Mix Examples-ITF Questions in English

(C) We use the formula $\tan^{-1}(iz) = i \tanh^{-1}(z) = \frac{i}{2} \log\left(\frac{1+z}{1-z}\right)$.
Here,$z = \frac{5}{3}$.
$\tan^{-1}\left(\frac{5i}{3}\right) = i \tanh^{-1}\left(\frac{5}{3}\right) = \frac{i}{2} \log\left(\frac{1 + 5/3}{1 - 5/3}\right)$.
$= \frac{i}{2} \log\left(\frac{8/3}{-2/3}\right) = \frac{i}{2} \log(-4)$.
Since $\log(-4) = \log(4) + i(\pi + 2k\pi)$,the expression becomes $\frac{i}{2} (\log 4 + i\pi) = \frac{i}{2} \log 4 - \frac{\pi}{2}$.
However,using the standard principal value branch for $\tanh^{-1}(z) = \frac{1}{2} \log\left(\frac{1+z}{1-z}\right)$,we have $\tan^{-1}(iz) = \frac{i}{2} \log\left(\frac{1+z}{1-z}\right)$.
For $z = 5/3$,$\tan^{-1}(5i/3) = \frac{i}{2} \log\left(\frac{8/3}{-2/3}\right) = \frac{i}{2} \log(-4)$.
Taking the principal value,$\log(-4) = \ln 4 + i\pi$.
Thus,$\tan^{-1}(5i/3) = \frac{i}{2} \ln 4 - \frac{\pi}{2}$.
The imaginary part is $\frac{1}{2} \ln 4 = \ln 2$.

(A) Given that $x, y, z$ are in $A.P.$,we have $2y = x + z$.
Since $\tan^{-1}x, \tan^{-1}y, \tan^{-1}z$ are in $A.P.$,we have $2\tan^{-1}y = \tan^{-1}x + \tan^{-1}z$.
Using the formula $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$,we get $\tan^{-1}\left(\frac{2y}{1-y^2}\right) = \tan^{-1}\left(\frac{x+z}{1-xz}\right)$.
This implies $\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$.
Substituting $x+z = 2y$,we get $\frac{2y}{1-y^2} = \frac{2y}{1-xz}$.
This equation holds if $2y = 0$ or $1-y^2 = 1-xz$.
If $2y = 0$,then $y=0$,which implies $x+z=0$ or $x=-z$.
If $1-y^2 = 1-xz$,then $y^2 = xz$.
Since $x, y, z$ are in $A.P.$ ($y^2 = xz$ implies $G.P.$),the only way for numbers to be in both $A.P.$ and $G.P.$ is if $x = y = z$.

(A) Given that $p, q, r$ are in $G.P.$,we have $q^2 = pr$.
Also,$\tan^{-1} p, \tan^{-1} q, \tan^{-1} r$ are in $A.P.$,which implies $2 \tan^{-1} q = \tan^{-1} p + \tan^{-1} r$.
Using the formula $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)$,we get $2 \tan^{-1} q = \tan^{-1} \left( \frac{p+r}{1-pr} \right)$.
Taking $\tan$ on both sides,$\frac{2q}{1-q^2} = \frac{p+r}{1-pr}$.
Since $q^2 = pr$,the denominator $1-q^2 = 1-pr$.
Thus,$2q = p+r$,which means $p, q, r$ are in $A.P.$
Since $p, q, r$ are in both $A.P.$ and $G.P.$,they must be equal,i.e.,$p = q = r$.

(A) Given the quadratic equation $6x^2 - 5x + 1 = 0$.
By comparing with $ax^2 + bx + c = 0$,we have $a = 6, b = -5, c = 1$.
The sum of roots $\alpha + \beta = -b/a = 5/6$.
The product of roots $\alpha \beta = c/a = 1/6$.
Using the formula $\tan^{-1}\alpha + \tan^{-1}\beta = \tan^{-1}\left(\frac{\alpha + \beta}{1 - \alpha \beta}\right)$,
Substitute the values: $\tan^{-1}\left(\frac{5/6}{1 - 1/6}\right) = \tan^{-1}\left(\frac{5/6}{5/6}\right) = \tan^{-1}(1)$.
Since $\tan^{-1}(1) = \pi / 4$,the final value is $\pi / 4$.

(D) Using the formula $\tan^{-1}(A) + \tan^{-1}(B) = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$,we have:
$\tan^{-1}\left(\frac{\frac{x-1}{x+1} + \frac{2x-1}{2x+1}}{1 - \left(\frac{x-1}{x+1}\right)\left(\frac{2x-1}{2x+1}\right)}\right) = \tan^{-1}\left(\frac{23}{36}\right)$
Simplifying the argument:
$\frac{(x-1)(2x+1) + (2x-1)(x+1)}{(x+1)(2x+1) - (x-1)(2x-1)} = \frac{23}{36}$
$\frac{2x^2 + x - 2x - 1 + 2x^2 + 2x - x - 1}{2x^2 + x + 2x + 1 - (2x^2 - x - 2x + 1)} = \frac{23}{36}$
$\frac{4x^2 - 2}{6x} = \frac{23}{36}$
$\frac{2x^2 - 1}{3x} = \frac{23}{36}$
$24x^2 - 12 = 69x$
$24x^2 - 69x - 12 = 0$
Dividing by $3$: $8x^2 - 23x - 4 = 0$
$(8x + 1)(x - 4) = 0$
Thus,$x = 4$ or $x = -\frac{1}{8}$.
Since none of the options match,the correct answer is $(d)$.

(A) Let ${\sin ^{ - 1}}a = A, {\sin ^{ - 1}}b = B, {\sin ^{ - 1}}c = C$.
Then $a = \sin A, b = \sin B, c = \sin C$ and $A + B + C = \pi$.
We know that in a triangle where $A + B + C = \pi$,the identity $\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C$ holds.
This can be written as $2\sin A \cos A + 2\sin B \cos B + 2\sin C \cos C = 4\sin A \sin B \sin C$.
Dividing both sides by $2$,we get $\sin A \cos A + \sin B \cos B + \sin C \cos C = 2\sin A \sin B \sin C$.
Since $\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - a^2}$,$\cos B = \sqrt{1 - b^2}$,and $\cos C = \sqrt{1 - c^2}$,we substitute these values:
$a\sqrt{1 - a^2} + b\sqrt{1 - b^2} + c\sqrt{1 - c^2} = 2abc$.

(C) We know that $\cos^{-1}(\cos x) = x - 2n\pi$ if $x \in [2n\pi, (2n+1)\pi]$. For $x = 12$,since $3\pi \approx 9.42$ and $4\pi \approx 12.56$,we have $3\pi < 12 < 4\pi$. Thus,$\cos^{-1}(\cos 12) = 4\pi - 12$.
We know that $\sin^{-1}(\sin x) = x - 2n\pi$ if $x \in [2n\pi - \pi/2, 2n\pi + \pi/2]$. For $x = 14$,since $4\pi \approx 12.56$ and $4\pi + \pi/2 \approx 14.13$,we have $4\pi - \pi/2 < 14 < 4\pi + \pi/2$. Thus,$\sin^{-1}(\sin 14) = 14 - 4\pi$.
Therefore,$\cos^{-1}(\cos 12) - \sin^{-1}(\sin 14) = (4\pi - 12) - (14 - 4\pi) = 8\pi - 26$.

(C) Given equation: $\cos^{-1} x + \cos^{-1} 2x + \pi = 0$
$\Rightarrow \cos^{-1} x + \cos^{-1} 2x = -\pi$
Since the range of $\cos^{-1} \theta$ is $[0, \pi]$,the minimum value of $\cos^{-1} x + \cos^{-1} 2x$ is $0 + 0 = 0$.
However,the equation requires the sum to be $-\pi$.
Since the sum of two non-negative values cannot be negative,there is no real value of $x$ that satisfies this equation.
Therefore,the number of real solutions is $0$.

