: \[(15-r) - 2r = 0 \Rightarrow r = 5 \] \[T_6 = \binom{15}{5}\,3^{10}\,(-2)^5 = -3003\cdot 59049\cdot 32 \] \[T_6 = -5\,674\,372\,704 \] Answer: \(\displaystyle -\binom{15}{5}\,3^{10}\,2^{5} = -5\,674\,372\,704\)

  5. Exercise 5

    Find the middle term (terms) in the expansion of
    (i)
    (xa−ax)10\displaystyle \left(\frac{x}{a}-\frac{a}{x}\right)^{10}
    (ii)
    (3x−x36)9\displaystyle \left(3 x-\frac{x^3}{6}\right)^9

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    NCERT’s answer
    (i)
    -$\displaystyle 252$
    (ii)
    \(\displaystyle \frac{189}{8} x^{17} ; \frac{-21}{16} x^{19}\)
    (i)
    \(\displaystyle n = 10\) is even, so there is one middle term, \(\displaystyle T_6\):
    \[T_6 = \binom{10}{5}\left(\frac{x}{a}\right)^{5}\left(-\frac{a}{x}\right)^{5} = -252 \]
    (ii)
    \(\displaystyle n = 9\) is odd, so the middle terms are \(\displaystyle T_5\) and \(\displaystyle T_6\):
    \[T_5 = \binom{9}{4}(3x)^{5}\left(-\frac{x^3}{6}\right)^{4} = 126\cdot 243\cdot\frac{1}{1296}\,x^{17} = \frac{189}{8}x^{17} \]
    \[T_6 = \binom{9}{5}(3x)^{4}\left(-\frac{x^3}{6}\right)^{5} = -126\cdot 81\cdot\frac{1}{7776}\,x^{19} = -\frac{21}{16}x^{19} \]
    Answer: (i) \(\displaystyle -252\); (ii) \(\displaystyle \dfrac{189}{8}x^{17}\) and \(\displaystyle -\dfrac{21}{16}x^{19}\)

  6. Exercise 6

    Find the coefficient of x15\displaystyle x^{15} in the expansion of (x−x2)10\displaystyle \left(x-x^2\right)^{10}.

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    NCERT’s answer
    -$\displaystyle 252$
    \[T_{r+1} = \binom{10}{r}x^{10-r}(-x^2)^{r} = (-1)^r\binom{10}{r}x^{10+r} \] \[10 + r = 15 \Rightarrow r = 5 \] \[\text{coefficient} = (-1)^5\binom{10}{5} = -252 \] Answer: \(\displaystyle -252\)

  7. Exercise 7

    Find the coefficient of 1x17\displaystyle \frac{1}{x^{17}} in the expansion of (x4−1x3)15\displaystyle \left(x^4-\frac{1}{x^3}\right)^{15}.

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    NCERT’s answer
    -$\displaystyle 1365$
    \[T_{r+1} = \binom{15}{r}(x^4)^{15-r}\left(-\frac{1}{x^3}\right)^{r} = (-1)^r\binom{15}{r}x^{60-7r} \] \[60 - 7r = -17 \Rightarrow r = 11 \] \[\text{coefficient} = (-1)^{11}\binom{15}{11} = -\binom{15}{4} = -1365 \] Answer: \(\displaystyle -1365\)

  8. Exercise 8

    Find the sixth term of the expansion (y12+x13)n\displaystyle \left(y^{\frac{1}{2}}+x^{\frac{1}{3}}\right)^n, if the binomial coefficient of the third term from the end is 45. [Hint: Binomial coefficient of third term from the end = Binomial coefficient of third term from beginning =nC2\displaystyle ={ }^n \mathrm{C}_2.]

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    NCERT’s answer
    \(\displaystyle 252 y^{\frac{5}{2}} x^{\frac{5}{3}}\)
    Third term from the end has the same binomial coefficient as the third term from the beginning: \[\binom{n}{2} = 45 \Rightarrow n(n-1) = 90 \Rightarrow n = 10 \] \[T_6 = \binom{10}{5}\left(y^{1/2}\right)^{5}\left(x^{1/3}\right)^{5} = 252\,y^{5/2}x^{5/3} \] Answer: \(\displaystyle 252\,x^{5/3}y^{5/2}\)

  9. Exercise 9

    Find the value of r\displaystyle r, if the coefficients of (2r+4)th \displaystyle (2 r+4)^{\text {th }} and (r−2)th \displaystyle (r-2)^{\text {th }} terms in the expansion of (1+x)18\displaystyle (1+x)^{18} are equal.

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    NCERT’s answer
    \(\displaystyle r=6\)
    The \(\displaystyle k\)th term of \(\displaystyle (1+x)^{18}\) has coefficient \(\displaystyle \binom{18}{k-1}\).\[\binom{18}{2r+3} = \binom{18}{r-3} \]\(\displaystyle \binom{n}{a}=\binom{n}{b}\) gives \(\displaystyle a=b\) or \(\displaystyle a+b=n\).\[2r+3=r-3 \Rightarrow r=-6 \quad \text{(rejected, } r-2\ge 1\text{)} \]\[(2r+3)+(r-3)=18 \]\[3r=18 \Rightarrow r=6 \]Check: 16th and 4th terms, \(\displaystyle \binom{18}{15}=\binom{18}{3}\).Answer: \(\displaystyle r=6\)

  10. Exercise 10

    If the coefficient of second, third and fourth terms in the expansion of (1+x)2n\displaystyle (1+x)^{2 n} are in A.P. Show that 2n2−9n+7=0\displaystyle 2 n^2-9 n+7=0.

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    Coefficients of the 2nd, 3rd, 4th terms: \(\displaystyle \binom{2n}{1},\ \binom{2n}{2},\ \binom{2n}{3}\). They are in A.P.\[2\binom{2n}{2}=\binom{2n}{1}+\binom{2n}{3} \]\[2n(2n-1)=2n+\frac{2n(2n-1)(2n-2)}{6} \]Divide by \(\displaystyle 2n\).\[2n-1=1+\frac{(2n-1)(2n-2)}{6} \]\[12n-6=6+4n^2-6n+2 \]\[4n^2-18n+14=0 \]\[2n^2-9n+7=0 \]Answer: \(\displaystyle 2n^2-9n+7=0\)