The correct option is B - II only.
[as per provisional answerkey]Solution
We are given the range: 1/3<x<y<2. We need to check if the statements are always correct.
Evaluating Statement I: x+1/x<y+1/y
Let f(t)=t+1/t. To see if f(x)<f(y) when x<y, we check if the function is increasing.
The derivative f′(t)=1−1/t2.
- If t>1, f′(t)>0 (Function is increasing).
- If t<1, f′(t)<0 (Function is decreasing).
Since our range (1/3 to 2) includes values both below and above 1, the inequality will not always hold.
Example: Let x=1/2 and y=1.
x+1/x=0.5+2=2.5
y+1/y=1+1=2
Here, 2.5<2 is False. Thus, Statement I is not always correct.
Evaluating Statement II: 1+y2/y<1+x2/x
Let g(t)=1+t2/t. We can rewrite this as g(t)=(1+t2)/t2=1/t2+1.
As t increases:
1. t^2 increases.
2. 1/t2 decreases.
3. (1/t2+1) decreases.
4. 1/t2+1 decreases.
Since g(t) is a strictly decreasing function for all t>0, if x<y, then g(x)>g(y).
This matches the statement: g(y)<g(x).
Therefore, Statement II is always correct.
Why the other options are incorrect
- Option (a) - I only: This is incorrect because Statement I fails when x and y are chosen such that x is significantly less than 1 (where the function is decreasing) and y is closer to 1.
- Option (c) - Both I and II: This is incorrect because, as demonstrated above, Statement I is not universally true for the given interval.
- Option (d) - Neither I nor II: This is incorrect because Statement II is mathematically certain due to the strictly decreasing nature of the function 1/t2+1 for positive values.
Key Concept
Monotonicity of functions: A function f(x) is increasing if its derivative is positive, and decreasing if its derivative is negative, determining the direction of inequalities.