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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
If $\text{f}(\text{x})=\Big(\frac{\text{x}^\text{l}}{\text{x}^\text{m}}\Big)^{\text{l}+\text{m}}\Big(\frac{\text{x}^\text{m}}{\text{x}^\text{n}}\Big)^{\text{m}+\text{n}}\Big(\frac{\text{x}^\text{n}}{\text{x}^\text{l}}\Big)^{\text{n}+1},$ the $f\ '(x)$ is equal to:
  • A
    $1$
  • ✓
    $0$
  • C
    $x^{1+m+n}$
  • D
    None of these.
✓

Answer

Correct option: B.
$0$
We have $\text{f}(\text{x})=\Big(\frac{\text{x}^\text{l}}{\text{x}^\text{m}}\Big)^{\text{l}+\text{m}}\Big(\frac{\text{x}^\text{m}}{\text{x}^\text{n}}\Big)^{\text{m}+\text{n}}\Big(\frac{\text{x}^\text{n}}{\text{x}^\text{l}}\Big)^{\text{n}+1}$
$\Rightarrow\text{f}(\text{x})=\text{x}^{(\text{l}-\text{m})(\text{l}+\text{m})}\times\text{x}^{(\text{m}-\text{n})(\text{m}+\text{n})}\times\text{x}^{(\text{n}-\text{l})(\text{n}-\text{l})}$
$\Rightarrow\text{f}(\text{x})=\text{x}^{\text{l}^2-\text{m}^2}\times\text{x}^{\text{m}^2-\text{n}^2}\times\text{x}^{\text{n}^2-\text{l}^2}$
$\Rightarrow\text{f}(\text{x})=\text{x}^{(\text{l}^2-\text{m}^2+\text{m}^2-\text{n}^2+\text{n}^2-\text{l}^2)}$
$\Rightarrow\text{f}(\text{x})=\text{x}^0$
$\Rightarrow\text{f}(\text{x})=1$
$\Rightarrow\text{f}\ '(\text{x})=0$

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