(B) Given that $\tan (\cos ^{ - 1}x) = \sin (\cot ^{ - 1}\frac{1}{2})$.
Let $\cot ^{ - 1}\frac{1}{2} = \phi$,then $\cot \phi = \frac{1}{2}$.
We know that $\sin \phi = \frac{1}{\sqrt{1 + \cot ^2 \phi}} = \frac{1}{\sqrt{1 + (1/2)^2}} = \frac{1}{\sqrt{5/4}} = \frac{2}{\sqrt{5}}$.
Let $\cos ^{ - 1}x = \theta$,then $\cos \theta = x$.
We know that $\tan \theta = \frac{\sqrt{1 - \cos ^2 \theta}}{\cos \theta} = \frac{\sqrt{1 - x^2}}{x}$.
Substituting these into the original equation:
$\frac{\sqrt{1 - x^2}}{x} = \frac{2}{\sqrt{5}}$.
Squaring both sides:
$\frac{1 - x^2}{x^2} = \frac{4}{5}$.
$5(1 - x^2) = 4x^2$.
$5 - 5x^2 = 4x^2$.
$9x^2 = 5$.
$x^2 = \frac{5}{9}$.
$x = \pm \frac{\sqrt{5}}{3}$.

(C) The given equation is $\tan^{-1}\sqrt{x(x + 1)} + \sin^{-1}\sqrt{x^2 + x + 1} = \frac{\pi}{2}$.
For $\tan^{-1}\sqrt{x(x + 1)}$ to be defined,we must have $x(x + 1) \ge 0$.
For $\sin^{-1}\sqrt{x^2 + x + 1}$ to be defined,the argument must satisfy $0 \le \sqrt{x^2 + x + 1} \le 1$,which implies $0 \le x^2 + x + 1 \le 1$.
Since $x^2 + x + 1 = (x + \frac{1}{2})^2 + \frac{3}{4}$,the minimum value of $x^2 + x + 1$ is $\frac{3}{4}$.
Thus,the condition $0 \le x^2 + x + 1 \le 1$ simplifies to $\frac{3}{4} \le x^2 + x + 1 \le 1$,which implies $x^2 + x \le 0$.
Combining $x(x + 1) \ge 0$ and $x(x + 1) \le 0$,we must have $x(x + 1) = 0$.
This gives $x = 0$ or $x = -1$.
Checking $x = 0$: $\tan^{-1}(0) + \sin^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$. This is a solution.
Checking $x = -1$: $\tan^{-1}(0) + \sin^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$. This is a solution.
Therefore,there are $2$ real solutions.

(C) Let ${\cos ^{ - 1}}x = y$. As $x \to 1$,$y \to 0$. Also,$x = \cos y$.
$\mathop {\lim }\limits_{x \to 1} \frac{{1 - \sqrt x }}{{{{({{\cos }^{ - 1}}x)}^2}}} = \mathop {\lim }\limits_{y \to 0} \frac{{1 - \sqrt {\cos y} }}{{{y^2}}}$
Rationalizing the numerator:
$= \mathop {\lim }\limits_{y \to 0} \frac{{(1 - \sqrt {\cos y} )(1 + \sqrt {\cos y} )}}{{{y^2}(1 + \sqrt {\cos y} )}} = \mathop {\lim }\limits_{y \to 0} \frac{{1 - \cos y}}{{{y^2}(1 + \sqrt {\cos y} )}}$
Using the standard limit $\mathop {\lim }\limits_{y \to 0} \frac{{1 - \cos y}}{{{y^2}}} = \frac{1}{2}$:
$= \left( \mathop {\lim }\limits_{y \to 0} \frac{{1 - \cos y}}{{{y^2}}} \right) \times \left( \mathop {\lim }\limits_{y \to 0} \frac{1}{{1 + \sqrt {\cos y} }} \right)$
$= \frac{1}{2} \times \frac{1}{{1 + \sqrt 1 }} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

(D) Given $f(x) = \cot^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right)$ and $g(x) = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$.
Let $x = \tan \theta$. For $0 < a < \frac{1}{\sqrt{3}}$,we have $0 < \theta < \frac{\pi}{6}$.
$f(x) = \cot^{-1}(\tan 3\theta) = \cot^{-1}(\cot(\frac{\pi}{2} - 3\theta)) = \frac{\pi}{2} - 3\theta = \frac{\pi}{2} - 3\tan^{-1}x$.
Thus,$f'(x) = -\frac{3}{1 + x^2}$.
$g(x) = \cos^{-1}(\cos 2\theta) = 2\theta = 2\tan^{-1}x$.
Thus,$g'(x) = \frac{2}{1 + x^2}$.
Using $L$'$H$ôpital's rule or the definition of the derivative:
$\lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f'(a)}{g'(a)} = \frac{-\frac{3}{1 + a^2}}{\frac{2}{1 + a^2}} = -\frac{3}{2}$.

(B) Let $y = \tan^{-1} \left[ \frac{3a^2x - x^3}{a(a^2 - 3x^2)} \right]$.
Substitute $x = a \tan \theta$,which implies $\theta = \tan^{-1}(\frac{x}{a})$.
Then,$y = \tan^{-1} \left[ \frac{3a^3 \tan \theta - a^3 \tan^3 \theta}{a(a^2 - 3a^2 \tan^2 \theta)} \right] = \tan^{-1} \left[ \frac{a^3(3 \tan \theta - \tan^3 \theta)}{a^3(1 - 3 \tan^2 \theta)} \right]$.
Using the trigonometric identity $\tan(3\theta) = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$,we get $y = \tan^{-1}(\tan 3\theta) = 3\theta$.
Since $\theta = \tan^{-1}(\frac{x}{a})$,we have $y = 3 \tan^{-1}(\frac{x}{a})$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 3 \cdot \frac{1}{1 + (x/a)^2} \cdot \frac{1}{a} = 3 \cdot \frac{a^2}{a^2 + x^2} \cdot \frac{1}{a} = \frac{3a}{a^2 + x^2}$.
At $x = 0$,$\frac{dy}{dx} = \frac{3a}{a^2 + 0} = \frac{3}{a}$.

(C) Given $u = \tan^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right)$ and $v = 2 \tan^{-1} x$.
Let $x = \tan \theta$. Then $\theta = \tan^{-1} x$.
Substituting $x = \tan \theta$ in $u$:
$u = \tan^{-1} \left( \frac{\sqrt{1 + \tan^2 \theta} - 1}{\tan \theta} \right) = \tan^{-1} \left( \frac{\sec \theta - 1}{\tan \theta} \right)$
$u = \tan^{-1} \left( \frac{1 - \cos \theta}{\sin \theta} \right) = \tan^{-1} \left( \frac{2 \sin^2(\theta/2)}{2 \sin(\theta/2) \cos(\theta/2)} \right)$
$u = \tan^{-1} (\tan(\theta/2)) = \theta/2 = \frac{1}{2} \tan^{-1} x$.
Now,$v = 2 \tan^{-1} x$.
We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.
$\frac{du}{dx} = \frac{1}{2} \cdot \frac{1}{1 + x^2}$.
$\frac{dv}{dx} = 2 \cdot \frac{1}{1 + x^2}$.
Therefore,$\frac{du}{dv} = \frac{\frac{1}{2(1 + x^2)}}{\frac{2}{1 + x^2}} = \frac{1}{4}$.

(B) Let $p = {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$.
Using the substitution $x = \tan \theta$,we get $p = {\sin ^{ - 1}}(\sin 2\theta) = 2\theta = 2{\tan ^{ - 1}}x$.
Now,let $q = {\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$.
Using the same substitution $x = \tan \theta$,we get $q = {\cos ^{ - 1}}(\cos 2\theta) = 2\theta = 2{\tan ^{ - 1}}x$.
Since $p = q$,the derivative of $p$ with respect to $q$ is $\frac{{dp}}{{dq}} = \frac{d}{{dq}}(q) = 1$.

