GATE CSE 2000 | Question: 2.3

Question

Let $S = \sum_{i=3}^{100} i \log_{2} i$, and $T = \int_{2}^{100} x \log_{2}x dx$.

Which of the following statements is true?

• A. $S > T$
• B. $S = T$
• C. $S < T$ and $2S > T$
• D. $2S ≤ T$

Comments

useful resources

https://en.wikipedia.org/wiki/Riemann_sum

https://en.wikipedia.org/wiki/Summation

https://gateoverflow.in/96898/tifr2016-a-5

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https://en.wikipedia.org/wiki/Summation#:~:text=of%20n.-,Approximation,-by%20definite%20integrals

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For an increasing function, the right-endpoint Riemann sum is an overestimate of the definite integral.,

\[
\underbrace{\sum_{i=0}^{n-1} f(x_i)\Delta x}_{L_n}
<
\int_a^b f(x)\,dx
<
\underbrace{\sum_{i=1}^{n} f(x_i)\Delta x}_{R_n}
\]

Since $x\log_2 x$ is increasing function and S represents the Right End point Riemann sum $R_n$

So S > T, Option A is correct.

Answer 1

$x\log_2 x$ is a continuously increasing function, and for a continuously increasing function $f(x)$,

$$\sum_{x=a}^{b} f(x) > \int_a^b f(x)dx$$

But in question, summation of L.H.S. above, $a=3$ and in R.H.S, $a=2$, so we don't know whether $S > T$. So we compute some initial values :

$\sum_{x=3}^{4} x\log_2 x \approx 12.754$, and $\int_2^4 x\log_2 x = 11$

Since $\sum_{x=3}^{4} x\log_2 x > \int_2^4 x\log_2 x$, and since we already know that

$\sum_{x=5}^{100} x\log_2 x > \int_5^{100} x\log_2 x$

So $\sum_{x=3}^{100} x\log_2 x > \int_2^{100} x\log_2 x$

So S > T, and option (A) is correct.

Comments

@ankitgupta.1729 thanx for the reference!

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@Happy Mittal how are you implying the So statment ? Atleast for the right hand side ? how does integration from x=2 to x=4 and from x= 5 to x=100 lead up to the final integration ? why is x= 4 to x=5 missing ?

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Graph by @Happy Mittal is wrong

Graphs by @ANKUSH KUMAR and @ankitgupta.1729 are correct

Answer 2

answer = option A

$x \log_2(x)$ is a continuously increasing function for the interval [2, 100]

We need to compare the numerical values 
S=$\sum_{i=3}^{100} i \log_{2} i$
T=$\int_{2}^{100} x \log_{2}x dx$

initially. 
$\sum_{x=3}^{4} x\log_2 x \approx 12.754$ and
$\int_2^4 x\log_2 x\ dx= 11$

we compute the actual values as:
S

 

T

where it is seen that S > T

Comments

Your answer is not getting loaded properly please check

Answer 3

A simpler way to look at the question could be to note that f(x) is increasing(strictly) in [2,100]. So, as long as we are within the "increasing domain", we can say \[\int_{i}^{i+1} f(x) \, dx \;\lt\; \int_{i}^{i+1} f(i+1) \, dx\]
Note that the RHS is just \[f(i+1)\]
\[\int_{2}^{3} f(x) \, dx \;\lt\;f(3) \,,\quad\int_{3}^{4} f(x) \, dx \;\lt\; f(4) \,,\quad\ldots,\quad\int_{99}^{100} f(x)\, dx \;\lt\; f(100) \, \].
Now, all that is left is to sum up the LHS and RHS to get \[\int_{2}^{100} f(x) \, dx \;\lt\; \sum_{3}^{100} f(k) \,\].
Answer is A.

Answer 4

T=2log₂2+S

T=2+S

i.e. S<T

and

if 2S>T

2S>2+S

S>2  (that is true )

so 2S>T

and answer is C

Comments

You assumed integral is same as summation which is not always true.

Also, your argument for second one is wrong. You assumed "2S > T" and went to get "S > 2" which is true and hence took "2S > T" as true. This is a wrong proof method. If you take an assumption, you should derive a "CONTRADICTION" and then you can say "ASSUMPTION" is false. You cannot derive a "TRUE" from assumption and then say assumption is TRUE.

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Also, here is the proof that S > T

http://www.wolframalpha.com/input/?i=%28sum+xlogx+from+x%3D3+to+100%29+-+%28integrate+xlogx+from+x%3D2+to+100%29+

The difference (S -T) there is approx 229.81