Properties of ITF Questions in English

(C) We have,$\sum_{n=1}^k \tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right) = \tan ^{-1} \alpha$.
Using the identity $\tan ^{-1} x - \tan ^{-1} y = \tan ^{-1}\left(\frac{x-y}{1+xy}\right)$,we can rewrite the term inside the summation:
$\frac{1}{n^2+3n+3} = \frac{(n+2)-(n+1)}{1+(n+2)(n+1)}$.
Thus,the summation becomes:
$\sum_{n=1}^k (\tan ^{-1}(n+2) - \tan ^{-1}(n+1)) = \tan ^{-1} \alpha$.
Expanding the sum:
$(\tan ^{-1} 3 - \tan ^{-1} 2) + (\tan ^{-1} 4 - \tan ^{-1} 3) + \dots + (\tan ^{-1}(k+2) - \tan ^{-1}(k+1)) = \tan ^{-1} \alpha$.
This is a telescoping series,so all intermediate terms cancel out:
$\tan ^{-1}(k+2) - \tan ^{-1} 2 = \tan ^{-1} \alpha$.
Applying the formula again:
$\tan ^{-1}\left(\frac{(k+2)-2}{1+(k+2)(2)}\right) = \tan ^{-1} \alpha$.
$\tan ^{-1}\left(\frac{k}{1+2k+4}\right) = \tan ^{-1} \alpha$.
Therefore,$\alpha = \frac{k}{2k+5}$.

(A) The given equation is $2 \tan^{-1} 2x = \sin^{-1} \left( \frac{4x}{1+4x^2} \right)$.
We know the identity $\sin^{-1} \left( \frac{2\theta}{1+\theta^2} \right) = 2 \tan^{-1} \theta$,which holds true when $-1 \leq \theta \leq 1$.
Here,let $\theta = 2x$. The equation becomes $2 \tan^{-1} 2x = \sin^{-1} \left( \frac{2(2x)}{1+(2x)^2} \right)$.
This identity is valid if and only if $-1 \leq 2x \leq 1$.
Dividing by $2$,we get $-\frac{1}{2} \leq x \leq \frac{1}{2}$.
Thus,the values of $x$ lie in the interval $[-\frac{1}{2}, \frac{1}{2}]$.

(D) Given,$y=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)+\tan ^{-1}\left(\frac{4 x-4 x^3}{1-6 x^2+x^4}\right)$.
Substitute $x=\tan \theta$,then $\theta = \tan^{-1} x$.
The expression becomes:
$y = \tan^{-1}(\tan 3\theta) + \tan^{-1}(\tan 4\theta)$.
This simplifies to $y = 3\theta + 4\theta = 7\theta$.
Substituting back,$y = 7 \tan^{-1} x$.
Differentiating with respect to $x$:
$\frac{d y}{d x} = 7 \times \frac{1}{1+x^2} = \frac{7}{1+x^2}$.

(D) We know that the range of $\sin ^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Since $\sin ^{-1} x = 2 \sin ^{-1} 2a$,the value of $2 \sin ^{-1} 2a$ must lie in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
$\Rightarrow -\frac{\pi}{2} \leq 2 \sin ^{-1} 2a \leq \frac{\pi}{2}$
$\Rightarrow -\frac{\pi}{4} \leq \sin ^{-1} 2a \leq \frac{\pi}{4}$
Taking sine on all sides,we get:
$\sin(-\frac{\pi}{4}) \leq 2a \leq \sin(\frac{\pi}{4})$
$\Rightarrow -\frac{1}{\sqrt{2}} \leq 2a \leq \frac{1}{\sqrt{2}}$
Dividing by $2$,we get:
$-\frac{1}{2\sqrt{2}} \leq a \leq \frac{1}{2\sqrt{2}}$
This is equivalent to $|a| \leq \frac{1}{2\sqrt{2}}$.

