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Question: The number of integral values of k for which the equation \(\sin^{-1}x +\tan^{-1}x=2k+1\) has a solu...

The number of integral values of k for which the equation sin⁡−1x+tan⁡−1x=2k+1\sin^{-1}x +\tan^{-1}x=2k+1 has a solution is:
(A) 1
(B) 2
(C) 3
(D) 4

Explanation

Solution

Range of a Function f(x): The range of a given function f is the set of all real values of y that you can get by putting real numbers into x.
Find the range of the function and then equate it to the value of the function, which will give the required values of k.

Complete step-by-step answer:
We are given sin⁡−1x+tan⁡−1x=2k+1\sin^{-1}x +\tan^{-1}x=2k+1.
Let f(x)=sin⁡−1x+tan⁡−1x=2k+1f(x)= \sin^{-1}x +\tan^{-1}x=2k+1
Now, the domain of this function f is [-1, 1].
Putting x = -1, we get,
f(−1)=sin⁡−1(−1)+tan⁡−1(−1)=−π2−π4=−3π4f(-1)=\sin^{-1}(-1)+\tan^{-1}(-1)=-\dfrac{\pi}{2}-\dfrac{\pi}{4}=-\dfrac{3\pi}{4}
Also, putting x = 1, we get,
f(1)=sin⁡−1(1)+tan⁡−1(1)=π2+π4=3π4f(1)=\sin^{-1}(1)+\tan^{-1}(1)=\dfrac{\pi}{2}+\dfrac{\pi}{4}=\dfrac{3\pi}{4}
Thus, the range of f is [−3π4,3π4]\left[\dfrac{-3\pi}{4},\dfrac{3\pi}{4}\right].
Now, according to the given condition 2k+1 must lie in this range. Thus,
−3π4≤2k+1≤3π4\dfrac{-3\pi}{4} \leq 2k+1 \leq \dfrac{3\pi}{4}
Subtracting 1 from all the terms, we get,
−3π4−1≤2k≤3π4−1\dfrac{-3\pi}{4}-1 \leq 2k \leq \dfrac{3\pi}{4}-1
Putting the value of π\pi, i.e. 3.14, we get,
−3×3.144−1≤2k≤3×3.144−1\dfrac{-3 \times 3.14}{4}-1 \leq 2k \leq \dfrac{3 \times 3.14}{4}-1
Simplifying, we get,
−3.35≤2k≤1.35-3.35 \leq 2k \leq 1.35
Dividing all the terms by 2, we get,
−1.67≤k≤0.67-1.67 \leq k \leq 0.67
From this, integral values of k are -1 and 0.
Thus, k can have 2 integral values.
Hence, option (B) is correct.

Note: Domain of sin⁡−1x\sin^{-1}x is [-1, 1] and its range is [−π2,π2]\left[\dfrac{-\pi}{2},\dfrac{\pi}{2}\right]. Remember the domain and range of all trigonometric functions for solving these types of questions .