(A) Given that $x, y, z$ are in arithmetic progression,we have $2y = x + z$.
Also,$\tan^{-1}x, \tan^{-1}y, \tan^{-1}z$ are in arithmetic progression,so $2\tan^{-1}y = \tan^{-1}x + \tan^{-1}z$.
Using the formula $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$,we get $\tan^{-1}\left(\frac{2y}{1-y^2}\right) = \tan^{-1}\left(\frac{x+z}{1-xz}\right)$.
This implies $\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$.
Since $x+z = 2y$,we have $\frac{2y}{1-y^2} = \frac{2y}{1-xz}$.
This equation holds if $2y = 0$ or $1-y^2 = 1-xz$.
If $2y = 0$,then $y = 0$,which implies $x = -z$.
If $1-y^2 = 1-xz$,then $y^2 = xz$.
Since $x, y, z$ are in arithmetic progression,$y^2 = xz$ implies $x = y = z$.
Thus,the condition $x = y = z$ satisfies the given requirements.

(A) Given $x, y, z$ are in $A.P.$,we have $2y = x + z$ ....$(1)$
Since $\tan^{-1} x, \tan^{-1} y, \tan^{-1} z$ are in $A.P.$,we have $2 \tan^{-1} y = \tan^{-1} x + \tan^{-1} z$.
Using the formula $\tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a+b}{1-ab} \right)$,we get $2 \tan^{-1} y = \tan^{-1} \left( \frac{x+z}{1-xz} \right)$.
Taking $\tan$ on both sides,$\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$.
Substituting $x+z = 2y$ from $(1)$,we get $\frac{2y}{1-y^2} = \frac{2y}{1-xz}$.
This implies either $2y = 0$ (which leads to $x=y=z=0$) or $\frac{1}{1-y^2} = \frac{1}{1-xz}$,which means $y^2 = xz$.
If $x, y, z$ are in both $A.P.$ and $G.P.$,then $x = y = z$.

(A) Let $s^2 = \frac{a+b+c}{abc}$. Then the expression becomes $\theta = \tan^{-1}\sqrt{a^2 s^2} + \tan^{-1}\sqrt{b^2 s^2} + \tan^{-1}\sqrt{c^2 s^2}$.
Since $a, b, c$ are positive,$\theta = \tan^{-1}(as) + \tan^{-1}(bs) + \tan^{-1}(cs)$.
Using the formula $\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \tan^{-1}\left(\frac{x+y+z-xyz}{1-(xy+yz+zx)}\right)$:
$\tan \theta = \frac{as+bs+cs - (as)(bs)(cs)}{1 - (asbs + bscs + csas)} = \frac{s(a+b+c) - s^3(abc)}{1 - s^2(ab+bc+ca)}$.
Substitute $s^2 = \frac{a+b+c}{abc}$,which implies $s^2(abc) = a+b+c$:
$\tan \theta = \frac{s(a+b+c) - s(s^2 abc)}{1 - s^2(ab+bc+ca)} = \frac{s(a+b+c) - s(a+b+c)}{1 - s^2(ab+bc+ca)} = \frac{0}{1 - s^2(ab+bc+ca)} = 0$.
Alternatively,let $a=b=c=1$. Then $\theta = \tan^{-1}\sqrt{3} + \tan^{-1}\sqrt{3} + \tan^{-1}\sqrt{3} = 3 \times 60^\circ = 180^\circ = \pi$. Thus,$\tan \theta = \tan \pi = 0$.

(A) Given equation: ${\tan ^{ - 1}}x + {\cos ^{ - 1}}\frac{y}{{\sqrt {1 + {y^2}} }} = {\sin ^{ - 1}}\frac{3}{{\sqrt {10} }}$
We know that ${\cos ^{ - 1}}\frac{y}{{\sqrt {1 + {y^2}} }} = {\tan ^{ - 1}}\frac{1}{y}$ and ${\sin ^{ - 1}}\frac{3}{{\sqrt {10} }} = {\tan ^{ - 1}}3$.
Substituting these,we get: ${\tan ^{ - 1}}x + {\tan ^{ - 1}}\frac{1}{y} = {\tan ^{ - 1}}3$.
Using the formula ${\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\frac{A+B}{1-AB}$,we have: $\frac{x + \frac{1}{y}}{1 - \frac{x}{y}} = 3$.
Simplifying: $\frac{xy + 1}{y - x} = 3 \implies xy + 1 = 3y - 3x \implies xy + 3x - 3y + 1 = 0$.
Adding $8$ to both sides to factorize: $x(y + 3) - 3(y + 3) + 10 = 0 \implies (x - 3)(y + 3) = -10 \implies (3 - x)(y + 3) = 10$.
Since $x, y$ are positive integers,we test factors of $10$: $(3-x)$ can be $1$ or $2$.
If $3 - x = 1 \implies x = 2$,then $y + 3 = 10 \implies y = 7$.
If $3 - x = 2 \implies x = 1$,then $y + 3 = 5 \implies y = 2$.
Thus,the pairs are $(1, 2)$ and $(2, 7)$.

(B) Given that $a_1, a_2, \dots, a_n$ is an $A.P.$ with common difference $d$,we have $a_{k+1} - a_k = d$ for all $k = 1, 2, \dots, n-1$.
Using the identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x - y}{1 + xy} \right)$,we can write each term as:
$\tan^{-1} \left( \frac{d}{1 + a_k a_{k+1}} \right) = \tan^{-1} \left( \frac{a_{k+1} - a_k}{1 + a_k a_{k+1}} \right) = \tan^{-1} a_{k+1} - \tan^{-1} a_k$.
Summing these terms from $k = 1$ to $n-1$ gives a telescoping series:
$\sum_{k=1}^{n-1} (\tan^{-1} a_{k+1} - \tan^{-1} a_k) = (\tan^{-1} a_2 - \tan^{-1} a_1) + (\tan^{-1} a_3 - \tan^{-1} a_2) + \dots + (\tan^{-1} a_n - \tan^{-1} a_{n-1})$.
All intermediate terms cancel out,leaving $\tan^{-1} a_n - \tan^{-1} a_1$.
Thus,the expression becomes $\tan \left( \tan^{-1} a_n - \tan^{-1} a_1 \right) = \tan \left( \tan^{-1} \left( \frac{a_n - a_1}{1 + a_n a_1} \right) \right) = \frac{a_n - a_1}{1 + a_1 a_n}$.
Since $a_n = a_1 + (n-1)d$,we have $a_n - a_1 = (n-1)d$.
Therefore,the result is $\frac{(n - 1)d}{1 + a_1 a_n}$.

(D) Let $\alpha = \cos^{-1} \frac{1}{5\sqrt{2}}$. Then $\cos \alpha = \frac{1}{5\sqrt{2}}$.
Since $\tan^2 \alpha = \sec^2 \alpha - 1 = (5\sqrt{2})^2 - 1 = 50 - 1 = 49$,we have $\tan \alpha = 7$.
Thus,$\alpha = \tan^{-1} 7$.
Let $\beta = \sin^{-1} \frac{4}{\sqrt{17}}$. Then $\sin \beta = \frac{4}{\sqrt{17}}$.
Since $\cos^2 \beta = 1 - \sin^2 \beta = 1 - \frac{16}{17} = \frac{1}{17}$,we have $\cos \beta = \frac{1}{\sqrt{17}}$.
Then $\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{4/\sqrt{17}}{1/\sqrt{17}} = 4$.
Thus,$\beta = \tan^{-1} 4$.
The expression becomes $\tan(\alpha - \beta) = \tan(\tan^{-1} 7 - \tan^{-1} 4)$.
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we get:
$\tan(\tan^{-1} 7 - \tan^{-1} 4) = \frac{7 - 4}{1 + (7)(4)} = \frac{3}{1 + 28} = \frac{3}{29}$.

