Question

Let $f(x) = \frac{1}{\pi}(sin^{-1}x + cos^{-1}x + tan^{-1}x) + \frac{(x+1)}{x^2+2x+10}$, if the absolute maximum value of the integral part of $\frac{1}{M}$ is ______.

The number of rational numbers in the domain of function

$y = sin^{-1}x + cos^{-1}x + tan^{-1}x + cosec^{-1}x + sec^{-1}x + cot^{-1}x$, is ______

If $sin^{-1}p + sin^{-1}q + sin^{-1}r = \frac{3\pi}{2}$, then $A = p^2 + q^2 + r^2$. If $(sin^{-1}x)^2 + (sin^{-1}y)^2 + (sin^{-1}z)^2 = \frac{3\pi^2}{4}$, then

$B = |(x+y+z)_{min}|$, then the value of (A + B) is ______.

Answer 1

S9: 1 S10: 2 S11: 6

Answer 2

Solution for S9:

The given function is $f(x) = \frac{1}{\pi}(\sin^{-1}x + \cos^{-1}x + \tan^{-1}x) + \frac{(x+1)}{x^2+2x+10}$.

1. Determine the domain of $f(x)$:
   The terms $\sin^{-1}x$ and $\cos^{-1}x$ are defined for $x \in [-1, 1]$.
   The term $\tan^{-1}x$ is defined for $x \in (-\infty, \infty)$.
   The term $\frac{(x+1)}{x^2+2x+10}$ is defined for all $x$ since the denominator $x^2+2x+10 = (x+1)^2+9$ is always positive.
   Therefore, the domain of $f(x)$ is the intersection of these domains, which is $x \in [-1, 1]$.

2. Simplify $f(x)$ using identities:
   We know that $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for $x \in [-1, 1]$.
   So, $f(x) = \frac{1}{\pi}\left(\frac{\pi}{2} + \tan^{-1}x\right) + \frac{(x+1)}{(x+1)^2+9}$.
   $f(x) = \frac{1}{2} + \frac{1}{\pi}\tan^{-1}x + \frac{x+1}{(x+1)^2+9}$.

3. Find the maximum value of $f(x)$ (denoted as $M$):
   To find the maximum value of $f(x)$ on $x \in [-1, 1]$, we analyze each component:

   • Let $g(x) = \frac{1}{\pi}\tan^{-1}x$. Since $\tan^{-1}x$ is an increasing function, its maximum value on $[-1, 1]$ occurs at $x=1$.
     $g(1) = \frac{1}{\pi}\tan^{-1}(1) = \frac{1}{\pi} \cdot \frac{\pi}{4} = \frac{1}{4}$.
   • Let $h(x) = \frac{x+1}{(x+1)^2+9}$. Let $u = x+1$. Since $x \in [-1, 1]$, $u \in [0, 2]$.
     So we need to find the maximum of $k(u) = \frac{u}{u^2+9}$ for $u \in [0, 2]$.
     To find the maximum, we compute the derivative $k'(u)$:
     $k'(u) = \frac{1 \cdot (u^2+9) - u \cdot (2u)}{(u^2+9)^2} = \frac{9-u^2}{(u^2+9)^2}$.
     For $u \in [0, 2]$, $u^2 \in [0, 4]$, so $9-u^2$ is always positive. Thus, $k'(u) > 0$ for $u \in [0, 2]$, meaning $k(u)$ is an increasing function on this interval.
     The maximum value of $k(u)$ occurs at $u=2$.
     $k(2) = \frac{2}{2^2+9} = \frac{2}{4+9} = \frac{2}{13}$.
     This corresponds to $x+1=2 \Rightarrow x=1$.

   Since both $g(x)$ and $h(x)$ attain their maximum values at $x=1$, the maximum value of $f(x)$ is:
   $M = f(1) = \frac{1}{2} + g(1) + h(1) = \frac{1}{2} + \frac{1}{4} + \frac{2}{13}$.
   $M = \frac{26}{52} + \frac{13}{52} + \frac{8}{52} = \frac{26+13+8}{52} = \frac{47}{52}$.

4. Calculate the integral part of $\frac{1}{M}$:
   $\frac{1}{M} = \frac{1}{\frac{47}{52}} = \frac{52}{47}$.
   $\frac{52}{47} = 1 + \frac{5}{47} \approx 1.106$.
   The integral part of $\frac{1}{M}$ is $\lfloor \frac{52}{47} \rfloor = 1$.

The absolute maximum value of the integral part of $\frac{1}{M}$ is 1.

