Question

Let A be the set containing all the integers in the range of

$f(x) = \sin^{-1}(1-2\sqrt{x}) + \tan^{-1}(\frac{\sqrt{2-1}-\sqrt{x}}{1+\sqrt{2x}-\sqrt{x}}).$

Question 48:

The number of relations defined on A, which are symmetric but not reflexive is $\lambda$. The value of $\frac{\lambda}{5}$ is equal to

Answer 1

11.2

Answer 2

Step 1: Determine the domain of the function $f(x)$. The function is $f(x) = \sin^{-1}(1-2\sqrt{x}) + \tan^{-1}(\frac{\sqrt{2}-1-\sqrt{x}}{1+(\sqrt{2}-1)\sqrt{x}})$.

For $\sin^{-1}(u)$ to be defined, $-1 \le u \le 1$. So, we must have: $-1 \le 1-2\sqrt{x} \le 1$ Subtracting 1 from all parts: $-2 \le -2\sqrt{x} \le 0$ Dividing by -2 and reversing inequalities: $0 \le \sqrt{x} \le 1$ Squaring all parts: $0 \le x \le 1$

Also, for $\sqrt{x}$ to be defined, $x \ge 0$. For the argument of $\tan^{-1}$, let $A = \sqrt{2}-1$ and $B = \sqrt{x}$. The argument is $\frac{A-B}{1+AB}$. Since $x \ge 0$, $\sqrt{x} \ge 0$. Since $\sqrt{2}-1 > 0$, the denominator $1+(\sqrt{2}-1)\sqrt{x}$ is always $\ge 1$, so it's never zero. The identity $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$ holds for $A, B > 0$. Here $A = \sqrt{2}-1 > 0$ and $B = \sqrt{x} \ge 0$. So the identity is valid.

Thus, the domain of $f(x)$ is $x \in [0, 1]$.

Step 2: Simplify the function $f(x)$. Using the identity for $\tan^{-1}$: $f(x) = \sin^{-1}(1-2\sqrt{x}) + \tan^{-1}(\sqrt{2}-1) - \tan^{-1}(\sqrt{x})$ We know that $\tan^{-1}(\sqrt{2}-1) = \frac{\pi}{8}$. So, $f(x) = \sin^{-1}(1-2\sqrt{x}) + \frac{\pi}{8} - \tan^{-1}(\sqrt{x})$.

Step 3: Find the range of $f(x)$. Let $t = \sqrt{x}$. Since $x \in [0, 1]$, $t \in [0, 1]$. The function becomes $g(t) = \sin^{-1}(1-2t) - \tan^{-1}(t) + \frac{\pi}{8}$ for $t \in [0, 1]$. Let $h(t) = \sin^{-1}(1-2t) - \tan^{-1}(t)$. We will find the range of $h(t)$ first. To find the range, we can analyze the derivative of $h(t)$: $h'(t) = \frac{d}{dt}(\sin^{-1}(1-2t)) - \frac{d}{dt}(\tan^{-1}(t))$ $h'(t) = \frac{-2}{\sqrt{1-(1-2t)^2}} - \frac{1}{1+t^2}$ $h'(t) = \frac{-2}{\sqrt{1-(1-4t+4t^2)}} - \frac{1}{1+t^2}$ $h'(t) = \frac{-2}{\sqrt{4t-4t^2}} - \frac{1}{1+t^2}$ $h'(t) = \frac{-2}{2\sqrt{t(1-t)}} - \frac{1}{1+t^2}$ $h'(t) = \frac{-1}{\sqrt{t(1-t)}} - \frac{1}{1+t^2}$

For $t \in (0, 1)$, $t(1-t) > 0$, so $\sqrt{t(1-t)} > 0$. Also $1+t^2 > 0$. Therefore, $h'(t)$ is always negative for $t \in (0, 1)$. This means $h(t)$ is a strictly decreasing function on $[0, 1]$. The maximum value of $h(t)$ occurs at $t=0$: $h(0) = \sin^{-1}(1-0) - \tan^{-1}(0) = \sin^{-1}(1) - 0 = \frac{\pi}{2}$. The minimum value of $h(t)$ occurs at $t=1$: $h(1) = \sin^{-1}(1-2) - \tan^{-1}(1) = \sin^{-1}(-1) - \frac{\pi}{4} = -\frac{\pi}{2} - \frac{\pi}{4} = -\frac{3\pi}{4}$. So, the range of $h(t)$ is $[-\frac{3\pi}{4}, \frac{\pi}{2}]$.