(C) Given $y = \tan^{-1} \left( \frac{\ln e - \ln x^2}{\ln e + \ln x^2} \right) + \tan^{-1} \left( \frac{3 + 2 \ln x}{1 - 6 \ln x} \right)$.
Since $\ln e = 1$ and $\ln x^2 = 2 \ln x$,we have $y = \tan^{-1} \left( \frac{1 - 2 \ln x}{1 + 2 \ln x} \right) + \tan^{-1} \left( \frac{3 + 2 \ln x}{1 - 6 \ln x} \right)$.
Using the formula $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A+B}{1-AB} \right)$,let $A = \frac{1 - 2 \ln x}{1 + 2 \ln x}$ and $B = \frac{3 + 2 \ln x}{1 - 6 \ln x}$.
Then $A + B = \frac{(1 - 2 \ln x)(1 - 6 \ln x) + (3 + 2 \ln x)(1 + 2 \ln x)}{(1 + 2 \ln x)(1 - 6 \ln x)} = \frac{1 - 8 \ln x + 12(\ln x)^2 + 3 + 8 \ln x + 4(\ln x)^2}{(1 + 2 \ln x)(1 - 6 \ln x)} = \frac{4 + 16(\ln x)^2}{(1 + 2 \ln x)(1 - 6 \ln x)}$.
And $1 - AB = 1 - \frac{(1 - 2 \ln x)(3 + 2 \ln x)}{(1 + 2 \ln x)(1 - 6 \ln x)} = \frac{(1 + 2 \ln x)(1 - 6 \ln x) - (3 - 4 \ln x - 4(\ln x)^2)}{(1 + 2 \ln x)(1 - 6 \ln x)} = \frac{1 - 4 \ln x - 12(\ln x)^2 - 3 + 4 \ln x + 4(\ln x)^2}{(1 + 2 \ln x)(1 - 6 \ln x)} = \frac{-2 - 8(\ln x)^2}{(1 + 2 \ln x)(1 - 6 \ln x)}$.
Thus,$\frac{A+B}{1-AB} = \frac{4(1 + 4(\ln x)^2)}{-2(1 + 4(\ln x)^2)} = -2$.
So $y = \tan^{-1}(-2)$,which is a constant.
Therefore,$\frac{dy}{dx} = 0$ and $\frac{d^2y}{dx^2} = 0$.

(A) Let $\cos \alpha = \frac{2}{\sqrt{13}}$ and $\sin \alpha = \frac{3}{\sqrt{13}}$. Then $\tan \alpha = \frac{3}{2}$.
The expression becomes:
$f(x) = \cos^{-1} (\cos \alpha \cos x - \sin \alpha \sin x) + \sin^{-1} (\cos \alpha \cos x + \sin \alpha \sin x)$
Using trigonometric identities $\cos(A+B) = \cos A \cos B - \sin A \sin B$ and $\cos(A-B) = \cos A \cos B + \sin A \sin B$:
$f(x) = \cos^{-1} (\cos(x + \alpha)) + \sin^{-1} (\cos(x - \alpha))$
Since $\sin^{-1}(\cos \theta) = \frac{\pi}{2} - \cos^{-1}(\cos \theta) = \frac{\pi}{2} - \theta$ (for appropriate range):
$f(x) = (x + \alpha) + \left( \frac{\pi}{2} - (x - \alpha) \right)$
$f(x) = x + \alpha + \frac{\pi}{2} - x + \alpha = \frac{\pi}{2} + 2\alpha$
Since $\alpha$ is a constant,the derivative $f'(x) = \frac{d}{dx} (\frac{\pi}{2} + 2\alpha) = 0$.
Thus,the derivative at $x = \frac{3}{4}$ is $0$.

(B) Let $x = \sin^2 \theta$. Since $x \in (0, 1/2)$,we have $\theta \in (0, \pi/4)$.
Then $\sqrt{1-x} = \sqrt{1-\sin^2 \theta} = \cos \theta$.
Also,$\sqrt{x(1-x)} = \sqrt{\sin^2 \theta \cos^2 \theta} = \sin \theta \cos \theta = \frac{1}{2} \sin 2\theta$.
Substituting these into $f(x)$:
$f(x) = 2 \sin^{-1}(\cos \theta) + \sin^{-1}(2 \cdot \frac{1}{2} \sin 2\theta)$
$f(x) = 2 \sin^{-1}(\sin(\pi/2 - \theta)) + \sin^{-1}(\sin 2\theta)$
Since $\theta \in (0, \pi/4)$,then $(\pi/2 - \theta) \in (\pi/4, \pi/2)$ and $2\theta \in (0, \pi/2)$.
Thus,$f(x) = 2(\pi/2 - \theta) + 2\theta = \pi - 2\theta + 2\theta = \pi$.
Since $f(x) = \pi$ is a constant function,its derivative $f'(x) = 0$.

(B) Let $\sin^{-1} x = \theta$. As $x \to 1/\sqrt{2}$,$\theta \to \pi/4$.
Substituting $x = \sin \theta$ into the expression:
$\lim_{\theta \to \pi/4} \frac{\sin \theta - \cos \theta}{1 - \tan \theta}$
$= \lim_{\theta \to \pi/4} \frac{\sin \theta - \cos \theta}{1 - \frac{\sin \theta}{\cos \theta}}$
$= \lim_{\theta \to \pi/4} \frac{\cos \theta (\sin \theta - \cos \theta)}{\cos \theta - \sin \theta}$
$= \lim_{\theta \to \pi/4} -\cos \theta$
$= -\cos(\pi/4) = -\frac{1}{\sqrt{2}}$

(A) First,evaluate the numerator: $\mathop {Limit}\limits_{x \to \infty } (\sqrt {x + 1} - \sqrt x) = \mathop {Limit}\limits_{x \to \infty } \frac{(x+1)-x}{\sqrt{x+1}+\sqrt{x}} = \mathop {Limit}\limits_{x \to \infty } \frac{1}{\sqrt{x+1}+\sqrt{x}} = 0$.
Thus,$\cot^{-1}(0) = \pi / 2$.
Next,evaluate the denominator: $\mathop {Limit}\limits_{x \to \infty } (\frac{2x+1}{x-1})^x = \mathop {Limit}\limits_{x \to \infty } (2 + \frac{3}{x-1})^x = \infty$.
Thus,$\sec^{-1}(\infty) = \pi / 2$.
Therefore,the limit is $\frac{\pi / 2}{\pi / 2} = 1$.

(D) Two functions are identical if they have the same domain and the same rule of correspondence.
$(A)$ For $y = \tan(\cos^{-1} x)$,the domain of $\cos^{-1} x$ is $[-1, 1]$. Since $\tan(\theta)$ is undefined at $\theta = \frac{\pi}{2}$,we must have $\cos^{-1} x \neq \frac{\pi}{2}$,which implies $x \neq 0$. Thus,the domain is $[-1, 0) \cup (0, 1]$. The expression simplifies to $\frac{\sqrt{1-x^2}}{x}$. Both functions share the same domain and expression.
$(B)$ For $y = \tan(\cot^{-1} x)$,since $\cot^{-1} x$ is defined for all $x \in \mathbb{R}$ and $\cot^{-1} x \in (0, \pi)$,$\tan(\cot^{-1} x)$ is defined for all $x \in \mathbb{R}$ except where $\cot^{-1} x = \frac{\pi}{2}$,i.e.,$x = 0$. Thus,the domain is $\mathbb{R} \setminus \{0\}$. The expression simplifies to $\frac{1}{x}$. Both functions are identical.
$(C)$ For $y = \sin(\tan^{-1} x)$,the domain of $\tan^{-1} x$ is $\mathbb{R}$. Since $\tan^{-1} x \in (-\frac{\pi}{2}, \frac{\pi}{2})$,$\sin(\tan^{-1} x)$ is defined for all $x \in \mathbb{R}$. The expression simplifies to $\frac{x}{\sqrt{1+x^2}}$. Both functions are identical.
Since $(A)$,$(B)$,and $(C)$ are all pairs of identical functions,the correct option is $(D)$.

(D) For two functions $f(x)$ and $g(x)$ to be equal,they must have the same domain and $f(x) = g(x)$ for all $x$ in the domain.
Check option $A$: Let $x = \tan \theta$. Then $f(x) = \cos(2\theta) = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{1 - x^2}{1 + x^2} = g(x)$. Both are defined for all $x \in \mathbb{R}$. Thus,$A$ is equal.
Check option $B$: Let $x = \cot \theta$. Then $g(x) = \sin(2\theta) = \frac{2\cot \theta}{1 + \cot^2 \theta} = \frac{2x}{1 + x^2} = f(x)$. Both are defined for all $x \in \mathbb{R}$. Thus,$B$ is equal.
Check option $C$: For $f(x)$,$\cot^{-1} x > 0$ for all $x \in \mathbb{R}$,so $\operatorname{sgn}(\cot^{-1} x) = 1$. Thus $f(x) = e^{\ln(1)} = 1$. For $g(x)$,$[1 + \{x\}] = 1$ because $0 \le \{x\} < 1$,so $1 \le 1 + \{x\} < 2$. Thus $g(x) = e^{\ln(1)} = 1$. Both are defined for all $x \in \mathbb{R}$. Thus,$C$ is equal.
Since all pairs are equal,the correct option is $D$.

