Animated Solution for Mathematics - Trigonometry: Let f:R→R be a function defined by
f(x)={x2sin(x2π)0;if x=0;if x=0
Then which of the following statements is TRUE?
Select Answer:
Visualized Solution
Analyzing the Function
We need to find the roots of f(x)=0.
Equating to Zero
For x>0, set x2sin(x2π)=0.
Since x=0, sin(x2π)=0.
General Solution of Sine
sin(θ)=0⟹θ=nπ
x2π=nπ
Finding the Roots
x2=n1⟹xn=n1
Since x>0, n∈{1,2,3,…}.
Checking Option A
Interval (10101,∞)
We need xn>10101⟹n1>10101
Evaluating Option A
n<1010⟹n<1020
The number of such integers is finite. Option A is FALSE.
Checking Option B
Interval [π1,∞)
We need xn≥π1⟹n1≥π1
Evaluating Option B
n≤π⟹n≤π2≈9.86
n∈{1,2,…,9}. Only 9 solutions exist. Option B is FALSE.
Checking Option C
Interval (0,10101)
We need 0<xn<10101⟹n>1020
Evaluating Option C
There are infinitely many integers n>1020. Option C is FALSE.
Checking Option D
Interval (π21,π1)
We need π21<n1<π1
Solving the Inequality
Take reciprocals: π<n<π2
Square all terms: π2<n<π4
Approximating Pi Powers
π2≈9.8696
π4≈(9.8696)2≈97.4
So, 9.86<n<97.4
Counting the Solutions
Valid integers: n∈{10,11,…,97}
Number of solutions =97−10+1=88
Final Conclusion
88 solutions exist, which is >25. Option D is TRUE.
00:00 / 00:00
The Sigma Insight: General Solution of Trigonometric Equations
Solution Diagram
Analyzing the Setup
The function under investigation is defined as:
f(x)={x2sin(x2π)0xeq0x=0
This function exhibits rapid oscillations as x approaches the origin. To analyze its behavior, we must determine the points where the function vanishes.
The Hunt for the Roots
To find the roots of f(x)=0 for x>0, we solve the equation:
x2sin(x2π)=0
Since $x^2
eq 0$ in this domain, the condition simplifies to sin(x2π)=0. Recalling the general solution for the sine function, we set the argument to an integer multiple of π:
x2π=nπ,n∈{1,2,3,…}
Canceling π and rearranging for x, we obtain the roots:
xn=n1
The Inequality Dance
We evaluate the interval (π21,π1) to determine the number of roots contained within it. We set up the following inequality:
π21<n1<π1
Taking the reciprocal of all parts reverses the inequality signs:
π<n<π2
Squaring the terms yields the range for the integer n:
π2<n<π4
Given the approximations π2≈9.87 and π4≈97.4, we seek integers n such that 9.87<n<97.4. This implies n∈{10,11,…,97}.
The total number of roots is calculated as 97−10+1=88. Since 88>25, the statement is true.
Why the Others Fail
For Option A, the interval (10101,∞) implies n1>10101, or n<1020. Because this set of integers is bounded, there are not infinitely many solutions.
For Option B, the interval [π1,∞) implies n1≥π1, leading to n≤π2≈9.86. There are exactly 9 solutions, contradicting the claim that there are no solutions.
Finally, for Option C, the interval (0,10101) requires n1<10101, or n>1020. Since there are infinitely many such integers, the set of solutions is infinite, rendering the claim of a finite set false.