(B) The $n^{\text{th}}$ term of the series is $t_{n} = \cot^{-1}(2n^2)$.
Using the identity $\cot^{-1} x = \tan^{-1} \frac{1}{x}$,we have $t_{n} = \tan^{-1} \frac{1}{2n^2}$.
We can rewrite this as $t_{n} = \tan^{-1} \frac{2}{4n^2} = \tan^{-1} \frac{(2n+1) - (2n-1)}{1 + (2n+1)(2n-1)}$.
Using the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy}$,we get $t_{n} = \tan^{-1}(2n+1) - \tan^{-1}(2n-1)$.
The sum $S_{n} = \sum_{k=1}^{n} t_{k} = (\tan^{-1} 3 - \tan^{-1} 1) + (\tan^{-1} 5 - \tan^{-1} 3) + \dots + (\tan^{-1}(2n+1) - \tan^{-1}(2n-1))$.
This is a telescoping series,so $S_{n} = \tan^{-1}(2n+1) - \tan^{-1} 1$.
Taking the limit as $n \rightarrow \infty$,$\lim_{n \rightarrow \infty} S_{n} = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

(C) We know that the range of $\sin ^{-1} \theta$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Given that $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2}$.
Since the maximum value of each term $\sin ^{-1} x, \sin ^{-1} y, \sin ^{-1} z$ is $\frac{\pi}{2}$,the sum can be $\frac{3 \pi}{2}$ only if $\sin ^{-1} x = \frac{\pi}{2}$,$\sin ^{-1} y = \frac{\pi}{2}$,and $\sin ^{-1} z = \frac{\pi}{2}$.
This implies $x = \sin(\frac{\pi}{2}) = 1$,$y = \sin(\frac{\pi}{2}) = 1$,and $z = \sin(\frac{\pi}{2}) = 1$.
Substituting these values into the expression:
$x^{9}+y^{9}+z^{9}-\frac{1}{x^{9} y^{9} z^{9}} = (1)^{9}+(1)^{9}+(1)^{9}-\frac{1}{(1)^{9}(1)^{9}(1)^{9}}$
$= 1+1+1-\frac{1}{1 \times 1 \times 1}$
$= 3-1 = 2$.

(D) We are given the inequation $\cos ^{-1} x < \sin ^{-1} x$.
We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,so $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$.
Substituting this into the inequation,we get:
$\frac{\pi}{2} - \sin ^{-1} x < \sin ^{-1} x$
$\frac{\pi}{2} < 2 \sin ^{-1} x$
$\sin ^{-1} x > \frac{\pi}{4}$
Taking the sine of both sides (since $\sin x$ is an increasing function in its domain):
$x > \sin\left(\frac{\pi}{4}\right)$
$x > \frac{1}{\sqrt{2}}$
Since the domain of $\sin ^{-1} x$ and $\cos ^{-1} x$ is $[-1, 1]$,we must have $x \le 1$.
Therefore,the solution set is $x \in \left(\frac{1}{\sqrt{2}}, 1\right]$.

(A) Given,$\sin ^{-1}\left(\frac{x}{13}\right)+\operatorname{cosec}^{-1}\left(\frac{13}{12}\right)=\frac{\pi}{2} \quad ...(i)$
We know that $\operatorname{cosec}^{-1}(z) = \sin^{-1}(\frac{1}{z})$.
Therefore,$\operatorname{cosec}^{-1}\left(\frac{13}{12}\right) = \sin^{-1}\left(\frac{12}{13}\right)$.
Substituting this into equation $(i)$,we get:
$\sin ^{-1}\left(\frac{x}{13}\right)+\sin ^{-1}\left(\frac{12}{13}\right)=\frac{\pi}{2}$.
We know the identity $\sin^{-1}(\theta) + \cos^{-1}(\theta) = \frac{\pi}{2}$.
Also,$\sin^{-1}(\frac{12}{13}) = \cos^{-1}(\sqrt{1 - (\frac{12}{13})^2}) = \cos^{-1}(\sqrt{1 - \frac{144}{169}}) = \cos^{-1}(\sqrt{\frac{25}{169}}) = \cos^{-1}(\frac{5}{13})$.
So,$\sin^{-1}(\frac{x}{13}) + \cos^{-1}(\frac{5}{13}) = \frac{\pi}{2}$.
Comparing this with $\sin^{-1}(\theta) + \cos^{-1}(\theta) = \frac{\pi}{2}$,we must have $\frac{x}{13} = \frac{5}{13}$.
Thus,$x = 5$.

(C) Given that $\cos ^{-1} \alpha+\cos ^{-1} \beta+\cos ^{-1} \gamma=3 \pi$.
We know that the range of $\cos ^{-1} x$ is $[0, \pi]$.
Since the sum of three values,each at most $\pi$,is $3 \pi$,each term must be equal to $\pi$.
Therefore,$\cos ^{-1} \alpha = \pi$,$\cos ^{-1} \beta = \pi$,and $\cos ^{-1} \gamma = \pi$.
This implies $\alpha = \cos(\pi) = -1$,$\beta = \cos(\pi) = -1$,and $\gamma = \cos(\pi) = -1$.
Now,we calculate the expression $\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$.
Substituting $\alpha = -1$,$\beta = -1$,and $\gamma = -1$:
$(-1)(-1-1) + (-1)(-1-1) + (-1)(-1-1) = (-1)(-2) + (-1)(-2) + (-1)(-2) = 2 + 2 + 2 = 6$.

