Skip to content
NCERT Exemplar · Q76

Q.If f(x)=1−x+x2−x3...−x99+x100f(x) = 1 - x + x^2 - x^3 ... - x^{99} + x^{100}, then f′(1)f'(1) is euqal to
(A) 150150
(B) −50-50
(C) −150-150
(D) 5050

Tripura TbseMCQ· 1mImportance★★★★★est
98% · 171/175 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Recognize the alternating geometric series, differentiate term-by-term, then substitute x=1x = 1 carefully; the derivative at x=1x = 1 is 5050.

Why this approach works

The derivative at a point measures the instantaneous rate of change of the function there. Here we have a polynomial with 101 terms in an alternating pattern. We could differentiate term-by-term and substitute x=1x = 1, but that's tedious. The key insight is to recognize the structure: this is a finite geometric series with alternating signs, which has a closed form. Once we have the closed form, differentiation becomes straightforward.

The series 1−x+x2−x3+…1 - x + x^2 - x^3 + \ldots is geometric with first term a=1a = 1 and common ratio r=−xr = -x.

Step-by-step solution

  1. Identify the series structure The function is

f(x)=1−x+x2−x3+…−x99+x100=∑k=0100(−x)kf(x) = 1 - x + x^2 - x^3 + \ldots - x^{99} + x^{100} = \sum_{k=0}^{100} (-x)^k

This is a finite geometric series with 101 terms, first term 11, and common ratio −x-x.

  1. Apply the geometric series formula For a geometric series ∑k=0nark=a⋅1−rn+11−r\sum_{k=0}^{n} ar^k = a \cdot \frac{1 - r^{n+1}}{1 - r} (when r≠1r \neq 1), we have

f(x)=1−(−x)1011−(−x)=1−(−1)101x1011+x=1+x1011+xf(x) = \frac{1 - (-x)^{101}}{1 - (-x)} = \frac{1 - (-1)^{101} x^{101}}{1 + x} = \frac{1 + x^{101}}{1 + x}

since (−1)101=−1(-1)^{101} = -1.

  1. Differentiate using the quotient rule

    We need f′(x)=ddx(1+x1011+x)f'(x) = \frac{d}{dx}\left(\frac{1 + x^{101}}{1 + x}\right).

    Using the quotient rule (uv)′=u′v−uv′v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}:

f′(x)=(101x100)(1+x)−(1+x101)(1)(1+x)2f'(x) = \frac{(101x^{100})(1 + x) - (1 + x^{101})(1)}{(1 + x)^2}

=101x100+101x101−1−x101(1+x)2= \frac{101x^{100} + 101x^{101} - 1 - x^{101}}{(1 + x)^2}

=101x100+100x101−1(1+x)2= \frac{101x^{100} + 100x^{101} - 1}{(1 + x)^2}

  1. Evaluate at x=1x = 1 …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.