Verified Solution • 2025–26 Session moderate

QJEEM20Ma1244: JEE Main 2020 Mathematics - Class 12 JEEM20Ma1244 Solution: Let {x} and [x] denote the fractional part of x and the…

Let {x} and [x] denote the fractional part of x and
the greatest integer ≤\le x respectively of a real
number x. If ∫0n{x}dx,∫0n[x]dx\int_0^n {\left\{ x \right\}dx} ,\int_0^n {\left[ x \right]dx} and 10(n2 – n),
\left( {n \in N,n > 1} \right) are three consecutive terms of a G.P., then n is equal to_____.

Step-by-Step Verified Solution

∫0n{x}dx=n∫01xdx=n(x22)01=n2\int\limits_0^n {\left\{ x \right\}} dx = n\int\limits_0^1 x dx = n\left( {{{{x^2}} \over 2}} \right)_0^1 = {n \over 2}

[As period of {x} = 1]

∫0n[x]dx=∫010dx+∫121dx+...+∫n−1n(n−1)dx\int\limits_0^n {\left[ x \right]} dx = \int\limits_0^1 0 dx + \int\limits_1^2 1 dx + ... + \int\limits_{n - 1}^n {\left( {n - 1} \right)} dx

= 1 + 2 + 3 + ....+ (n - 1)

= n(n−1)2{{n\left( {n - 1} \right)} \over 2}

As n2{n \over 2} , n(n−1)2{{n\left( {n - 1} \right)} \over 2} , 10(n2 – n) are in GP.

∴\therefore [n(n−1)2]2=n2×10(n2−n){\left[ {{{n\left( {n - 1} \right)} \over 2}} \right]^2} = {n \over 2} \times 10\left( {{n^2} - n} \right)

⇒\Rightarrow n2 = 21n

⇒\Rightarrow n = 21

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