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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
If $\frac{\text{d}}{\text{dx}}[\text{x}^\text{n}-\text{a}_1\text{x}^{\text{n}-1}+\text{a}_2\text{x}^{\text{n}-2}+...+(-1)^\text{n}\text{a}_\text{n }]\text{e}^\text{x}=\text{x}^\text{n}\ \text{e}^\text{x},$ then the value of $a, 0 < r <$ is equals to :
  • A
    $\frac{\text{n}!}{\text{r}!}$
  • B
    $\frac{(\text{n}-\text{r})!}{\text{r}!}$
  • ✓
    $\frac{\text{n}!}{(\text{n}-\text{r})!}$
  • D
    None of these
✓

Answer

Correct option: C.
$\frac{\text{n}!}{(\text{n}-\text{r})!}$
$\frac{\text{d}}{\text{dx}}[\text{x}^\text{n}-\text{a}_1\text{x}^{\text{n}-1}+\text{a}_2\text{x}^{\text{n}-2}+...+(-1)^\text{n}\text{a}_\text{n }]\text{e}^\text{x}=\text{x}^\text{n}\ \text{e}^\text{x},$
$\Rightarrow\text{e}^\text{x}(\text{nx}^{\text{n-1}}-\text{a}_1(\text{n}-1)\text{x}^{\text{n-2}}+\text{a}_2(\text{n}-2)\text{x}^{\text{n}-3}+...+(-1)^{\text{n}-1}\text{a}_{\text{n}-1}+\text{x}^\text{a}-\text{a}_1\text{x}^{\text{n-2}}+...+(-1)^\text{n}\text{a}_\text{n})=\text{x}^\text{n}\text{e}^\text{x}$
$\Rightarrow\text{e}^\text{x}(\text{x}^\text{n}+(\text{n}-\text{a}_1)\text{x}^{\text{n}-1}-(\text{a}_1(\text{n-1})-\text{a}_2)\text{x}^{\text{n}-2}$
$+(\text{a}_2(\text{n}-2)-\text{a}_3)\text{x}^{\text{n}-3}-...)=\text{x}^\text{n}\text{e}^\text{x}$
on comparing both sides we get
$\text{n}-\text{a}_1=0$
$\Rightarrow\text{a}_1=\text{n}$
$\text{a}_1(\text{n}-1)-\text{a}_2=0$
$\Rightarrow\text{a}_2=\text{a}_1(\text{n}-1)=\text{n}(\text{n}-1)$
$\text{a}_2(\text{n}-2)-\text{a}_3=0$
$\Rightarrow\text{a}_3=\text{a}_2(\text{n}-2)=\text{n}(\text{n}-1)(\text{n}-2)$
So,
$\text{a}_\text{r}=\text{n}(\text{n-1})(\text{n}-2)...(\text{n}-(\text{r}-1)=\frac{\text{n}!}{(\text{n}-\text{r})!}$

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