(D) Given expression: $2 \cot ^{-1} \frac{1}{2} - \cot ^{-1} \frac{4}{3}$
Using the property $\cot ^{-1} x = \tan ^{-1} \frac{1}{x}$ for $x > 0$:
$= 2 \tan ^{-1} 2 - \tan ^{-1} \frac{3}{4}$
Using the formula $2 \tan ^{-1} x = \pi + \tan ^{-1} \frac{2x}{1-x^2}$ for $x > 1$:
$= \pi + \tan ^{-1} \frac{2(2)}{1-2^2} - \tan ^{-1} \frac{3}{4}$
$= \pi + \tan ^{-1} \frac{4}{-3} - \tan ^{-1} \frac{3}{4}$
$= \pi - \tan ^{-1} \frac{4}{3} - \tan ^{-1} \frac{3}{4}$
$= \pi - (\tan ^{-1} \frac{4}{3} + \tan ^{-1} \frac{3}{4})$
Using the property $\tan ^{-1} x + \tan ^{-1} \frac{1}{x} = \frac{\pi}{2}$ for $x > 0$:
$= \pi - \frac{\pi}{2} = \frac{\pi}{2}$

(B) Let $f(x) = (sin^{-1}x)^{2} + (cos^{-1}x)^{2}$.
Since $cos^{-1}x = \frac{\pi}{2} - sin^{-1}x$,we have:
$f(x) = (sin^{-1}x)^{2} + (\frac{\pi}{2} - sin^{-1}x)^{2}$
$f(x) = (sin^{-1}x)^{2} + \frac{\pi^{2}}{4} - \pi sin^{-1}x + (sin^{-1}x)^{2}$
$f(x) = 2(sin^{-1}x)^{2} - \pi sin^{-1}x + \frac{\pi^{2}}{4}$
$f(x) = 2[(sin^{-1}x)^{2} - \frac{\pi}{2} sin^{-1}x] + \frac{\pi^{2}}{4}$
$f(x) = 2[(sin^{-1}x - \frac{\pi}{4})^{2} - \frac{\pi^{2}}{16}] + \frac{\pi^{2}}{4}$
$f(x) = 2(sin^{-1}x - \frac{\pi}{4})^{2} + \frac{\pi^{2}}{8}$.
Given $x \in [-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}]$,the range of $sin^{-1}x$ is $[-\frac{\pi}{3}, \frac{\pi}{4}]$.
To maximize $f(x)$,we choose the value of $sin^{-1}x$ furthest from $\frac{\pi}{4}$,which is $-\frac{\pi}{3}$.
Max value $= 2(-\frac{\pi}{3} - \frac{\pi}{4})^{2} + \frac{\pi^{2}}{8} = 2(-\frac{7\pi}{12})^{2} + \frac{\pi^{2}}{8} = 2(\frac{49\pi^{2}}{144}) + \frac{\pi^{2}}{8} = \frac{49\pi^{2}}{72} + \frac{9\pi^{2}}{72} = \frac{58\pi^{2}}{72} = \frac{29\pi^{2}}{36}$.
Thus,$m = 29$ and $n = 36$. Since $\gcd(29, 36) = 1$,$m+n = 29 + 36 = 65$.

(C) Given equation: $\tan^{-1}4x + \tan^{-1}6x = \frac{\pi}{6}$.
Using the formula $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$,we have:
$\tan^{-1}\left(\frac{4x+6x}{1-(4x)(6x)}\right) = \frac{\pi}{6}$.
$\frac{10x}{1-24x^2} = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$.
$10\sqrt{3}x = 1 - 24x^2 \implies 24x^2 + 10\sqrt{3}x - 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-10\sqrt{3} \pm \sqrt{(10\sqrt{3})^2 - 4(24)(-1)}}{2(24)} = \frac{-10\sqrt{3} \pm \sqrt{300 + 96}}{48} = \frac{-10\sqrt{3} \pm \sqrt{396}}{48} = \frac{-10\sqrt{3} \pm 6\sqrt{11}}{48} = \frac{-5\sqrt{3} \pm 3\sqrt{11}}{24}$.
The range is $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$,where $\frac{1}{2\sqrt{6}} \approx 0.204$.
$x_1 = \frac{-5\sqrt{3} + 3\sqrt{11}}{24} \approx \frac{-8.66 + 9.95}{24} \approx 0.054$ (In range).
$x_2 = \frac{-5\sqrt{3} - 3\sqrt{11}}{24} \approx \frac{-8.66 - 9.95}{24} \approx -0.775$ (Not in range).
Only one solution satisfies the condition.