Solution for S10:

The given function is $y = \sin^{-1}x + \cos^{-1}x + \tan^{-1}x + \csc^{-1}x + \sec^{-1}x + \cot^{-1}x$.

1. Determine the domain of each inverse trigonometric function:

   • Domain($\sin^{-1}x$): $x \in [-1, 1]$
   • Domain($\cos^{-1}x$): $x \in [-1, 1]$
   • Domain($\tan^{-1}x$): $x \in (-\infty, \infty)$
   • Domain($\csc^{-1}x$): $x \in (-\infty, -1] \cup [1, \infty)$
   • Domain($\sec^{-1}x$): $x \in (-\infty, -1] \cup [1, \infty)$
   • Domain($\cot^{-1}x$): $x \in (-\infty, \infty)$

2. Find the domain of $y$ by intersecting all individual domains:
   Domain($y$) = Domain($\sin^{-1}x$) $\cap$ Domain($\cos^{-1}x$) $\cap$ Domain($\tan^{-1}x$) $\cap$ Domain($\csc^{-1}x$) $\cap$ Domain($\sec^{-1}x$) $\cap$ Domain($\cot^{-1}x$).

   • $([-1, 1]) \cap ([-1, 1]) = [-1, 1]$
   • $([-1, 1]) \cap ((-\infty, \infty)) = [-1, 1]$
   • $([-1, 1]) \cap ((-\infty, -1] \cup [1, \infty))$: This intersection yields only the values $x=-1$ and $x=1$.
     If $x \in [-1, 1]$ and $x \le -1$, then $x=-1$.
     If $x \in [-1, 1]$ and $x \ge 1$, then $x=1$.
     So, the intersection is $\{-1, 1\}$.
   • $\{-1, 1\} \cap ((-\infty, \infty)) = \{-1, 1\}$.

   Thus, the domain of the function $y$ is $\{-1, 1\}$.

3. Count the number of rational numbers in the domain:
   The numbers in the domain are $-1$ and $1$. Both $-1$ and $1$ are rational numbers.
   Therefore, there are 2 rational numbers in the domain of the function.

The number of rational numbers in the domain of the function is 2.

Solution for S11:

Part 1: Find A
Given $\sin^{-1}p + \sin^{-1}q + \sin^{-1}r = \frac{3\pi}{2}$.
The range of $\sin^{-1}x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
The maximum value of each $\sin^{-1}$ term is $\frac{\pi}{2}$.
For the sum of three such terms to be $\frac{3\pi}{2}$, each term must individually attain its maximum value.
So, $\sin^{-1}p = \frac{\pi}{2}$, $\sin^{-1}q = \frac{\pi}{2}$, and $\sin^{-1}r = \frac{\pi}{2}$.
This implies $p = \sin(\frac{\pi}{2}) = 1$.
Similarly, $q = 1$ and $r = 1$.
Then $A = p^2 + q^2 + r^2 = 1^2 + 1^2 + 1^2 = 1+1+1 = 3$.

Part 2: Find B
Given $(\sin^{-1}x)^2 + (\sin^{-1}y)^2 + (\sin^{-1}z)^2 = \frac{3\pi^2}{4}$.
Let $a = \sin^{-1}x$, $b = \sin^{-1}y$, $c = \sin^{-1}z$.
Since the range of $\sin^{-1}t$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, the range of $(\sin^{-1}t)^2$ is $[0, (\frac{\pi}{2})^2] = [0, \frac{\pi^2}{4}]$.
So, $a^2 \in [0, \frac{\pi^2}{4}]$, $b^2 \in [0, \frac{\pi^2}{4}]$, $c^2 \in [0, \frac{\pi^2}{4}]$.
For their sum $a^2 + b^2 + c^2$ to be $\frac{3\pi^2}{4}$, each term must individually attain its maximum value.
So, $a^2 = \frac{\pi^2}{4}$, $b^2 = \frac{\pi^2}{4}$, $c^2 = \frac{\pi^2}{4}$.
This means:
$\sin^{-1}x = \pm \frac{\pi}{2} \Rightarrow x = \sin(\pm \frac{\pi}{2}) \Rightarrow x = \pm 1$.
$\sin^{-1}y = \pm \frac{\pi}{2} \Rightarrow y = \sin(\pm \frac{\pi}{2}) \Rightarrow y = \pm 1$.
$\sin^{-1}z = \pm \frac{\pi}{2} \Rightarrow z = \sin(\pm \frac{\pi}{2}) \Rightarrow z = \pm 1$.
So, $x, y, z \in \{-1, 1\}$.