Now, the range of $f(x)$ (which is $g(t)$) is $[-\frac{3\pi}{4} + \frac{\pi}{8}, \frac{\pi}{2} + \frac{\pi}{8}]$. Lower bound: $-\frac{3\pi}{4} + \frac{\pi}{8} = -\frac{6\pi}{8} + \frac{\pi}{8} = -\frac{5\pi}{8}$. Upper bound: $\frac{\pi}{2} + \frac{\pi}{8} = \frac{4\pi}{8} + \frac{\pi}{8} = \frac{5\pi}{8}$. So, the range of $f(x)$ is $[-\frac{5\pi}{8}, \frac{5\pi}{8}]$.

Step 4: Determine the set A. Set A contains all integers in the range of $f(x)$. We need to approximate the bounds: $\frac{5\pi}{8} \approx \frac{5 \times 3.14159}{8} \approx \frac{15.70795}{8} \approx 1.963$. So the range is approximately $[-1.963, 1.963]$. The integers in this range are $-1, 0, 1$. Thus, $A = \{-1, 0, 1\}$. The number of elements in set A is $|A| = n = 3$.

Step 5: Calculate the number of symmetric but not reflexive relations on A. Let $n = |A| = 3$. A relation $R$ on $A$ is a subset of $A \times A$. The total number of possible relations is $2^{n^2}$. For $n=3$, $n^2 = 9$. So there are $2^9 = 512$ relations.

A relation $R$ is symmetric if for every $(a, b) \in R$, then $(b, a) \in R$. The diagonal elements $(a, a)$ can be chosen independently. There are $n$ such elements. For off-diagonal elements $(a, b)$ where $a \ne b$, if $(a, b) \in R$, then $(b, a)$ must also be in $R$. There are $n^2 - n$ off-diagonal elements. These can be grouped into $\frac{n^2-n}{2}$ pairs of the form $\{(a, b), (b, a)\}$. For each pair, we can either include both elements in $R$ or exclude both. So there are $2^{\frac{n^2-n}{2}}$ ways to choose the off-diagonal elements. The total number of symmetric relations is $2^n \times 2^{\frac{n^2-n}{2}} = 2^{n + \frac{n^2-n}{2}} = 2^{\frac{n^2+n}{2}}$. For $n=3$, the number of symmetric relations is $2^{\frac{3^2+3}{2}} = 2^{\frac{9+3}{2}} = 2^{\frac{12}{2}} = 2^6 = 64$.

A relation $R$ is reflexive if for every $a \in A$, $(a, a) \in R$. For a relation to be reflexive, all $n$ diagonal elements must be in $R$. The number of symmetric and reflexive relations: Since all $n$ diagonal elements must be present, there is only $1$ way to choose them. The off-diagonal elements still must satisfy the symmetry condition. So there are $2^{\frac{n^2-n}{2}}$ ways to choose them. The number of symmetric and reflexive relations is $1 \times 2^{\frac{n^2-n}{2}}$. For $n=3$, the number of symmetric and reflexive relations is $2^{\frac{3^2-3}{2}} = 2^{\frac{6}{2}} = 2^3 = 8$.

The number of relations defined on A which are symmetric but not reflexive is the total number of symmetric relations minus the number of symmetric and reflexive relations. $\lambda = (\text{Number of symmetric relations}) - (\text{Number of symmetric and reflexive relations})$ $\lambda = 2^{\frac{n^2+n}{2}} - 2^{\frac{n^2-n}{2}}$ For $n=3$: $\lambda = 2^6 - 2^3 = 64 - 8 = 56$.

Step 6: Calculate $\frac{\lambda}{5}$. $\frac{\lambda}{5} = \frac{56}{5} = 11.2$.

The final answer is $\boxed{11.2}$.