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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY1 Mark
MCQ
If $\text{y}=\frac{1}{1+\text{x}^{\text{a}-\text{b}}+\text{x}^{\text{c}-\text{b}}}+\frac{1}{1+\text{x}^{\text{b}-\text{c}}+\text{x}^{\text{a}-\text{c}}}+\frac{1}{1+\text{x}^{\text{b}-\text{a}}+\text{x}^{\text{c}-\text{a}}},$ then $\frac{\text{dy}}{\text{dx}}$ is equal to:
  • A
    1
  • B
    $(\text{a}+\text{b}+\text{c})^{\text{x}^{\text{a}+\text{b}+\text{c}-1}}$
  • ✓
    0
  • D
    None of these.
✓

Answer

Correct option: C.
0
$\text{y}=\frac{1}{1+\text{x}^{\text{a}-\text{b}}+\text{x}^{\text{c}-\text{b}}}+\frac{1}{1+\text{x}^{\text{b}-\text{c}}+\text{x}^{\text{a}-\text{c}}}+\frac{1}{1+\text{x}^{\text{b}-\text{a}}+\text{x}^{\text{c}-\text{a}}}$
$\text{y}=\frac{1}{1+\frac{\text{x}^\text{a}}{\text{x}^\text{b}}+\frac{\text{x}^\text{c}}{\text{x}^\text{b}}}+\frac{1}{1+\frac{\text{x}^\text{b}}{\text{x}^\text{c}}+\frac{\text{x}^\text{a}}{\text{x}^\text{c}}}+\frac{1}{1+\frac{\text{x}^\text{b}}{\text{x}^\text{a}}+\frac{\text{x}^\text{c}}{\text{x}^\text{a}}}$
$\text{y}=\frac{\text{x}^\text{b}}{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}+\frac{\text{x}^\text{c}}{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}+\frac{\text{x}^\text{a}}{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}$
$\text{y}=\frac{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}{\text{x}^\text{a}+\text{x}^\text{b}+\text{x}^\text{c}}$
$\text{y}=1$
$\frac{\text{dy}}{\text{dx}}=0$

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