(C) For $\sin^{-1} \sqrt{x}$ to be defined,we must have $0 \leq \sqrt{x} \leq 1$,which implies $x \in [0, 1]$.
For $\cos^{-1} \sqrt{x^2 - 1}$ to be defined,we must have $0 \leq \sqrt{x^2 - 1} \leq 1$,which implies $0 \leq x^2 - 1 \leq 1$,so $1 \leq x^2 \leq 2$. Since $x \geq 0$,we have $x \in [1, \sqrt{2}]$.
Combining these,the only possible value for $x$ is $x = 1$.
Substituting $x = 1$ into the equation,we get $\sin^{-1}(1) + \cos^{-1}(0) + \tan^{-1}(\tan y) = a$.
This simplifies to $\frac{\pi}{2} + \frac{\pi}{2} + y = a$,where $y$ is any real number such that $\tan y$ is defined.
Thus,$a = \pi + y$. Since $\tan^{-1}(\tan y)$ is a periodic function with range $(-\frac{\pi}{2}, \frac{\pi}{2})$,we have $a \in (\pi - \frac{\pi}{2}, \pi + \frac{\pi}{2})$,which is $a \in (\frac{\pi}{2}, \frac{3\pi}{2})$.
Using $\pi \approx 3.14$,the interval is $(\frac{3.14}{2}, \frac{3 \times 3.14}{2}) \approx (1.57, 4.71)$.
The integral values of $a$ in this interval are $2, 3, 4$.
Therefore,there are $3$ integral values.

(B) Let $\sin^{-1} \sqrt{1-x^2} = \theta$. Then $\sin \theta = \sqrt{1-x^2}$.
This implies $\cos \theta = \sqrt{1 - (1-x^2)} = \sqrt{x^2} = |x|$. Since the domain of $\sqrt{\frac{2}{x}-1}$ requires $x > 0$,we have $\cos \theta = x$.
Thus,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{1-x^2}}{x} = \sqrt{\frac{1-x^2}{x^2}} = \sqrt{\frac{1}{x^2}-1}$.
Given $\tan \theta = \sqrt{\frac{2}{x}-1}$,we equate the two expressions:
$\sqrt{\frac{1}{x^2}-1} = \sqrt{\frac{2}{x}-1}$.
Squaring both sides: $\frac{1}{x^2} - 1 = \frac{2}{x} - 1$.
$\frac{1}{x^2} = \frac{2}{x} \implies x^2 = \frac{x}{2} \implies x(x - \frac{1}{2}) = 0$.
Since $x > 0$,we have $x = \frac{1}{2}$.
Here,$a = 1$ and $b = 2$,which are coprime.
Therefore,$a^2 + b^2 = 1^2 + 2^2 = 1 + 4 = 5$.

(C) Given $f(x) = \cot \left( \sin^{-1} \sqrt{\frac{2}{3 + \cos 2x}} \right)$.
We know that $\cos 2x = 2\cos^2 x - 1$,so $3 + \cos 2x = 3 + 2\cos^2 x - 1 = 2 + 2\cos^2 x = 2(1 + \cos^2 x)$.
Substituting this into the expression,we get $\sqrt{\frac{2}{2(1 + \cos^2 x)}} = \frac{1}{\sqrt{1 + \cos^2 x}}$.
However,let us simplify the argument of $\cot$ differently. Let $\theta = \sin^{-1} \sqrt{\frac{2}{3 + \cos 2x}}$. Then $\sin \theta = \sqrt{\frac{2}{3 + \cos 2x}}$.
$\sin^2 \theta = \frac{2}{3 + \cos 2x} \implies \csc^2 \theta = \frac{3 + \cos 2x}{2} = \frac{3 + 2\cos^2 x - 1}{2} = 1 + \cos^2 x$.
Since $\cot^2 \theta = \csc^2 \theta - 1$,we have $\cot^2 \theta = (1 + \cos^2 x) - 1 = \cos^2 x$.
Thus,$f(x) = \cot \theta = \sqrt{\cos^2 x} = |\cos x|$.
At $x = \frac{2\pi}{3}$,$\cos x = -\frac{1}{2}$,so $f(x) = |\cos x| = -\cos x$ in the neighborhood of $\frac{2\pi}{3}$.
$f'(x) = \frac{d}{dx}(-\cos x) = \sin x$.
$f'\left( \frac{2\pi}{3} \right) = \sin \left( \frac{2\pi}{3} \right) = \frac{\sqrt{3}}{2}$.

(B) Let $g(x) = |\tan^{-1}x|$ and $h(x) = \cot^{-1}x$.
We need to find $f(x) = \max\{g(x), h(x)\}$.
For $x < 0$,$g(x) = |\tan^{-1}x| = -\tan^{-1}x$ and $h(x) = \cot^{-1}x = \pi - \tan^{-1}x$. Since $\pi - \tan^{-1}x > -\tan^{-1}x$,$f(x) = \cot^{-1}x$ for $x < 0$.
For $x \ge 0$,$g(x) = \tan^{-1}x$ and $h(x) = \cot^{-1}x$. They intersect where $\tan^{-1}x = \cot^{-1}x$,i.e.,$x = 1$. At $x=1$,$f(1) = \frac{\pi}{4}$.
For $x \in [0, 1]$,$\cot^{-1}x \ge \tan^{-1}x$,so $f(x) = \cot^{-1}x$. For $x > 1$,$\tan^{-1}x > \cot^{-1}x$,so $f(x) = \tan^{-1}x$.
Statement $I$: The function is continuous everywhere but not differentiable at $x=1$ (where the functions switch) and potentially at $x=0$ (where $|\tan^{-1}x|$ has a corner). Thus,$I$ is wrong.
Statement $II$: The range is $[\frac{\pi}{4}, \pi)$. As $x \to -\infty$,$f(x) \to \pi$. As $x \to \infty$,$f(x) \to \frac{\pi}{2}$. The minimum value is $\frac{\pi}{4}$ at $x=1$. Thus,$II$ is wrong.
Statement $III$: Since the function is not monotonic and does not cover the entire codomain $R$,it is many-one into. Thus,$III$ is correct.
Only one statement is correct.

(C) Given $\cos ^{-1}(x) + \cos ^{-1}(2x) + \cos ^{-1}(3x) = \pi.$
Rearranging the terms,we get $\cos ^{-1}(2x) + \cos ^{-1}(3x) = \pi - \cos ^{-1}(x).$
Using the identity $\cos ^{-1}(A) + \cos ^{-1}(B) = \cos ^{-1}(AB - \sqrt{1-A^2}\sqrt{1-B^2})$ and $\pi - \cos ^{-1}(x) = \cos ^{-1}(-x),$
$\cos ^{-1}(6x^2 - \sqrt{1-4x^2}\sqrt{1-9x^2}) = \cos ^{-1}(-x).$
Equating the arguments,$6x^2 - \sqrt{1-4x^2}\sqrt{1-9x^2} = -x.$
Rearranging,$6x^2 + x = \sqrt{1-4x^2}\sqrt{1-9x^2}.$
Squaring both sides,$(6x^2 + x)^2 = (1-4x^2)(1-9x^2).$
$36x^4 + x^2 + 12x^3 = 1 - 13x^2 + 36x^4.$
Simplifying,$12x^3 + 14x^2 - 1 = 0.$
Comparing this with $ax^3 + bx^2 + cx - 1 = 0,$ we get $a = 12, b = 14, c = 0.$
Thus,$a + b + c = 12 + 14 + 0 = 26.$

(D) Given $x = \sin(2 \tan^{-1} 2)$. Let $\tan^{-1} 2 = \theta$,so $\tan \theta = 2$.
Since $\tan \theta = 2$,we have $\sin \theta = \frac{2}{\sqrt{5}}$ and $\cos \theta = \frac{1}{\sqrt{5}}$.
Then $x = \sin(2 \theta) = 2 \sin \theta \cos \theta = 2 \left( \frac{2}{\sqrt{5}} \right) \left( \frac{1}{\sqrt{5}} \right) = \frac{4}{5}$.
Thus,$1 - x = 1 - \frac{4}{5} = \frac{1}{5}$.
Now,given $y = \sin \left( \frac{1}{2} \tan^{-1} \frac{4}{3} \right)$. Let $\tan^{-1} \frac{4}{3} = \alpha$,so $\tan \alpha = \frac{4}{3}$.
Using the identity $\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}$,where $\cos \alpha = \frac{3}{5}$ (since $\tan \alpha = \frac{4}{3}$),we get $y = \sqrt{\frac{1 - 3/5}{2}} = \sqrt{\frac{2/5}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}$.
Therefore,$y^2 = \frac{1}{5}$.
Comparing the results,$y^2 = \frac{1}{5}$ and $1 - x = \frac{1}{5}$,so $y^2 = 1 - x$.

