Question

The number of solutions of the equation
tan-¹2x + tan-¹3x = ¶/ 4 is​

Answer 1

Given,

$\displaystyle\sf{\longrightarrow \tan^{-1} (2x)+\tan^{-1}(3x)=\dfrac {\pi}{4}}$

Case 1:-

We have,

$\displaystyle\sf{\longrightarrow \tan^{-1}a+\tan^{-1}b=\tan^{-1}\left (\dfrac {a+b}{1-ab}\right)}$

iff $\displaystyle\sf {a>0,\ b>0,\ ab<1.}$

Let,

$\displaystyle\sf {\longrightarrow2x>0}$

$\displaystyle\sf {\longrightarrow3x>0}$

$\displaystyle\sf {\Longrightarrow x\in(0,\ \infty)\quad\quad\dots (1.1)}$

And,

$\displaystyle\sf{\longrightarrow 6x^2<1.}$

$\displaystyle\sf{\Longrightarrow x\in\left (-\dfrac {1}{\sqrt 6},\ \dfrac {1}{\sqrt6}\right)\quad\quad\dots(1.2)}$

Then,

$\displaystyle\sf{\longrightarrow \tan^{-1}\left (\dfrac {2x+3x}{1-6x^2}\right)=\dfrac {\pi}{4}}$

$\displaystyle\sf{\longrightarrow \dfrac {5x}{1-6x^2}=\tan\left (\dfrac {\pi}{4}\right)}$

$\displaystyle\sf{\longrightarrow \dfrac {5x}{1-6x^2}=1}$

$\displaystyle\sf{\longrightarrow 6x^2+5x-1=0}$

$\displaystyle\sf{\longrightarrow 6x^2+6x-x-1=0}$

$\displaystyle\sf{\longrightarrow (x+1)(6x-1)=0}$

$\displaystyle\sf{\Longrightarrow x\in\left\{-1,\ \dfrac {1}{6}\right\}\quad\quad\dots (1.3)}$

Taking $\displaystyle\sf {(1.1)\land (1.2)\land(1.3),}$ we get,

$\displaystyle\sf{\longrightarrow x=\dfrac {1}{6}\quad\quad\dots (i)}$

Case 2:-

We have,

$\displaystyle\sf{\longrightarrow \tan^{-1}a+\tan^{-1}b=\pi+\tan^{-1}\left (\dfrac {a+b}{1-ab}\right)}$

iff $\displaystyle\sf {a>0,\ b>0,\ ab>1.}$

Let,

$\displaystyle\sf {\longrightarrow2x>0}$

$\displaystyle\sf {\longrightarrow3x>0}$

$\displaystyle\sf {\Longrightarrow x\in(0,\ \infty)\quad\quad\dots (2.1)}$

And,

$\displaystyle\sf{\longrightarrow 6x^2>1.}$

$\displaystyle\sf{\Longrightarrow x\in\left(-\infty,\ -\dfrac {1}{\sqrt6}\right)\cup\left (\dfrac {1}{\sqrt 6},\ \infty\right)\quad\quad\dots(2.2)}$

Then,

$\displaystyle\sf{\longrightarrow \pi+\tan^{-1}\left (\dfrac {2x+3x}{1-6x^2}\right)=\dfrac {\pi}{4}}$

$\displaystyle\sf{\longrightarrow \dfrac {5x}{1-6x^2}=-\dfrac {3\pi}{4}}$

But since $\displaystyle\sf {-\dfrac {3\pi}{4}\in\left (-\dfrac {\pi}{4},\ \dfrac {\pi}{4}\right),}$

$\displaystyle\sf{\Longrightarrow x\in\phi\quad\quad\dots(2.3)}$

Taking $\displaystyle\sf {(2.1)\land(2.2)\land(2.3),}$ we get,

$\displaystyle\sf{\longrightarrow x\in\phi\quad\quad\dots(ii)}$

Case 3:-

We have,

$\displaystyle\sf{\longrightarrow \tan^{-1}a+\tan^{-1}b=-\pi+\tan^{-1}\left (\dfrac {a+b}{1-ab}\right)}$

iff $\displaystyle\sf {a<0,\ b<0,\ ab>1.}$

Let,

$\displaystyle\sf {\longrightarrow2x<0}$

$\displaystyle\sf {\longrightarrow3x<0}$

$\displaystyle\sf {\Longrightarrow x\in(-\infty,\ 0)\quad\quad\dots (3.1)}$

And,

$\displaystyle\sf{\longrightarrow 6x^2>1.}$

$\displaystyle\sf{\Longrightarrow x\in\left(-\infty,\ -\dfrac {1}{\sqrt6}\right)\cup\left (\dfrac {1}{\sqrt 6},\ \infty\right)\quad\quad\dots(3.2)}$

