JEE Challenger
More from Application of Derivatives

Number of Real Roots of Trigonometric Transcendental Equation

Find the total number of points in the interval (−∞,∞)(-\infty, \infty) that satisfy the equation x2−xsin⁡x−cos⁡x=0x^2 - x \sin x - \cos x = 0.

Options

A

6

B

4

C

2

Correct
D

0

Step-by-Step Solution

To find the total number of real roots of the equation f(x)=x2−xsin⁡x−cos⁡x=0f(x) = x^2 - x \sin x - \cos x = 0 in the interval (−∞,∞)(-\infty, \infty), we analyze the behavior and properties of the function f(x)f(x).

Step 1: Symmetry of the Function

First, we check the symmetry of f(x)f(x): f(−x)=(−x)2−(−x)sin⁡(−x)−cos⁡(−x)=x2−(−x)(−sin⁡x)−cos⁡x=x2−xsin⁡x−cos⁡x=f(x)f(-x) = (-x)^2 - (-x)\sin(-x) - \cos(-x) = x^2 - (-x)(-\sin x) - \cos x = x^2 - x\sin x - \cos x = f(x) Thus, f(x)f(x) is an even function, which means its graph is symmetric with respect to the yy-axis. Any positive root will have a corresponding negative root of the same magnitude.

Step 2: Critical Points and Monotonicity

Differentiating f(x)f(x) with respect to xx: f′(x)=ddx(x2−xsin⁡x−cos⁡x)f'(x) = \frac{d}{dx}\left(x^2 - x \sin x - \cos x\right) f′(x)=2x−(sin⁡x+xcos⁡x)−(−sin⁡x)f'(x) = 2x - (\sin x + x \cos x) - (-\sin x) f′(x)=2x−sin⁡x−xcos⁡x+sin⁡xf'(x) = 2x - \sin x - x \cos x + \sin x f′(x)=2x−xcos⁡x=x(2−cos⁡x)f'(x) = 2x - x \cos x = x(2 - \cos x)

Since −1≤cos⁡x≤1-1 \le \cos x \le 1 for all x∈Rx \in \mathbb{R}, we have: 1≤2−cos⁡x≤31 \le 2 - \cos x \le 3 Thus, (2−cos⁡x)>0(2 - \cos x) > 0 for all x∈Rx \in \mathbb{R}.

This implies that the sign of f′(x)f'(x) is completely determined by xx:

  • For x>0x > 0, f′(x)>0  ⟹  f(x)f'(x) > 0 \implies f(x) is strictly increasing on [0,∞)[0, \infty).
  • For x<0x < 0, f′(x)<0  ⟹  f(x)f'(x) < 0 \implies f(x) is strictly decreasing on (−∞,0](-\infty, 0].
  • At x=0x = 0, f′(0)=0f'(0) = 0, representing a global minimum.

Step 3: Determining the Number of Roots

Evaluating f(x)f(x) at x=0x = 0: f(0)=02−0⋅sin⁡(0)−cos⁡(0)=−1<0f(0) = 0^2 - 0 \cdot \sin(0) - \cos(0) = -1 < 0

Taking the limit as x→∞x \to \infty: lim⁡x→∞f(x)=lim⁡x→∞(x2−xsin⁡x−cos⁡x)=∞(since f(x)≥x2−x−1)\lim_{x \to \infty} f(x) = \lim_{x \to \infty} (x^2 - x\sin x - \cos x) = \infty \quad (\text{since } f(x) \ge x^2 - x - 1)

  • Since f(0)=−1<0f(0) = -1 < 0, lim⁡x→∞f(x)=∞\lim_{x \to \infty} f(x) = \infty, and f(x)f(x) is strictly increasing and continuous on [0,∞)[0, \infty), by the Intermediate Value Theorem, there exists exactly one root in the interval (0,∞)(0, \infty).
  • Since f(x)f(x) is an even function, there exists exactly one root in the interval (−∞,0)(-\infty, 0).
  • At x=0x = 0, f(0)=−1≠0f(0) = -1 \neq 0, so x=0x = 0 is not a root.

Therefore, the total number of points satisfying the equation is: 1+1=21 + 1 = 2

Correct Answer: (C) 2