Let \[f(x)=\dfrac{\sin 4\pi \left[ x \right]}{1+{{\left[ x \right]}^{2}}}\] where \[\left[ x \right]\] greatest integer less than or equal to x, then
A. f(x) is not differentiable at some points.
B .f(x) exists but in different from zero
C. LHD(at x=0) = 0 and RHD (at x=1) = 0
Answer
574.8k+ views
Hint: To be able to solve this question you must first know what [x] that is the greatest integer less than or equal to less means. Once you are sure that you understand the logic you can solve this question. You must first try by seeing what values of f(x) will we get for the different values of x we put in the greater integer less than or equal to the bracket. Once we know what values we can get for f(x) we can check with the answers if any answer is True or Not. We after that will also take its differentiation and based on that check the options to find the correct one
Complete step-by-step answer:
Now first we know what f(x) is therefore we can see that
\[f(x)=\dfrac{\sin 4\pi \left[ x \right]}{1+{{\left[ x \right]}^{2}}}\]
Now we know that [x] is the greater integer less than or equal to x as said in the question therefore we can see that whatever input we put on x we will get an integer as the output. Making this easier for us to understand. Therefore we can write the numerator to be
\[\sin 4\pi n=0\] , Since we know that \[\sin \pi n\] is always equal to zero no matter what the n is.
Similarly for denominators we can say and see that \[1+{{\left[ x \right]}^{2}}\] will always be positive. So the denominator too therefore will never be equal to zero.
Now we can also see that since we get f(x) to be zero no matter what therefore the differentiation of f(x) that is f’(x) will also always be zero no matter whatever value of x we check at therefore we can say the left and right hand derivative will be zero.
\[LHD(x=0)=0\]
\[RHD(x=1)=0\]
Hence we can say that the option C is correct
So, the correct answer is “Option C”.
Note: [x] can also be said to be the floor function where the output we get whatever we put is an integer. If we put in decimals it will give the greater integer after it and when you give in an integer as input you get the same integer as output.
Complete step-by-step answer:
Now first we know what f(x) is therefore we can see that
\[f(x)=\dfrac{\sin 4\pi \left[ x \right]}{1+{{\left[ x \right]}^{2}}}\]
Now we know that [x] is the greater integer less than or equal to x as said in the question therefore we can see that whatever input we put on x we will get an integer as the output. Making this easier for us to understand. Therefore we can write the numerator to be
\[\sin 4\pi n=0\] , Since we know that \[\sin \pi n\] is always equal to zero no matter what the n is.
Similarly for denominators we can say and see that \[1+{{\left[ x \right]}^{2}}\] will always be positive. So the denominator too therefore will never be equal to zero.
Now we can also see that since we get f(x) to be zero no matter what therefore the differentiation of f(x) that is f’(x) will also always be zero no matter whatever value of x we check at therefore we can say the left and right hand derivative will be zero.
\[LHD(x=0)=0\]
\[RHD(x=1)=0\]
Hence we can say that the option C is correct
So, the correct answer is “Option C”.
Note: [x] can also be said to be the floor function where the output we get whatever we put is an integer. If we put in decimals it will give the greater integer after it and when you give in an integer as input you get the same integer as output.
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