Test Index

BITSAT 2012 Paper

Section: Mathematics

Question : 122 of 150

Marks: +1, -0

\frac{x^{100}}{100} + \frac{x^{99}}{99}

\frac{x^2}{2}

f '(1) = mf ' (0), where m is equal to

Solution:

Given,

f (x) =

\frac{x^{100}}{100} + \frac{x^{99}}{99}

+ ... +

\frac{x^2}{2}

+ x + 1

⇒ f ' (x) =

\frac{100x^{99}}{100} + \frac{99x^{98}}{99}

+ ... +

\frac{2x}{2}

+ 1 + 0

⇒ f ' (x) =

x^{99}+x^{98}

+ ... + x + 1 ... (i)

Putting x = 1, we get

f ' (1) =

\underset{100 \times}{(1)^{99} + 1^{98} + \dots + 1 + 1}

\underset{100 \times}{1+1+1+\dots+1+1}

⇒ 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) = 100f ' (0)

Hence, m = 100

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