BITSAT 2012 Paper

Given,
f (x) =

x100

100

+

x99

99

+ ... +

x2

2

+ x + 1
⇒ f ' (x) =

100x99

100

+

99x98

99

+ ... +

2x

2

+ 1 + 0
⇒ f ' (x) = x99+x98 + ... + x + 1 ... (i)
Putting x = 1, we get
f ' (1) =

(1)99+198+...+1+1

100‌times

=

1+1+1+...+1+1

100‌times

⇒ 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