Then,

$\displaystyle\sf{\longrightarrow -\pi+\tan^{-1}\left (\dfrac {2x+3x}{1-6x^2}\right)=\dfrac {\pi}{4}}$

$\displaystyle\sf{\longrightarrow \dfrac {5x}{1-6x^2}=\dfrac {5\pi}{4}}$

But since $\displaystyle\sf {\dfrac {5\pi}{4}\in\left (-\dfrac {\pi}{4},\ \dfrac {\pi}{4}\right),}$

$\displaystyle\sf{\Longrightarrow x\in\phi\quad\quad\dots(3.3)}$

Taking $\displaystyle\sf {(3.1)\land(3.2)\land(3.3),}$ we get,

$\displaystyle\sf{\longrightarrow x\in\phi\quad\quad\dots(iii)}$

Finally, taking $\displaystyle\sf {(i)\lor(ii)\lor(iii),}$

$\displaystyle\sf {\longrightarrow\underline{\underline {x=\dfrac {1}{6}}}}$

Thus the equation has only one solution.

Answer 2

Answer:

$Given,</p><p></p><p>\displaystyle\sf{\longrightarrow \tan^{-1} (2x)+\tan^{-1}(3x)=\dfrac {\pi}{4}}⟶tan−1(2x)+tan−1(3x)=4π</p><p></p><p>Case 1:-</p><p></p><p>We have,</p><p></p><p>\displaystyle\sf{\longrightarrow \tan^{-1}a+\tan^{-1}b=\tan^{-1} (\dfrac {a+b}{1-ab})}⟶tan−1a+tan−1b=tan−1(1−aba+b)</p><p></p><p>iff \displaystyle\sf {a > 0,\ b > 0,\ ab < 1.}a>0, b>0, ab<1.</p><p></p><p>Let,</p><p></p><p>\displaystyle\sf {\longrightarrow2x > 0}⟶2x>0</p><p></p><p>\displaystyle\sf {\longrightarrow3x > 0}⟶3x>0</p><p></p><p>\displaystyle\sf {\Longrightarrow x\in(0,\ \infty)\quad\quad\dots (1.1)}⟹x∈(0, ∞)…(1.1)</p><p></p><p>And,</p><p></p><p>\displaystyle\sf{\longrightarrow 6x^2 < 1.}⟶6x2<1.</p><p></p><p>\displaystyle\sf{\Longrightarrow x\in (-\dfrac {1}{\sqrt 6},\ \dfrac {1}{\sqrt6})\quad\quad\dots(1.2)}⟹x∈(−61, 61)…(1.2)</p><p></p><p>Then,</p><p></p><p>\displaystyle\sf{\longrightarrow \tan^{-1} (\dfrac {2x+3x}{1-6x^2})=\dfrac {\pi}{4}}⟶tan−1(1−6x22x+3x)=4π</p><p></p><p>\displaystyle\sf{\longrightarrow \dfrac {5x}{1-6x^2}=\tan (\dfrac {\pi}{4})}⟶1−6x25x=tan(4π)</p><p></p><p>\displaystyle\sf{\longrightarrow \dfrac {5x}{1-6x^2}=1}⟶1−6x25x=1</p><p></p><p>\displaystyle\sf{\longrightarrow 6x^2+5x-1=0}⟶6x2+5x−1=0</p><p></p><p>\displaystyle\sf{\longrightarrow 6x^2+6x-x-1=0}⟶6x2+6x−x−1=0</p><p></p><p>\displaystyle\sf{\longrightarrow (x+1)(6x-1)=0}⟶(x+1)(6x−1)=0</p><p></p><p>\displaystyle\sf{\Longrightarrow x\in\{-1,\ \dfrac {1}{6}\}\quad\quad\dots (1.3)}⟹x∈{−1, 61}…(1.3)</p><p></p><p>Taking \displaystyle\sf {(1.1)\land (1.2)\land(1.3),}(1.1)∧(1.2)∧(1.3), we get,</p><p></p><p>\displaystyle\sf{\longrightarrow x=\dfrac {1}{6}\quad\quad\dots (i)}⟶x=61…(i)</p><p></p><p>Case 2:-</p><p></p><p>We have,</p><p></p><p>\displaystyle\sf{\longrightarrow \tan^{-1}a+\tan^{-1}b=\pi+\tan^{-1} (\dfrac {a+b}{1-ab})}⟶tan−1a+tan−1b=π+tan−1(1−aba+b)</p><p></p><p>iff \displaystyle\sf {a > 0,\ b > 0,\ ab > 1.}a>0, b>0, ab>1.