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NCERT Exemplar · Class 9 Mathematics Number Systems

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EXERCISE 1.3 11–14 (part 5 of 6)

  1. Exercise 11

    Find the values of a\displaystyle a and b\displaystyle b in each of the following: (i) 5+237+43=a−63\displaystyle \frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a-6 \sqrt{3} (ii) 3−53+25=a5−1911\displaystyle \frac{3-\sqrt{5}}{3+2 \sqrt{5}}=a \sqrt{5}-\frac{19}{11} (iii) 2+332−23=2−b6\displaystyle \frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}=2-b \sqrt{6} (iv) 7+57−5−7−57+5=a+7115b\displaystyle \frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{11} \sqrt{5} b

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    NCERT’s answer
    (i)
    $\displaystyle a=11$ (ii) $\displaystyle a=\frac{9}{11}$ (iii) $\displaystyle b=\frac{-5}{6}$ (iv) $\displaystyle a=0, b=1$
    (i)
    \[\frac{5+2\sqrt3}{7+4\sqrt3}\times\frac{7-4\sqrt3}{7-4\sqrt3} = \frac{11-6\sqrt3}{49-48}=11-6\sqrt3 \]
    \[11-6\sqrt3=a-6\sqrt3 \implies a=11 \]
    (ii)
    \[\frac{3-\sqrt5}{3+2\sqrt5}\times\frac{3-2\sqrt5}{3-2\sqrt5} = \frac{19-9\sqrt5}{9-20}=\frac{9\sqrt5-19}{11} \]
    \[\frac{9}{11}\sqrt5-\frac{19}{11}=a\sqrt5-\frac{19}{11} \implies a=\frac{9}{11} \]
    (iii)
    \[\frac{\sqrt2+\sqrt3}{3\sqrt2-2\sqrt3}\times\frac{3\sqrt2+2\sqrt3}{3\sqrt2+2\sqrt3} = \frac{12+5\sqrt6}{18-12}=2+\frac{5}{6}\sqrt6 \]
    \[2+\frac56\sqrt6=2-b\sqrt6 \implies b=-\frac56 \]
    (iv)
    \[\frac{7+\sqrt5}{7-\sqrt5}-\frac{7-\sqrt5}{7+\sqrt5} = \frac{(7+\sqrt5)^2-(7-\sqrt5)^2}{49-5}=\frac{28\sqrt5}{44}=\frac{7}{11}\sqrt5 \]
    \[0+\frac{7}{11}\sqrt5\cdot1 = a+\frac{7}{11}\sqrt5\,b \implies a=0,\ b=1 \]
    Answer: (i) \(\displaystyle a=11\) (ii) \(\displaystyle a=\tfrac9{11}\) (iii) \(\displaystyle b=-\tfrac56\) (iv) \(\displaystyle a=0,\ b=1\)

  2. Exercise 12

    If a=2+3\displaystyle a=2+\sqrt{3}, then find the value of a−1a\displaystyle a-\frac{1}{a}.

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    NCERT’s answer
    $\displaystyle 2 \sqrt{3}$
    \[a=2+\sqrt3 \] \[\frac1a = \frac{1}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}=\frac{2-\sqrt3}{4-3}=2-\sqrt3 \] \[a-\frac1a = (2+\sqrt3)-(2-\sqrt3)=2\sqrt3 \]Answer: \(\displaystyle 2\sqrt3\)

  3. Exercise 13

    Rationalise the denominator in each of the following and hence evaluate by taking 2=1.414,3=1.732\displaystyle \sqrt{2}=1.414, \sqrt{3}=1.732 and 5=2.236\displaystyle \sqrt{5}=2.236, upto three places of decimal. (i) 43\displaystyle \frac{4}{\sqrt{3}} (ii) 66\displaystyle \frac{6}{\sqrt{6}} (iii) 10−52\displaystyle \frac{\sqrt{10}-\sqrt{5}}{2} (iv) 22+2\displaystyle \frac{\sqrt{2}}{2+\sqrt{2}} (v) 13+2\displaystyle \frac{1}{\sqrt{3}+\sqrt{2}}