(A) Let $x = \sin\theta$. Since $x \in [-1/2, 1/2]$,we have $\theta \in [-\pi/6, \pi/6]$.
The expression becomes $y = 3\sin^{-1}(\sin\theta) + \sin^{-1}(\sin(3\theta))$.
Since $\theta \in [-\pi/6, \pi/6]$,we have $3\theta \in [-\pi/2, \pi/2]$.
Therefore,$\sin^{-1}(\sin\theta) = \theta$ and $\sin^{-1}(\sin(3\theta)) = 3\theta$.
Substituting these,we get $y = 3\theta + 3\theta = 6\theta$.
Given $\theta \in [-\pi/6, \pi/6]$,multiplying by $6$ gives $6\theta \in [-\pi, \pi]$.
Thus,$y \in [-\pi, \pi]$.

(B) Given the equation $\tan^{-1}(1 - \alpha) + \tan^{-1}(1 - \beta) = \frac{\pi}{4}$.
Using the formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$,we get $\frac{(1-\alpha)+(1-\beta)}{1-(1-\alpha)(1-\beta)} = \tan\left(\frac{\pi}{4}\right) = 1$.
This simplifies to $2 - (\alpha + \beta) = 1 - (1 - \alpha - \beta + \alpha\beta)$.
$2 - \alpha - \beta = \alpha + \beta - \alpha\beta \Rightarrow 2 = 2(\alpha + \beta) - \alpha\beta$.
Substitute $\beta = \frac{1}{3\alpha}$ into the equation:
$2 = 2(\alpha + \frac{1}{3\alpha}) - \alpha(\frac{1}{3\alpha}) = 2\alpha + \frac{2}{3\alpha} - \frac{1}{3}$.
$2 + \frac{1}{3} = 2\alpha + \frac{2}{3\alpha} \Rightarrow \frac{7}{3} = \frac{6\alpha^2 + 2}{3\alpha}$.
$7\alpha = 6\alpha^2 + 2 \Rightarrow 6\alpha^2 - 7\alpha + 2 = 0$.
Factoring the quadratic: $(2\alpha - 1)(3\alpha - 2) = 0$.
Thus,$\alpha = \frac{1}{2}$ or $\alpha = \frac{2}{3}$.
If $\alpha = \frac{1}{2}$,then $\beta = \frac{1}{3(1/2)} = \frac{2}{3}$.
If $\alpha = \frac{2}{3}$,then $\beta = \frac{1}{3(2/3)} = \frac{1}{2}$.
In both cases,$\alpha + \beta = \frac{1}{2} + \frac{2}{3} = \frac{7}{6}$.
Therefore,$6(\alpha + \beta) = 6 \times \frac{7}{6} = 7$.

(A) We use the identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)$.
Given the term $\tan^{-1} \left( \frac{2^{p-1}}{1+2^{2p-1}} \right)$,we can rewrite it as $\tan^{-1} \left( \frac{2^p - 2^{p-1}}{1 + 2^p \cdot 2^{p-1}} \right) = \tan^{-1}(2^p) - \tan^{-1}(2^{p-1})$.
Now,sum the telescoping series: $\sum_{p=1}^{11} (\tan^{-1}(2^p) - \tan^{-1}(2^{p-1})) = (\tan^{-1}(2^1) - \tan^{-1}(2^0)) + (\tan^{-1}(2^2) - \tan^{-1}(2^1)) + \dots + (\tan^{-1}(2^{11}) - \tan^{-1}(2^{10}))$.
This simplifies to $\tan^{-1}(2^{11}) - \tan^{-1}(2^0) = \tan^{-1}(2048) - \frac{\pi}{4}$.
Substituting this back into the original equation: $\frac{\pi}{4} + (\tan^{-1}(2048) - \frac{\pi}{4}) = \tan^{-1}(2048)$.
Thus,$\tan^{-1} \alpha = \tan^{-1}(2048)$,which implies $\alpha = 2048$.
Therefore,$\tan \alpha = \tan(2048)$.