For the function f(x)= x^ 100 100 + x^ 99 99 + x^ 2 2 +x+1 ^ (1)=mf^ (0), where m is equal to

Correct answer: C. 100 Given, f(x)= x^ 100 100 + x^ 99 99 + + x^ 2 2 +x+1 f^ (x)= 100 x^ 99 100 + 99 x^ 98 99 + +(2 x)/(2)+1+0[ f(x)=x^ n f^ (x)=nx^ n-1 ] f^ (x)=x^ 99 +x^ 98 + +x+1 ...(i) Putting x=1 , we get f^ (1)= (1)^ 99 +1^ 98 + +1+1 _ 100 times = 1+1+1 +1+1 _ 100 times f^ (1)=100 ...(ii) Again, putting x=0 , we get f^ (0)=0+0+ +0+1 f^ (0)=1 ...(iii) From eqs. (ii) and(iii), we get; f^ (1)=100 f^ (0) Hence, m=100

BITSAT 2012

Exam: BITSAT
Year: 2012
Subject: Mathematics
Chapter: Differentiation

BITSAT 2012 · Mathematics · Differentiation

For the function f(x)= x^ 100 100 + x^ 99 99 + x^ 2 2 +x+1 ^...

For the function

f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots \frac{x^{2}}{2}+x+1\mathrmf^{\prime}(1)=\mathrm{mf}^{\prime}(0),

where

\mathrm{m}

is equal to

Options

A50
B0
C100
D200

Correct answer

C. 100

Step-by-step solution

Given,

f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots+\frac{x^{2}}{2}+x+1\Rightarrow \quad f^{\prime}(x)=\frac{100 x^{99}}{100}+\frac{99 x^{98}}{99}+\ldots+\frac{2 x}{2}+1+0\left[\because \mathrm{f}(\mathrm{x})=\mathrm{x}^{\mathrm{n}} \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{nx}^{\mathrm{n}-1}\right]\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^{99}+\mathrm{x}^{98}+\ldots+\mathrm{x}+1

...(i) Putting

\mathrm{x}=1

, we get

f^{\prime}(1)=\underbrace{(1)^{99}+1^{98}+\ldots+1+1}_{100 \text{ times }}=\underbrace{1+1+1 \ldots+1+1}_{100 \text{ times }}\Rightarrow \quad \mathrm{f}^{\prime}(1)=100

...(ii) Again, putting

\mathrm{x}=0

, we get

\mathrm{f}^{\prime}(0)=0+0+\ldots+0+1 \quad \Rightarrow \mathrm{f}^{\prime}(0)=1

...(iii) From eqs. (ii) and(iii), we get;

\mathrm{f}^{\prime}(1)=100 \mathrm{f}^{\prime}(0)

Hence,

m=100

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