(C) Let $S = \sum\limits_{\lambda = 1}^{10} {{{\sin }^{ - 1}}\left( {\sin \left( {\lambda \pi - \frac{\pi }{6}} \right)} \right)} $.
For even $\lambda = 2n$ (where $n = 1, 2, 3, 4, 5$):
$\sin(2n\pi - \frac{\pi }{6}) = \sin(-\frac{\pi }{6}) = -\frac{1}{2}$.
Thus,${{\sin }^{ - 1}}(\sin(2n\pi - \frac{\pi }{6})) = -\frac{\pi }{6}$.
There are $5$ such terms,so the sum is $5 \times (-\frac{\pi }{6}) = -\frac{5\pi }{6}$.
For odd $\lambda = 2n-1$ (where $n = 1, 2, 3, 4, 5$):
$\sin((2n-1)\pi - \frac{\pi }{6}) = \sin(\pi - \frac{\pi }{6}) = \sin(\frac{\pi }{6}) = \frac{1}{2}$.
Thus,${{\sin }^{ - 1}}(\sin((2n-1)\pi - \frac{\pi }{6})) = \frac{\pi }{6}$.
There are $5$ such terms,so the sum is $5 \times \frac{\pi }{6} = \frac{5\pi }{6}$.
Adding these together: $S = -\frac{5\pi }{6} + \frac{5\pi }{6} = 0$.

(C) The domain of the function $f(x) = \cos^{-1}x + 2\cot^{-1}x - 2x^3 - 4x$ is $[-1, 1]$.
First,we find the derivative of $f(x)$:
$f'(x) = \frac{-1}{\sqrt{1-x^2}} - \frac{2}{1+x^2} - 6x^2 - 4$.
Since $\frac{-1}{\sqrt{1-x^2}} < 0$,$\frac{-2}{1+x^2} < 0$,and $-6x^2 - 4 < 0$ for all $x \in (-1, 1)$,$f'(x) < 0$ for all $x$ in the domain.
Thus,$f(x)$ is a strictly decreasing function on $[-1, 1]$.
The maximum value occurs at the left endpoint $x = -1$:
$M = f(-1) = \cos^{-1}(-1) + 2\cot^{-1}(-1) - 2(-1)^3 - 4(-1) = \pi + 2(\frac{3\pi}{4}) + 2 + 4 = \pi + \frac{3\pi}{2} + 6 = \frac{5\pi}{2} + 6$.
The minimum value occurs at the right endpoint $x = 1$:
$m = f(1) = \cos^{-1}(1) + 2\cot^{-1}(1) - 2(1)^3 - 4(1) = 0 + 2(\frac{\pi}{4}) - 2 - 4 = \frac{\pi}{2} - 6$.
The sum of the minimum and maximum values is:
$m + M = (\frac{\pi}{2} - 6) + (\frac{5\pi}{2} + 6) = \frac{6\pi}{2} = 3\pi$.

(A) Let $\alpha = \cos^{-1} \left( \frac{3}{5} \right)$,then $\cos \alpha = \frac{3}{5}$. Since $\cos^2 \alpha + \sin^2 \alpha = 1$,we have $\sin \alpha = \sqrt{1 - (\frac{3}{5})^2} = \frac{4}{5}$. Thus,$\alpha = \sin^{-1} \left( \frac{4}{5} \right)$.
Let $\beta = \tan^{-1} 2$,then $\tan \beta = 2$. Using the identity $\sin \beta = \frac{\tan \beta}{\sqrt{1 + \tan^2 \beta}}$,we get $\sin \beta = \frac{2}{\sqrt{1 + 2^2}} = \frac{2}{\sqrt{5}}$. Thus,$\beta = \sin^{-1} \left( \frac{2}{\sqrt{5}} \right)$.
The expression becomes $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$.
We have $\sin \alpha = \frac{4}{5}$,$\cos \alpha = \frac{3}{5}$,$\sin \beta = \frac{2}{\sqrt{5}}$,and $\cos \beta = \frac{1}{\sqrt{1 + 2^2}} = \frac{1}{\sqrt{5}}$.
Substituting these values: $\sin(\alpha + \beta) = (\frac{4}{5} \times \frac{1}{\sqrt{5}}) + (\frac{3}{5} \times \frac{2}{\sqrt{5}}) = \frac{4}{5\sqrt{5}} + \frac{6}{5\sqrt{5}} = \frac{10}{5\sqrt{5}} = \frac{2}{\sqrt{5}}$.

(C) The range of the function $f(x, y) = \tan^{-1} x - \tan^{-1} y$ is $(-\pi, \pi)$. Therefore,the maximum value of $\lfloor \tan^{-1} x - \tan^{-1} y \rfloor$ is $\lfloor \pi^- \rfloor = 3$.
The range of the function $g(u, v) = \sin^{-1} u - \sin^{-1} v$ is $[-\pi, \pi]$. Therefore,the minimum value of $\lfloor \sin^{-1} u - \sin^{-1} v \rfloor$ is $\lfloor -\pi \rfloor = -4$.
To maximize the expression $\lfloor \tan^{-1} x - \tan^{-1} y \rfloor - \lfloor \sin^{-1} u - \sin^{-1} v \rfloor$,we take the maximum value of the first term and subtract the minimum value of the second term.
Maximum value $= 3 - (-4) = 7$.

(D) Let $f(x) = \cos (\tan ^{-1}(\sin (\cos ^{-1} x))) + \sin (\cot ^{-1}(\cos (\sin ^{-1} x)))$.
Since $\sin (\cos ^{-1} x) = \sqrt{1-x^2}$ and $\cos (\sin ^{-1} x) = \sqrt{1-x^2}$,we have:
$f(x) = \cos (\tan ^{-1}(\sqrt{1-x^2})) + \sin (\cot ^{-1}(\sqrt{1-x^2}))$.
Let $\theta = \tan ^{-1}(\sqrt{1-x^2})$,then $\tan \theta = \sqrt{1-x^2}$.
Then $\cos \theta = \frac{1}{\sqrt{1+\tan^2 \theta}} = \frac{1}{\sqrt{1+(1-x^2)}} = \frac{1}{\sqrt{2-x^2}}$.
Similarly,let $\phi = \cot ^{-1}(\sqrt{1-x^2})$,then $\cot \phi = \sqrt{1-x^2}$.
Then $\sin \phi = \frac{1}{\sqrt{1+\cot^2 \phi}} = \frac{1}{\sqrt{1+(1-x^2)}} = \frac{1}{\sqrt{2-x^2}}$.
Thus,$f(x) = \frac{1}{\sqrt{2-x^2}} + \frac{1}{\sqrt{2-x^2}} = \frac{2}{\sqrt{2-x^2}}$.
Since $x \in [-1, 1]$,$x^2 \in [0, 1]$,so $2-x^2 \in [1, 2]$.
Thus,$\sqrt{2-x^2} \in [1, \sqrt{2}]$,and $f(x) \in [\sqrt{2}, 2]$.
So $m = \sqrt{2}$ and $M = 2$.
The equation is $\operatorname{sgn} (|x - 1| - 2) = \ln |x - 2|$.
By analyzing the graph of $y = \operatorname{sgn} (|x - 1| - 2)$ and $y = \ln |x - 2|$,we find that the curves intersect at $4$ points.

