
The period of \[f\left( x \right) = {10^{\left\{ {{{\cos }^2}\pi x + x - \left\{ x \right\}} \right\}}} + {\sin ^2}\pi x\] is
A. 1
B. 2
C. 3
D. \[\pi \]
Answer
233.1k+ views
Hint: A function's period is the length of time after which it repeats itself or the numbers that may be written as integrals of algebraic differential forms over algebraically specified domains, that is its algebraic geometry.
Formula Used: {x} = x – [x] and f(x) + n = f(x), where, f is a function, x is any real number, {x} is its fractional part, [x] is its greatest integer and n is an integer.
Complete step-by-step solution:
We have the given function as \[f\left( x \right) = {10^{\left\{ {{{\cos }^2}\pi x + x - \left\{ x \right\}} \right\}}} + {\sin ^2}\pi x\] whose period we have to find. We
We can say that the period of \[{10^{\left\{ {{{\cos }^2}\pi x + x - \left\{ x \right\}} \right\}}}\] will be the same as that of the exponential part of 10, that is \[{\cos ^2}\pi x + x - \left\{ x \right\}\], as the exponent just increases the amplitude of the function for different but it repeats itself in the same interval.
We can find the period of exponent as,
\[
{\cos ^2}\pi x + x - \left\{ x \right\} = {\cos ^2}\pi x + x - \left( {x - \left[ x \right]} \right) \\
= {\cos ^2}\pi x + \left[ x \right] \\
= {\cos ^2}\pi x \\
= 1
\]
We neglected the term [x] as it won’t affect the final result of our solution.
Since period of \[{\sin ^2}\pi x\] is same as period of \[{\cos ^2}\pi x\], the period of \[{\sin ^2}\pi x\] is also 1.
So, we can say that the overall period of the function is also 1.
So, option A, the period of \[f\left( x \right) = {10^{\left\{ {{{\cos }^2}\pi x + x - \left\{ x \right\}} \right\}}} + {\sin ^2}\pi x\] is 1, is the required solution.
Additional Information: The variable {x} is the fractional part of any real number x, which is the decimal part of x, and [x] is the greatest integer of x which is the integer that is just less than or equal to x.
Note: Remember that whenever a function is repeating itself by any integer, the period of the function remains unchanged. The period of trigonometric even functions of sin and cos, tan and cot, and, sec and cosec are always the same.
Formula Used: {x} = x – [x] and f(x) + n = f(x), where, f is a function, x is any real number, {x} is its fractional part, [x] is its greatest integer and n is an integer.
Complete step-by-step solution:
We have the given function as \[f\left( x \right) = {10^{\left\{ {{{\cos }^2}\pi x + x - \left\{ x \right\}} \right\}}} + {\sin ^2}\pi x\] whose period we have to find. We
We can say that the period of \[{10^{\left\{ {{{\cos }^2}\pi x + x - \left\{ x \right\}} \right\}}}\] will be the same as that of the exponential part of 10, that is \[{\cos ^2}\pi x + x - \left\{ x \right\}\], as the exponent just increases the amplitude of the function for different but it repeats itself in the same interval.
We can find the period of exponent as,
\[
{\cos ^2}\pi x + x - \left\{ x \right\} = {\cos ^2}\pi x + x - \left( {x - \left[ x \right]} \right) \\
= {\cos ^2}\pi x + \left[ x \right] \\
= {\cos ^2}\pi x \\
= 1
\]
We neglected the term [x] as it won’t affect the final result of our solution.
Since period of \[{\sin ^2}\pi x\] is same as period of \[{\cos ^2}\pi x\], the period of \[{\sin ^2}\pi x\] is also 1.
So, we can say that the overall period of the function is also 1.
So, option A, the period of \[f\left( x \right) = {10^{\left\{ {{{\cos }^2}\pi x + x - \left\{ x \right\}} \right\}}} + {\sin ^2}\pi x\] is 1, is the required solution.
Additional Information: The variable {x} is the fractional part of any real number x, which is the decimal part of x, and [x] is the greatest integer of x which is the integer that is just less than or equal to x.
Note: Remember that whenever a function is repeating itself by any integer, the period of the function remains unchanged. The period of trigonometric even functions of sin and cos, tan and cot, and, sec and cosec are always the same.
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