Let f be any function defined on R and let it satisfy the conditions:

|f(x)f(y)||(xy)2|,(x,y)R

If f (0) = 1, then

Option 1 - <p>f (x) &gt; 0,&nbsp;<math><mrow><mo>∀</mo><mtext> </mtext><mtext> </mtext><mi>x</mi><mo>∈</mo></mrow></math>&nbsp;R</p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>0</mn> <mo>,</mo> <mo>∀</mo> <mtext> </mtext> <mtext> </mtext> <mi>x</mi> <mo>∈</mo> <mi>R</mi> </mrow> </math> </span></p>
Option 3 - <p>f (x) can take any value in R&nbsp;</p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>&lt;</mo> <mn>0</mn> <mo>,</mo> <mo>∀</mo> <mtext> </mtext> <mtext> </mtext> <mi>x</mi> <mo>∈</mo> <mi>R</mi> </mrow> </math> </span></p>
2 Views|Posted 7 months ago
Asked by Shiksha User
1 Answer
P
7 months ago
Correct Option - 1
Detailed Solution:

|f (x)f (y)|| (xy)2|

|f (x)f (y)xy|xy

Taking the limit y x on both sides

Ltyx|f (x)f (y)xy|Ltyx (xy)

|f' (x)|0

Hence, modulus cannot be zero. Hence f' (x) = 0. Integrating, we get f (x) = c

at x = 0, f (0) = c = 1

f (x)=1>0, xR

Hence, option (A) is correct option.

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