(A) We know that $\cot x - 2 \cot 2x = \cot x - 2 \left( \frac{\cot^2 x - 1}{2 \cot x} \right) = \cot x - \frac{\cot^2 x - 1}{\cot x} = \frac{\cot^2 x - \cot^2 x + 1}{\cot x} = \frac{1}{\cot x} = \tan x$.
Thus,$f(x) = \tan^{-1}(\tan x)$.
For $x \in [1, 7]$,we evaluate $f(r) = \tan^{-1}(\tan r)$:
$f(1) = 1$ (since $1 \in (-\pi/2, \pi/2)$)
$f(2) = 2 - \pi$ (since $2 \in (\pi/2, 3\pi/2)$)
$f(3) = 3 - \pi$
$f(4) = 4 - \pi$
$f(5) = 5 - 2\pi$ (since $5 \in (3\pi/2, 5\pi/2)$)
$f(6) = 6 - 2\pi$
$f(7) = 7 - 2\pi$
Summing these values: $\sum_{r=1}^7 f(r) = (1 + 2 + 3 + 4 + 5 + 6 + 7) - (3\pi + 6\pi) = 28 - 9\pi$.
Since $\pi \approx 3.14159$,$9\pi \approx 28.2743$.
Therefore,$[28 - 9\pi] = [28 - 28.2743] = [-0.2743] = -1$.

(D) Given the quadratic equation $6x^2 + 10x + 1 = 0$.
By Vieta's formulas,the sum of roots $p + q = -\frac{10}{6} = -\frac{5}{3}$ and the product of roots $pq = \frac{1}{6}$.
Since $p+q < 0$ and $pq > 0$,both roots $p$ and $q$ are negative.
Let $p = -a$ and $q = -b$,where $a, b > 0$.
Then $\tan^{-1} p + \tan^{-1} q = -(\tan^{-1} a + \tan^{-1} b)$.
Using the formula $\tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a+b}{1-ab} \right)$ for $ab < 1$:
$\tan^{-1} p + \tan^{-1} q = -\tan^{-1} \left( \frac{p+q}{1-pq} \right) = -\tan^{-1} \left( \frac{-5/3}{1-1/6} \right) = -\tan^{-1} \left( \frac{-5/3}{5/6} \right) = -\tan^{-1} (-2) = \tan^{-1} 2$.
Wait,re-evaluating: $\tan^{-1} p + \tan^{-1} q = -\tan^{-1} \left( \frac{a+b}{1-ab} \right)$ where $a+b = 5/3$ and $ab = 1/6$.
$\tan^{-1} p + \tan^{-1} q = -\tan^{-1} \left( \frac{5/3}{1-1/6} \right) = -\tan^{-1} \left( \frac{5/3}{5/6} \right) = -\tan^{-1} (2)$.
Since $\tan^{-1} (1) < \tan^{-1} (2) < \tan^{-1} (\sqrt{3})$,we have $\frac{\pi}{4} < \tan^{-1} 2 < \frac{\pi}{3}$.
Thus,$-\frac{\pi}{3} < -\tan^{-1} 2 < -\frac{\pi}{4}$.
Since $-1.047 < -\tan^{-1} 2 < -0.785$,the greatest integer $[-\tan^{-1} 2]$ is $-1$.

(D) We know that $31\pi \approx 31 \times 3.14159 = 97.389$ and $32\pi \approx 32 \times 3.14159 = 100.53$. Thus,$31\pi < 100 < 32\pi$.
$1$. For $\sin ^{-1}(\sin x)$,since $31\pi < 100 < 32\pi$,we use the identity $\sin ^{-1}(\sin x) = x - 32\pi$ (as $100$ is in the interval $[31.5\pi, 32.5\pi]$).
So,$\sin ^{-1}(\sin 100) = 100 - 32\pi$.
$2$. For $\cos ^{-1}(\cos x)$,since $31\pi < 100 < 32\pi$,we use the identity $\cos ^{-1}(\cos x) = 2n\pi - x$. Here $n=16$,so $32\pi - 100$.
$3$. For $\tan ^{-1}(\tan x)$,since $31\pi < 100 < 32\pi$,we use the identity $\tan ^{-1}(\tan x) = x - 31\pi$.
So,$\tan ^{-1}(\tan 100) = 100 - 31\pi$.
$4$. For $\cot ^{-1}(\cot x)$,since $31\pi < 100 < 32\pi$,we use the identity $\cot ^{-1}(\cot x) = x - 31\pi$.
So,$\cot ^{-1}(\cot 100) = 100 - 31\pi$.
Summing these: $(100 - 32\pi) + (32\pi - 100) + (100 - 31\pi) + (100 - 31\pi) = 200 - 62\pi$.

(A) Let $y = \sec^{-1}x$. The domain of $\sec^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$,so $y \in [0, \pi/2) \cup (\pi/2, \pi]$.
Given inequality: $(y - 4)(y - 1)(y - 2) \ge 0$.
Since $y \in [0, \pi] \approx [0, 3.14]$,the term $(y - 4)$ is always negative.
Dividing by $(y - 4)$,the inequality reverses: $(y - 1)(y - 2) \le 0$.
This holds for $y \in [1, 2]$.
Since $y = \sec^{-1}x$,we have $1 \le \sec^{-1}x \le 2$.
Applying the secant function (which is increasing on $[0, \pi/2)$ and $[1, \pi]$),we get $\sec 1 \le x \le \sec 2$.
However,we must respect the domain of $\sec^{-1}x$. Since $1 \le y \le 2$,$x$ must be in the range of $\sec y$. For $y \in [1, 2]$,$x \in [\sec 1, \sec 2]$ is valid.

(D) For the equation to be defined,the domains of the inverse trigonometric functions must be satisfied:
$1) -1 \leq x - 1 \leq 1 \Rightarrow 0 \leq x \leq 2$
$2) -1 \leq x - 3 \leq 1 \Rightarrow 2 \leq x \leq 4$
$3) -x^2 + 2 \neq 0 \Rightarrow x^2 \neq 2 \Rightarrow x \neq \pm\sqrt{2}$
Taking the intersection of the domains from $(1)$ and $(2)$,we get $x = 2$.
Substituting $x = 2$ into the original equation:
$m = \sin^{-1}(2 - 1) + \cos^{-1}(2 - 3) + \tan^{-1}\left(\frac{2}{-2^2 + 2}\right)$
$m = \sin^{-1}(1) + \cos^{-1}(-1) + \tan^{-1}\left(\frac{2}{-2}\right)$
$m = \frac{\pi}{2} + \pi + \tan^{-1}(-1)$
$m = \frac{3\pi}{2} - \frac{\pi}{4} = \frac{6\pi - \pi}{4} = \frac{5\pi}{4}$

