To find the total number of real roots of the equation
f(x)=x2−xsinx−cosx=0
in the interval (−∞,∞), we analyze the behavior and properties of the function f(x).
Step 1: Symmetry of the Function
First, we check the symmetry of f(x):
f(−x)=(−x)2−(−x)sin(−x)−cos(−x)=x2−(−x)(−sinx)−cosx=x2−xsinx−cosx=f(x)
Thus, f(x) is an even function, which means its graph is symmetric with respect to the y-axis. Any positive root will have a corresponding negative root of the same magnitude.
Step 2: Critical Points and Monotonicity
Differentiating f(x) with respect to x:
f′(x)=dxd(x2−xsinx−cosx)
f′(x)=2x−(sinx+xcosx)−(−sinx)
f′(x)=2x−sinx−xcosx+sinx
f′(x)=2x−xcosx=x(2−cosx)
Since −1≤cosx≤1 for all x∈R, we have:
1≤2−cosx≤3
Thus, (2−cosx)>0 for all x∈R.
This implies that the sign of f′(x) is completely determined by x:
- For x>0, f′(x)>0⟹f(x) is strictly increasing on [0,∞).
- For x<0, f′(x)<0⟹f(x) is strictly decreasing on (−∞,0].
- At x=0, f′(0)=0, representing a global minimum.
Step 3: Determining the Number of Roots
Evaluating f(x) at x=0:
f(0)=02−0⋅sin(0)−cos(0)=−1<0
Taking the limit as x→∞:
limx→∞f(x)=limx→∞(x2−xsinx−cosx)=∞(since f(x)≥x2−x−1)
- Since f(0)=−1<0, limx→∞f(x)=∞, and f(x) is strictly increasing and continuous on [0,∞), by the Intermediate Value Theorem, there exists exactly one root in the interval (0,∞).
- Since f(x) is an even function, there exists exactly one root in the interval (−∞,0).
- At x=0, f(0)=−1=0, so x=0 is not a root.
Therefore, the total number of points satisfying the equation is:
1+1=2
Correct Answer: (C) 2