JEE MAIN - Mathematics (2020 - 4th September Evening Slot - No. 13)
Let {x} and [x] denote the fractional part of x and
the greatest integer $$ \le $$ x respectively of a real
number x. If $$\int_0^n {\left\{ x \right\}dx} ,\int_0^n {\left[ x \right]dx} $$ and 10(n2 – n),
$$\left( {n \in N,n > 1} \right)$$ are three consecutive terms of a G.P., then n is equal to_____.
the greatest integer $$ \le $$ x respectively of a real
number x. If $$\int_0^n {\left\{ x \right\}dx} ,\int_0^n {\left[ x \right]dx} $$ and 10(n2 – n),
$$\left( {n \in N,n > 1} \right)$$ are three consecutive terms of a G.P., then n is equal to_____.
Answer
21
Explanation
$$\int\limits_0^n {\left\{ x \right\}} dx = n\int\limits_0^1 x dx = n\left( {{{{x^2}} \over 2}} \right)_0^1 = {n \over 2}$$
[As period of {x} = 1]
$$\int\limits_0^n {\left[ x \right]} dx = \int\limits_0^1 0 dx + \int\limits_1^2 1 dx + ... + \int\limits_{n - 1}^n {\left( {n - 1} \right)} dx$$
= 1 + 2 + 3 + ....+ (n - 1)
= $${{n\left( {n - 1} \right)} \over 2}$$
As $${n \over 2}$$, $${{n\left( {n - 1} \right)} \over 2}$$, 10(n2 – n) are in GP.
$$ \therefore $$ $${\left[ {{{n\left( {n - 1} \right)} \over 2}} \right]^2} = {n \over 2} \times 10\left( {{n^2} - n} \right)$$
$$ \Rightarrow $$ n2 = 21n
$$ \Rightarrow $$ n = 21
[As period of {x} = 1]
$$\int\limits_0^n {\left[ x \right]} dx = \int\limits_0^1 0 dx + \int\limits_1^2 1 dx + ... + \int\limits_{n - 1}^n {\left( {n - 1} \right)} dx$$
= 1 + 2 + 3 + ....+ (n - 1)
= $${{n\left( {n - 1} \right)} \over 2}$$
As $${n \over 2}$$, $${{n\left( {n - 1} \right)} \over 2}$$, 10(n2 – n) are in GP.
$$ \therefore $$ $${\left[ {{{n\left( {n - 1} \right)} \over 2}} \right]^2} = {n \over 2} \times 10\left( {{n^2} - n} \right)$$
$$ \Rightarrow $$ n2 = 21n
$$ \Rightarrow $$ n = 21
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