Rajasthan BoardEnglish MediumSTD 9MATHSNumber systems1 Mark
Question
If $\text{x}^{-2}=64,$ then $\text{x}^{\frac{1}{3}}+\text{x}^0=$
✓
Answer
$\frac{3}{2}$ Solution: We have to find the value of $\text{x}^{\frac{1}{3}}+\text{x}^0$ if $\text{x}^{-2}=64$ Consider, $\text{x}^{-2}=2^6$ $\frac{1}{\text{x}^2}=2^6$ Multiply $\frac{1}{2}$ on both sides of powers we get $\frac{1}{\text{x}^{2\times\frac{1}{2}}}=2^{6\times\frac{1}{6}}$ $\frac{1}{\text{x}}=2^3$ $\frac{1}{\text{x}}=\frac{8}{1}$ By taking reciprocal on both sides we get, $\frac{1}{8}=\text{x}$ Substituting $\frac{1}{8}$ in $\text{x}^{\frac{1}{3}}+\text{x}^0$ we get $=\Big(\frac{1}{8}\Big)^{\frac{1}{3}}+\Big(\frac{1}{8}\Big)^0$ $=\Big(\frac{1}{2^2}\Big)^{\frac{1}{3}}+\Big(\frac{1}{8}\Big)^0$ $=\frac{1}{2^{3\times\frac{1}{3}}}+1$ $=\frac{1}{2^1}+1$ $=\frac{1}{2}+1$ By taking least common multiply we get $=\frac{1}{2}+\frac{1\times2}{1\times2}$ $=\frac{1}{2}+\frac{2}{2}$ $=\frac{1+2}{2}$ $=\frac{3}{2}$ Hence the correct choice is c.
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