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

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

@Happy Mittal I think your intuition works only for left Riemann sums(only top left corners of rectangles touch the curve). If we use a right Riemann sum (only top right corners of rectangles touch the curve), each rectangle will have some area which is above the curve (only for continuously increasing functions). Thus the sum of the areas of rectangles in right Riemann sum will be greater than area under the curve. Please correct me if I am wrong. Link which explains left and right Riemann sums https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-2/a/left-and-right-riemann-sums

---

Summation takes discrete values and integration takes continuous values .

So, I think this graph need a modification

 

Curve must intersect the discrete values

---

@ankitgupta.1729 thanx for the reference!

Answer 2

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.

---

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

Answer 3

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