</p><p></p><p>Let,</p><p></p><p>\displaystyle\sf {\longrightarrow2x > 0}⟶2x>0</p><p></p><p>\displaystyle\sf {\longrightarrow3x > 0}⟶3x>0</p><p></p><p>\displaystyle\sf {\Longrightarrow x\in(0,\ \infty)\quad\quad\dots (2.1)}⟹x∈(0, ∞)…(2.1)</p><p></p><p>And,</p><p></p><p>\displaystyle\sf{\longrightarrow 6x^2 > 1.}⟶6x2>1.</p><p></p><p>\displaystyle\sf{\Longrightarrow x\in(-\infty,\ -\dfrac {1}{\sqrt6})\cup (\dfrac {1}{\sqrt 6},\ \infty)\quad\quad\dots(2.2)}⟹x∈(−∞, −61)∪(61, ∞)…(2.2)</p><p></p><p>Then,</p><p></p><p>\displaystyle\sf{\longrightarrow \pi+\tan^{-1} (\dfrac {2x+3x}{1-6x^2})=\dfrac {\pi}{4}}⟶π+tan−1(1−6x22x+3x)=4π</p><p></p><p>\displaystyle\sf{\longrightarrow \dfrac {5x}{1-6x^2}=-\dfrac {3\pi}{4}}⟶1−6x25x=−43π</p><p></p><p>But since \displaystyle\sf {-\dfrac {3\pi}{4}\in (-\dfrac {\pi}{4},\ \dfrac {\pi}{4}),}−43π∈(−4π, 4π),</p><p></p><p>\displaystyle\sf{\Longrightarrow x\in\phi\quad\quad\dots(2.3)}⟹x∈ϕ…(2.3)</p><p></p><p>Taking \displaystyle\sf {(2.1)\land(2.2)\land(2.3),}(2.1)∧(2.2)∧(2.3), we get,</p><p></p><p>\displaystyle\sf{\longrightarrow x\in\phi\quad\quad\dots(ii)}⟶x∈ϕ…(ii)</p><p></p><p>Case 3:-</p><p></p><p>We have,</p><p></p><p>\displaystyle\sf{\longrightarrow \tan^{-1}a+\tan^{-1}b=-\pi+\tan^{-1} (\dfrac {a+b}{1-ab})}⟶tan−1a+tan−1b=−π+tan−1(1−aba+b)</p><p></p><p>iff \displaystyle\sf {a < 0,\ b < 0,\ ab > 1.}a<0, b<0, ab>1.</p><p></p><p>Let,</p><p></p><p>\displaystyle\sf {\longrightarrow2x < 0}⟶2x<0</p><p></p><p>\displaystyle\sf {\longrightarrow3x < 0}⟶3x<0</p><p></p><p>\displaystyle\sf {\Longrightarrow x\in(-\infty,\ 0)\quad\quad\dots (3.1)}⟹x∈(−∞, 0)…(3.1)</p><p></p><p>And,</p><p></p><p>\displaystyle\sf{\longrightarrow 6x^2 > 1.}⟶6x2>1.</p><p></p><p>\displaystyle\sf{\Longrightarrow x\in(-\infty,\ -\dfrac {1}{\sqrt6})\cup (\dfrac {1}{\sqrt 6},\ \infty)\quad\quad\dots(3.2)}⟹x∈(−∞, −61)∪(61, ∞)…(3.2)</p><p></p><p>Then,</p><p></p><p>\displaystyle\sf{\longrightarrow -\pi+\tan^{-1} (\dfrac {2x+3x}{1-6x^2})=\dfrac {\pi}{4}}⟶−π+tan−1(1−6x22x+3x)=4π</p><p></p><p>\displaystyle\sf{\longrightarrow \dfrac {5x}{1-6x^2}=\dfrac {5\pi}{4}}⟶1−6x25x=45π</p><p></p><p>But since \displaystyle\sf {\dfrac {5\pi}{4}\in (-\dfrac {\pi}{4},\ \dfrac {\pi}{4}),}45π∈(−4π, 4π),</p><p></p><p>\displaystyle\sf{\Longrightarrow x\in\phi\quad\quad\dots(3.3)}⟹x∈ϕ…(3.3)</p><p></p><p>Taking \displaystyle\sf {(3.1)\land(3.2)\land(3.3),}(3.1)∧(3.2)∧(3.3), we get,</p><p></p><p>\displaystyle\sf{\longrightarrow x\in\phi\quad\quad\dots(iii)}⟶x∈ϕ…(iii)</p><p></p><p>Finally, taking \displaystyle\sf {(i)\lor(ii)\lor(iii),}(i)∨(ii)∨(iii),</p><p></p><p>\displaystyle\sf {\longrightarrow\underline{\underline {x=\dfrac {1}{6}}}}⟶x=61</p><p></p><p>Thus the equation has</p><p></p><p>$