Question

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Answer 1

Question :–

$\\ \bf \to \: If \: \: f(x) = 1 - x + {x}^{2} - {x}^{3} + ........ - {x}^{99} + {x}^{100} ,then \: \: f'(1) \: \: is \: \: equal \: \: to \: -$

$\\ \bf (A) \: 150$

$\\ \bf (B) \: -50$

$\\ \bf (C) \: -150$

$\\ \bf (D) \: 50$

ANSWER :–

GIVEN :–

• A function $\\ \bf \: \: f(x) = 1 - x + {x}^{2} - {x}^{3} + ........ - {x}^{99} + {x}^{100}$

TO FIND :–

• f'(x) = ?

SOLUTION :–

$\\ \implies \bf f(x) = 1 - x + {x}^{2} - {x}^{3} + ........ - {x}^{99} + {x}^{100}$

• Using identity –

$\\ \implies \bf \dfrac{d({x}^{n})}{dx} = n {x}^{n - 1}$

• Now Differentiate with respect to 'x' –

$\\ \bf \implies f'(x) = \dfrac{d(1)}{dx} - \dfrac{d(x)}{dx} +\dfrac{d( {x}^{2} )}{dx}-\dfrac{d({x}^{3})}{dx}+ ........ - \dfrac{d( {x}^{99} )}{dx}+\dfrac{d( {x}^{100} )}{dx}$

$\\ \bf \implies f'(x) = 0-1+2x-3 {x}^{2} + ........ -99 {x}^{98} +100 {x}^{99}$

• Now put x = 1 –

$\\ \bf \implies f'(1) = -1+2(1)-3 {(1)}^{2} + ........ -99 {(1)}^{98} +100 {(1)}^{99}$

$\\ \bf \implies f'(1) = -1+2-3+ ........ -99+100$

$\\ \bf \implies f'(1) = (-1+2) + (-3+4) + ........ + (-99+100)$

$\\ \bf \implies f'(1) = 1+1 + ........ + 1$

$\\ \bf \: \: \because \: \: only \: \: 50 \: \: terms \: \: remaining.$

• So that –

$\\ \large \implies{ \boxed{ \bf f'(1) =50}}$

• Hence , Option (D) is correct.

Answer 2

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