Solveeit Logo
Login

Question

Question: The range of the function $f(x) = \sin^{-1}[x^2+\frac{1}{2}] + \cos^{-1}[x^2-\frac{1}{2}]$, where [....

The range of the function f(x)=sin⁡−1[x2+12]+cos⁡−1[x2−12]f(x) = \sin^{-1}[x^2+\frac{1}{2}] + \cos^{-1}[x^2-\frac{1}{2}], where [.] is the greatest integer function, is

A

{π2,π}\{\frac{\pi}{2}, \pi\}

B

{0,−12}\{0, -\frac{1}{2}\}

C

{π}\{\pi\}

D

{0,π2}\{0, \frac{\pi}{2}\}

Answer

{π}

Explanation

Solution

To find the range of the function f(x)=sin⁡−1[x2+12]+cos⁡−1[x2−12]f(x) = \sin^{-1}[x^2+\frac{1}{2}] + \cos^{-1}[x^2-\frac{1}{2}], we first need to determine the domain of the function.

Let u=[x2+12]u = [x^2+\frac{1}{2}] and v=[x2−12]v = [x^2-\frac{1}{2}]. For sin⁡−1(u)\sin^{-1}(u) and cos⁡−1(v)\cos^{-1}(v) to be defined, their arguments uu and vv must lie in the interval [−1,1][-1, 1]. Since uu and vv are outputs of the greatest integer function, they must be integers. So, u∈{−1,0,1}u \in \{-1, 0, 1\} and v∈{−1,0,1}v \in \{-1, 0, 1\}.

Also, we know that x2≥0x^2 \ge 0.

Let's analyze u=[x2+12]u = [x^2+\frac{1}{2}]: Since x2≥0x^2 \ge 0, we have x2+12≥12x^2+\frac{1}{2} \ge \frac{1}{2}. Therefore, [x2+12]≥[12]=0[x^2+\frac{1}{2}] \ge [\frac{1}{2}] = 0. Combining this with u∈{−1,0,1}u \in \{-1, 0, 1\}, we find that uu can only be 00 or 11.

Let's analyze v=[x2−12]v = [x^2-\frac{1}{2}]: Since x2≥0x^2 \ge 0, we have x2−12≥−12x^2-\frac{1}{2} \ge -\frac{1}{2}. Therefore, [x2−12]≥[−12]=−1[x^2-\frac{1}{2}] \ge [-\frac{1}{2}] = -1. Combining this with v∈{−1,0,1}v \in \{-1, 0, 1\}, we find that vv can be −1-1, 00, or 11.

Now, let's consider the possible values for x2x^2 that satisfy these conditions.

Case 1: u=[x2+12]=0u = [x^2+\frac{1}{2}] = 0 This implies 0≤x2+12<10 \le x^2+\frac{1}{2} < 1. Subtracting 12\frac{1}{2} from all parts, we get −12≤x2<12-\frac{1}{2} \le x^2 < \frac{1}{2}. Since x2≥0x^2 \ge 0, the valid range for x2x^2 in this case is 0≤x2<120 \le x^2 < \frac{1}{2}.

For this range of x2x^2, let's find v=[x2−12]v = [x^2-\frac{1}{2}]: If 0≤x2<120 \le x^2 < \frac{1}{2}, then −12≤x2−12<0-\frac{1}{2} \le x^2-\frac{1}{2} < 0. So, v=[x2−12]=−1v = [x^2-\frac{1}{2}] = -1. This value v=−1v=-1 is valid for cos⁡−1\cos^{-1}. In this case, f(x)=sin⁡−1(0)+cos⁡−1(−1)=0+π=πf(x) = \sin^{-1}(0) + \cos^{-1}(-1) = 0 + \pi = \pi.

Case 2: u=[x2+12]=1u = [x^2+\frac{1}{2}] = 1 This implies 1≤x2+12<21 \le x^2+\frac{1}{2} < 2. Subtracting 12\frac{1}{2} from all parts, we get 12≤x2<32\frac{1}{2} \le x^2 < \frac{3}{2}.

For this range of x2x^2, let's find v=[x2−12]v = [x^2-\frac{1}{2}]: If 12≤x2<32\frac{1}{2} \le x^2 < \frac{3}{2}, then 0≤x2−12<10 \le x^2-\frac{1}{2} < 1. So, v=[x2−12]=0v = [x^2-\frac{1}{2}] = 0. This value v=0v=0 is valid for cos⁡−1\cos^{-1}. In this case, f(x)=sin⁡−1(1)+cos⁡−1(0)=π2+π2=πf(x) = \sin^{-1}(1) + \cos^{-1}(0) = \frac{\pi}{2} + \frac{\pi}{2} = \pi.

Consider if x2≥32x^2 \ge \frac{3}{2}: If x2≥32x^2 \ge \frac{3}{2}, then x2+12≥32+12=2x^2+\frac{1}{2} \ge \frac{3}{2}+\frac{1}{2} = 2. So, [x2+12]≥2[x^2+\frac{1}{2}] \ge 2. However, the argument of sin⁡−1\sin^{-1} must be less than or equal to 11. Thus, x2≥32x^2 \ge \frac{3}{2} is not possible for the function to be defined.

Combining the possible ranges for x2x^2: The domain of the function f(x)f(x) is 0≤x2<120 \le x^2 < \frac{1}{2} (from Case 1) combined with 12≤x2<32\frac{1}{2} \le x^2 < \frac{3}{2} (from Case 2). This means the function is defined for 0≤x2<320 \le x^2 < \frac{3}{2}. For all values of xx such that 0≤x2<320 \le x^2 < \frac{3}{2}, the function f(x)f(x) consistently evaluates to π\pi.

Therefore, the range of the function f(x)f(x) is {π}\{\pi\}.