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    NCERT’s answer
    (i)
    2.$\displaystyle 309$ (ii) $\displaystyle 2.449$ (iii) $\displaystyle 0.463$ (iv) $\displaystyle 0.414$ (v) $\displaystyle 0.318$
    (i)
    \[\frac{4}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}=\frac{4\sqrt3}{3} \]
    \[\frac{4(1.732)}{3}=2.309 \]
    (ii)
    \[\frac{6}{\sqrt6}=\sqrt6=\sqrt2\cdot\sqrt3 \]
    \[(1.414)(1.732)=2.449 \]
    (iii)
    \[\sqrt{10}=\sqrt2\cdot\sqrt5=(1.414)(2.236)=3.162 \]
    \[\frac{\sqrt{10}-\sqrt5}{2}=\frac{3.162-2.236}{2}=0.463 \]
    (iv)
    \[\frac{\sqrt2}{2+\sqrt2}\times\frac{2-\sqrt2}{2-\sqrt2}=\frac{2\sqrt2-2}{2}=\sqrt2-1 \]
    \[1.414-1=0.414 \]
    (v)
    \[\frac{1}{\sqrt3+\sqrt2}\times\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}=\sqrt3-\sqrt2 \]
    \[1.732-1.414=0.318 \]
    Answer: (i) \(\displaystyle 2.309\) (ii) \(\displaystyle 2.449\) (iii) \(\displaystyle 0.463\) (iv) \(\displaystyle 0.414\) (v) \(\displaystyle 0.318\)

  4. Exercise 14

    Simplify: (i) (13+23+33)12\displaystyle \left(1^{3}+2^{3}+3^{3}\right)^{\frac{1}{2}} (ii) (35)4×(85)−12×(325)6\displaystyle \left(\frac{3}{5}\right)^{4}\times\left(\frac{8}{5}\right)^{-12}\times\left(\frac{32}{5}\right)^{6} (iii) (127)−23\displaystyle \left(\frac{1}{27}\right)^{-\frac{2}{3}} (iv) [(625−12)−14]2\displaystyle \left[\left(625^{-\frac{1}{2}}\right)^{-\frac{1}{4}}\right]^{2} (v) 913×27−12316×3−23\displaystyle \frac{9^{\frac{1}{3}} \times 27^{-\frac{1}{2}}}{3^{\frac{1}{6}} \times 3^{-\frac{2}{3}}} (vi) 64−13(6413−6423)\displaystyle 64^{-\frac{1}{3}}\left(64^{\frac{1}{3}}-64^{\frac{2}{3}}\right) (vii) 813×161332−13\displaystyle \frac{8^{\frac{1}{3}} \times 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}

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    NCERT’s answer
    (i)
    $\displaystyle 6$ (ii) $\displaystyle \frac{2025}{64}$ (iii) $\displaystyle 9$ (iv) $\displaystyle 5$ (v) $\displaystyle 3^{-\frac{1}{3}}$ (vi) -$\displaystyle 3$ (vii) $\displaystyle 16$
    (i)
    \[1^3+2^3+3^3 = 1+8+27=36 \]
    \[(36)^{1/2}=6 \]
    (ii)
    \[\left(\frac35\right)^4\left(\frac85\right)^{-12}\left(\frac{32}5\right)^6 = \frac{3^4}{5^4}\cdot\frac{5^{12}}{2^{36}}\cdot\frac{2^{30}}{5^6} \]
    \[= 3^4\cdot5^{12-4-6}\cdot2^{30-36} = 3^4\cdot5^2\cdot2^{-6} = \frac{81\times25}{64}=\frac{2025}{64} \]
    (iii)
    \[\left(\frac1{27}\right)^{-2/3} = 27^{2/3}=(3^3)^{2/3}=3^2=9 \]
    (iv)
    \[\left[\left(625^{-1/2}\right)^{-1/4}\right]^2 = 625^{(-1/2)(-1/4)(2)} = 625^{1/4}=(5^4)^{1/4}=5 \]
    (v)
    \[9^{1/3}=3^{2/3},\quad 27^{-1/2}=3^{-3/2} \]
    \[3^{1/6}\times3^{-2/3}=3^{-1/2} \]
    \[\frac{3^{2/3-3/2}}{3^{-1/2}}=\frac{3^{-5/6}}{3^{-1/2}}=3^{-1/3}=\frac{\sqrt[3]9}{3} \]
    (vi)
    \[64^{1/3}=4,\quad 64^{2/3}=16,\quad 64^{-1/3}=\frac14 \]
    \[64^{-1/3}\left(64^{1/3}-64^{2/3}\right)=\frac14(4-16)=\frac14(-12)=-3 \]
    (vii)
    \[8^{1/3}\times16^{1/3}=2^{1}\times2^{4/3}=2^{7/3} \]
    \[32^{-1/3}=2^{-5/3} \]
    \[\frac{2^{7/3}}{2^{-5/3}}=2^{4}=16 \]
    Answer: (i) \(\displaystyle 6\) (ii) \(\displaystyle \dfrac{2025}{64}\) (iii) \(\displaystyle 9\) (iv) \(\displaystyle 5\) (v) \(\displaystyle \dfrac{\sqrt[3]9}{3}\) (vi) \(\displaystyle -3\) (vii) \(\displaystyle 16\).