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Differentiation question

2003 · Shift 0 · Q116

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JEE Main
Mathematics
Differentiation
2003 · Shift 0 · Q116

Differentiation question

2003 · Shift 0 · Q116

JEE MainMathematicsDifferentiationMCQ+4 / −1

If

f\left( x \right) = {x^n},

then the value of

f\left( 1 \right) - {{f'\left( 1 \right)} \over {1!}} + {{f''\left( 1 \right)} \over {2!}} - {{f'''\left( 1 \right)} \over {3!}} + ..........{{{{\left( { - 1} \right)}^n}{f^n}\left( 1 \right)} \over {n!}}

is

A
$1$
B
${{2^n}}$
C
${{2^n} - 1}$
D
$0$

View written solutionFree

Correct answer: D

We are given $f(x)=x^n.$ We need to evaluate $f(1)-\frac{f'(1)}{1!}+\frac{f''(1)}{2!}-\frac{f'''(1)}{3!}+\cdots+\frac{(-1)^n f^{(n)}(1)}{n!}.$
First, find the derivatives of $f(x)=x^n$ .

For $k\le n$ , $f^{(k)}(x)=n(n-1)(n-2)\cdots(n-k+1)x^{n-k}.$ So at $x=1$ , $f^{(k)}(1)=n(n-1)(n-2)\cdots(n-k+1)=\frac{n!}{(n-k)!}.$

Therefore, $\frac{f^{(k)}(1)}{k!}=\frac{n!}{(n-k)!k!}=\binom{n}{k}.$

Substitute into the given expression:

=\sum_{k=0}^{n}(-1)^k\binom{n}{k}.$$ 4. Now use the binomial theorem: $$\sum_{k=0}^{n}\binom{n}{k}(-1)^k=(1-1)^n=0^n=0,$$ for positive integer $n$. Hence the required value is $$0.$$ 5. Checking options: - A: $1$ ❌ - B: $2^n$ ❌ - C: $2^n-1$ ❌ - D: $0$ ✅ So the correct option is **D**.

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