(C) We know that $\sum_{k=1}^n k = \frac{n(n+1)}{2}$.
Substituting this into the expression,we get:
$\cot^{-1} \left( \frac{2 \cdot \frac{n(n+1)}{2} - 1}{3} \right) = \cot^{-1} \left( \frac{n^2+n-1}{3} \right) = \tan^{-1} \left( \frac{3}{n^2+n-1} \right)$.
We can rewrite the argument as $\frac{(n+2)-(n-1)}{1+(n+2)(n-1)} = \frac{3}{1+(n^2+n-2)}$.
Thus,$T_n = \tan^{-1}(n+2) - \tan^{-1}(n-1)$.
The sum is $S_N = \sum_{n=1}^N (\tan^{-1}(n+2) - \tan^{-1}(n-1))$.
Expanding the sum: $S_N = (\tan^{-1} 3 - \tan^{-1} 0) + (\tan^{-1} 4 - \tan^{-1} 1) + (\tan^{-1} 5 - \tan^{-1} 2) + \dots + (\tan^{-1}(N+2) - \tan^{-1}(N-1))$.
Most terms cancel out,leaving $S_N = \tan^{-1}(N+2) + \tan^{-1}(N+1) - \tan^{-1} 1 - \tan^{-1} 0$.
As $N \to \infty$,$S = \frac{\pi}{2} + \frac{\pi}{2} - \frac{\pi}{4} = \frac{3\pi}{4}$.
Since $\frac{3\pi}{4} = \frac{\pi}{2} + \frac{\pi}{4} = \frac{\pi}{2} + \cot^{-1} 1$,or using the options,$\frac{3\pi}{4} = \frac{\pi}{2} + \tan^{-1} 1$. Note that $\tan^{-1} 1 = \cot^{-1} 1$. Checking the options,the sum is $\frac{3\pi}{4}$. Since $\cot^{-1} 2 + \tan^{-1} 2 = \frac{\pi}{2}$,we have $\frac{3\pi}{4} = \frac{\pi}{4} + \frac{\pi}{2} = \cot^{-1} 1 + \frac{\pi}{2}$. Option $A$ is $\frac{3\pi}{4} + \cot^{-1} 2$,which is not correct. Re-evaluating the sum: $S = \lim_{N \to \infty} (\tan^{-1}(N+2) + \tan^{-1}(N+1) - \frac{\pi}{4}) = \frac{\pi}{2} + \frac{\pi}{2} - \frac{\pi}{4} = \frac{3\pi}{4}$. The correct value is $\frac{3\pi}{4}$.

(C) We know the principal value ranges: $\sin^{-1}(\sin x) = x - n\pi$ or $(n+1)\pi - x$,$\cos^{-1}(\cos x) = |x - 2n\pi|$,$\tan^{-1}(\tan x) = x - n\pi$,and $\cot^{-1}(\cot x) = x - n\pi$.
For $x = 100$,we note that $31\pi \approx 97.389$ and $32\pi \approx 100.53$.
Since $31\pi < 100 < 32\pi$:
$1$. $\sin^{-1}(\sin 100) = 100 - 32\pi$ (as $100$ is in the range where $\sin x$ is negative and decreasing).
$2$. $\cos^{-1}(\cos 100) = 32\pi - 100$ (since $100$ is in the interval $(31\pi, 32\pi)$).
$3$. $\tan^{-1}(\tan 100) = 100 - 32\pi$ (since $100$ is in the interval $(31\pi, 32\pi)$).
$4$. $\cot^{-1}(\cot 100) = 100 - 31\pi$ (since $100$ is in the interval $(31\pi, 32\pi)$).
Summing these: $(100 - 32\pi) + (32\pi - 100) + (100 - 32\pi) + (100 - 31\pi) = 200 - 63\pi$.

(A) Given equation: $2\tan^{-1}(\cos^2 x) = \tan^{-1}(2\csc^2 x)$.
Using the formula $2\tan^{-1} \theta = \tan^{-1} \left( \frac{2\theta}{1 - \theta^2} \right)$,we have:
$\tan^{-1} \left( \frac{2\cos^2 x}{1 - \cos^4 x} \right) = \tan^{-1} \left( \frac{2}{\sin^2 x} \right)$.
Equating the arguments:
$\frac{2\cos^2 x}{1 - \cos^4 x} = \frac{2}{\sin^2 x}$.
Since $1 - \cos^4 x = (1 - \cos^2 x)(1 + \cos^2 x) = \sin^2 x(1 + \cos^2 x)$,the equation becomes:
$\frac{2\cos^2 x}{\sin^2 x(1 + \cos^2 x)} = \frac{2}{\sin^2 x}$.
Assuming $\sin^2 x \neq 0$,we cancel $2/\sin^2 x$ from both sides:
$\frac{\cos^2 x}{1 + \cos^2 x} = 1$.
$\cos^2 x = 1 + \cos^2 x$,which implies $0 = 1$,a contradiction.
Thus,there are no real values of $x$ that satisfy the equation.
Therefore,$m = 0$.
Since $m = 0$,the condition $m \le 1$ is satisfied.

(C) Let $u = \tan^{-1} x$. Then $\cot^{-1} x = \frac{\pi}{2} - u$.
Given inequality: $u(\frac{\pi}{2} - u) - u(1 + \frac{\pi}{2}) - 2(\frac{\pi}{2} - u) + 2(1 + \frac{\pi}{2}) > \lim_{x \to \infty} [\sec^{-1} x - \frac{\pi}{2}]$.
As $x \to \infty$,$\sec^{-1} x \to \frac{\pi}{2}$,so the limit is $[\frac{\pi}{2} - \frac{\pi}{2}] = [0] = 0$.
Simplifying the left side: $\frac{\pi}{2}u - u^2 - u - \frac{\pi}{2}u - \pi + 2u + 2 + \pi > 0$.
$-u^2 + u + 2 > 0 \Rightarrow u^2 - u - 2 < 0$.
$(u - 2)(u + 1) < 0 \Rightarrow -1 < u < 2$.
Since $u = \tan^{-1} x$,we have $-1 < \tan^{-1} x < 2$.
Taking tangent on all sides: $\tan(-1) < x < \tan 2$.
Thus,$-\tan 1 < x < \tan 2$.

(D) Given the equation: $\tan^{-1} (x^2 + 3|x|-4) = \tan^{-1} (4\pi + \sin^{-1}(\sin 14))$.
Since $\tan^{-1}$ is a one-to-one function,we have $x^2 + 3|x| - 4 = 4\pi + \sin^{-1}(\sin 14)$.
We know that $\sin^{-1}(\sin 14) = 14 - 4\pi$ because $4\pi < 14 < \frac{9\pi}{2}$ (since $12.56 < 14 < 14.13$).
Substituting this into the equation: $x^2 + 3|x| - 4 = 4\pi + (14 - 4\pi) = 14$.
So,$x^2 + 3|x| - 18 = 0$.
Let $t = |x|$,where $t \ge 0$. Then $t^2 + 3t - 18 = 0$.
$(t+6)(t-3) = 0$. Since $t \ge 0$,we have $t = 3$,so $|x| = 3$.
We need to find $\cos^{-1}(\cos 3|x|) = \cos^{-1}(\cos 9)$.
Since $2\pi < 9 < 3\pi$ (approx $6.28 < 9 < 9.42$),we use the property $\cos^{-1}(\cos \theta) = 2\pi - \theta$ if $2\pi \le \theta \le 3\pi$ is not applicable here,rather $\cos^{-1}(\cos \theta) = \theta - 2\pi$ if $2\pi \le \theta \le 3\pi$.
Thus,$\cos^{-1}(\cos 9) = 9 - 2\pi$.

(B) Given the equation: $2[\cos^{-1}x] + 6[\text{sgn}(\sin x)] = 3$ for $x \in [0, 1]$.
Step $1$: Analyze the range of $\cos^{-1}x$ for $x \in [0, 1]$.
Since $x \in [0, 1]$,$\cos^{-1}x \in [0, \pi/2]$.
Thus,$[\cos^{-1}x]$ can take values $0$ or $1$ (since $\pi/2 \approx 1.57$).
Step $2$: Analyze the range of $\text{sgn}(\sin x)$ for $x \in [0, 1]$.
For $x \in (0, 1]$,$\sin x > 0$,so $\text{sgn}(\sin x) = 1$.
For $x = 0$,$\sin 0 = 0$,so $\text{sgn}(0) = 0$.
Step $3$: Test the cases.
Case $1$: If $x = 0$,then $[\cos^{-1}0] = [\pi/2] = 1$ and $\text{sgn}(\sin 0) = 0$.
The equation becomes $2(1) + 6(0) = 2 \neq 3$.
Case $2$: If $x \in (0, 1]$,then $\text{sgn}(\sin x) = 1$,so $[\text{sgn}(\sin x)] = 1$.
The equation becomes $2[\cos^{-1}x] + 6(1) = 3$,which simplifies to $2[\cos^{-1}x] = -3$,or $[\cos^{-1}x] = -1.5$.
Since the greatest integer function must result in an integer,there is no solution for this case.
Conclusion: There are no values of $x$ that satisfy the equation. Therefore,the number of solutions is $0$.