Question

If $\left[ x \right]$ donates the integral part of x for the real value of x, then the value of
$\left[ \dfrac{1}{4} \right]+\left[ \dfrac{1}{4}+\dfrac{1}{200} \right]+\left[ \dfrac{1}{4}+\dfrac{1}{100} \right]+\left[ \dfrac{1}{4}+\dfrac{3}{200} \right]+......................+\left[ \dfrac{1}{4}+\dfrac{199}{200} \right]$ is $10a$. Find a.

Answer 1

Hint: $\left[ x \right]$ donates the integral part x for real values of x mean that if $x=n+f$ where f is the fractional part and n is the integral part.
$\left[ x \right]=n$
For example $\left[ 1.5 \right]=1$
For $0\le x<1\text{ }\left[ x \right]=0$
.And \[1\le x<2\text{ }\left[ x \right]=1\text{ }............\text{ and so on }......\].

Complete step-by-step answer:
\[\left[ \dfrac{1}{4} \right]+\left[ \dfrac{1}{4}+\dfrac{1}{200} \right].................\left[ \dfrac{1}{4}+\dfrac{199}{200} \right]=10a\]now we have,
\[\left[ \dfrac{1}{4}+\dfrac{3}{4} \right]=1\Rightarrow \left[ \dfrac{1}{4}+\dfrac{150}{200} \right]=1\]
Similarly, \[\left[ \dfrac{1}{4}+\dfrac{151}{200} \right]=\left[ 1.005 \right]=1\]
So we can write nth term of the series in the form \[\left[ \dfrac{1}{4}+\dfrac{m}{200} \right]\text{ 0}\le \text{m}\le 199\text{ m}\varepsilon \text{N}\]
Here, when \[m<150\]
\[\left[ \dfrac{1}{4}+\dfrac{m}{200} \right]\text{ = 0}\] because \[\dfrac{1}{4}+\dfrac{m}{200}<1\]
When \[m>150\]
\[\left[ \dfrac{1}{4}+\dfrac{m}{200} \right]\text{ = 1}\] as \[1\le \dfrac{1}{4}+\dfrac{M}{200}<2\]
So $\left[ \dfrac{1}{4} \right]+\left[ \dfrac{1}{4}+\dfrac{1}{200} \right]+..............\left[ \dfrac{1}{4}+\dfrac{150}{200} \right]+\left[ \dfrac{1}{4}+\dfrac{151}{200} \right]+..........+\left[ \dfrac{1}{4}+\dfrac{199}{200} \right]\Rightarrow $
$=0+0+0.............+1+1+........1$
$=50$ as then will be $50$term form $\left[ \dfrac{1}{4}+\dfrac{150}{200} \right]$ to $\left[ \dfrac{1}{4}+\dfrac{191}{200} \right]$
But it is given that the sum of above term is $10a$
$10a=50$
$\Rightarrow a=5$

Note: Thus one certain properties of $\left[ x \right]$where it donates the integral part of x
$\left[ {{x}_{1}} \right]+\left[ {{x}_{2}} \right]\ne \left[ {{x}_{1}}+{{x}_{2}} \right]$
For example: ${{x}_{1}}=1.5$
${{x}_{2}}=2.5$
$\left[ {{x}_{1}} \right]=\left[ 1.5 \right]=1$
$\left[ {{x}_{2}} \right]=\left[ 2.5 \right]=2$
$\left[ {{x}_{1}}+{{x}_{2}} \right]=3$
But $\left[ {{x}_{1}}+{{x}_{2}} \right]=\left[ 1.5+2.5 \right]=4$
Hence it does not follow $f(a)+f(6)=f(a+6)$
$\therefore $ it is not a function.