We need to find $B = |(x+y+z)_{min}|$.
To minimize the sum $(x+y+z)$, we must choose each variable to be its minimum possible value.
So, $(x+y+z)_{min} = (-1) + (-1) + (-1) = -3$.
Then $B = |(x+y+z)_{min}| = |-3| = 3$.

Part 3: Find (A + B)
$A = 3$ and $B = 3$.
$A+B = 3+3 = 6$.

The value of (A + B) is 6.

The question asks for the answer to S9, S10, and S11.
S9 answer: 1
S10 answer: 2
S11 answer: 6

The format of the question implies a single answer for each S. I will provide the final answers for each as requested.

Final Answer:
S9: 1
S10: 2
S11: 6

The question asks for "the absolute maximum value of the integral part of $\frac{1}{M}$ is ______." This is a fill-in-the-blank type question.
The final answer for S9 is 1.

The question asks for "The number of rational numbers in the domain of function ... is ______". This is a fill-in-the-blank type question.
The final answer for S10 is 2.

The question asks for "the value of (A + B) is ______". This is a fill-in-the-blank type question.
The final answer for S11 is 6.

Since the question asks for a single numerical answer for each, I'll provide the final numerical value.

S9 Answer: 1
S10 Answer: 2
S11 Answer: 6

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Explanation of the solution:

For S9:

1. Simplify $f(x)$ using $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ and rewrite the quadratic denominator.
2. Identify the domain of $f(x)$ as $[-1, 1]$.
3. Find the maximum value of each term in $f(x)$ within the domain $[-1, 1]$.
   • For $\frac{1}{\pi}\tan^{-1}x$, the maximum is at $x=1$, giving $\frac{1}{4}$.
   • For $\frac{x+1}{(x+1)^2+9}$, let $u=x+1$, so $u \in [0,2]$. The function $\frac{u}{u^2+9}$ is increasing on $[0,2]$, so its maximum is at $u=2$ (i.e., $x=1$), giving $\frac{2}{13}$.
4. Since both terms are maximized at $x=1$, the maximum value $M = f(1) = \frac{1}{2} + \frac{1}{4} + \frac{2}{13} = \frac{47}{52}$.
5. Calculate $\frac{1}{M} = \frac{52}{47} \approx 1.106$.
6. The integral part of $\frac{1}{M}$ is $\lfloor 1.106 \rfloor = 1$.

For S10:

1. List the domains for each inverse trigonometric function:
   • $\sin^{-1}x, \cos^{-1}x$: $[-1, 1]$
   • $\tan^{-1}x, \cot^{-1}x$: $(-\infty, \infty)$
   • $\csc^{-1}x, \sec^{-1}x$: $(-\infty, -1] \cup [1, \infty)$
2. The domain of the sum function is the intersection of all individual domains.
3. The intersection of $[-1, 1]$ and $(-\infty, -1] \cup [1, \infty)$ is $\{-1, 1\}$. This is the most restrictive part. All other domains include $\{-1, 1\}$.
4. The domain of $y$ is $\{-1, 1\}$.
5. Both $-1$ and $1$ are rational numbers. So, there are 2 rational numbers in the domain.

For S11:

1. For A: Given $\sin^{-1}p + \sin^{-1}q + \sin^{-1}r = \frac{3\pi}{2}$. Since the maximum value of each $\sin^{-1}$ term is $\frac{\pi}{2}$, this equality holds only if $p=q=r=1$. Thus, $A = 1^2+1^2+1^2 = 3$.
2. For B: Given $(\sin^{-1}x)^2 + (\sin^{-1}y)^2 + (\sin^{-1}z)^2 = \frac{3\pi^2}{4}$. Since the maximum value of $(\sin^{-1}t)^2$ is $(\frac{\pi}{2})^2 = \frac{\pi^2}{4}$, this equality holds only if $(\sin^{-1}x)^2 = (\sin^{-1}y)^2 = (\sin^{-1}z)^2 = \frac{\pi^2}{4}$. This implies $\sin^{-1}x = \sin^{-1}y = \sin^{-1}z = \pm \frac{\pi}{2}$, so $x, y, z \in \{-1, 1\}$.
3. To find $(x+y+z)_{min}$, choose $x=y=z=-1$. So $(x+y+z)_{min} = -3$.
4. $B = |(x+y+z)_{min}| = |-3| = 3$.
5. Finally, $A+B = 3+3 = 6$.

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Answer: S9: 1 S10: 2 S11: 6

Subject: Mathematics Chapter: Inverse Trigonometric Functions Topic: Properties of Inverse Trigonometric Functions, Domain and Range of Inverse Trigonometric Functions, Maxima and Minima of Functions.

Difficulty Level: Medium

Question Type: fill_in_the_blank