To solve this problem, we need to match each function from List-I with the number of integers in its range from List-II.
Let's analyze each function step-by-step:
### (P) [tex]\( \sin^{-1}x - \cos^{-1}x, x \in [-1,1] \)[/tex]
For [tex]\( \sin^{-1}x \)[/tex] and [tex]\( \cos^{-1}x \)[/tex] within the interval [tex]\( x \in [-1, 1] \)[/tex]:
- The function [tex]\( \sin^{-1} x \)[/tex] (arcsine) has a range of [tex]\([- \frac{\pi}{2}, \frac{\pi}{2}] \)[/tex].
- The function [tex]\( \cos^{-1} x \)[/tex] (arccosine) has a range of [tex]\([0, \pi] \)[/tex].
Thus, the range of [tex]\( \sin^{-1}x - \cos^{-1} x \)[/tex] would be [tex]\([- \pi, \pi]\)[/tex]. The integers in this range are [tex]\(-3, -2, -1, 0, 1, 2\)[/tex], totaling 6 integers.
Therefore, [tex]\( (P) \rightarrow (3) \)[/tex].
### (Q) [tex]\( \sin^{-1} x \cdot \cos^{-1}x, x \in [0,1] \)[/tex]
In [tex]\( [0, 1] \)[/tex]:
- [tex]\(\sin^{-1} x\)[/tex] varies from 0 to [tex]\( \frac{\pi}{2} \)[/tex].
- [tex]\(\cos^{-1} x\)[/tex] varies from [tex]\( \frac{\pi}{2} \)[/tex] to 0.
The product [tex]\( \sin^{-1} x \cdot \cos^{-1} x \)[/tex] can be 0 when either [tex]\( x = 0 \)[/tex] or [tex]\( x = 1 \)[/tex]. Thus, the range contains only integer 0.
Therefore, [tex]\( (Q) \rightarrow (1) \)[/tex].
### (R) [tex]\( \frac{\sin^{-1} x}{\cos^{-1} x}, x \neq 1 \)[/tex]
For [tex]\( x \neq 1 \)[/tex]:
- As [tex]\( x \)[/tex] approaches 1, [tex]\( \sin^{-1} x \)[/tex] approaches [tex]\( \frac{\pi}{2} \)[/tex] and [tex]\( \cos^{-1} x \)[/tex] approaches 0 but remains positive.
Thus the range [tex]\( \frac{\sin^{-1} x}{\cos^{-1} x} \)[/tex] essentially can take any positive value, implying infinitely many values.
Therefore, [tex]\( (R) \rightarrow (4) \)[/tex].
### (S) [tex]\( \frac{\cos^{-1} x}{\sin^{-1} x}, x \neq 0 \)[/tex]
For [tex]\( x \neq 0 \)[/tex]:
- As [tex]\( x \)[/tex] approaches 0, [tex]\( \cos^{-1} x \)[/tex] approaches [tex]\( \frac{\pi}{2} \)[/tex] and [tex]\( \sin^{-1} x \)[/tex] approaches 0 but remains positive.
Thus the range [tex]\( \frac{\cos^{-1} x}{\sin^{-1} x} \)[/tex] can take any positive value and can be infinitely large.
Therefore, [tex]\( (S) \rightarrow (4) \)[/tex].
Combining our results with the correct option:
- [tex]\( (P) \rightarrow (3) \)[/tex]
- [tex]\( (Q) \rightarrow (1) \)[/tex]
- [tex]\( (R) \rightarrow (4) \)[/tex]
- [tex]\( (S) \rightarrow (4) \)[/tex]
Thus, the correct solution is:
C. [tex]\( (P) \rightarrow (3) ;(Q) \rightarrow (1) ;(R) \rightarrow(4) ;(S) \rightarrow (4) \)[/tex]
