MCQ
If $\cos (\alpha - \beta ) = 1$ and $\cos (\alpha + \beta ) = \frac{1}{e}$, $ - \pi < \alpha ,\beta < \pi $, then total number of ordered pair of $(\alpha ,\beta )$ is
  • A
    $0$
  • B
    $1$
  • C
    $2$
  • $4$

Answer

Correct option: D.
$4$
d
(d) $ - 2\pi < \alpha - \beta < 2\pi $

$\cos (\alpha - \beta ) = 1$

==> $\alpha - \beta = 0$

==> $\alpha = \beta $$\cos 2\alpha = \frac{1}{e}$

and $ - 2\pi < 2\alpha < 2\pi $

Hence, there will be four solutions.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

$\mathop {\lim }\limits_{x \to \pi /6} \frac{{{{\cot }^2}\theta - 3}}{{{\rm{cosec}}\theta - 2}} = $
A line through $A( - 5, - \;4)$ meets the lines $x + 3y + 2 = 0,$ $2x + y + 4 = 0$ and $x - y - 5 = 0$ at $B, \,C$ and $D$ respectively. If ${\left( {\frac{{15}}{{AB}}} \right)^2} + {\left( {\frac{{10}}{{AC}}} \right)^2} = {\left( {\frac{6}{{AD}}} \right)^2},$ then the equation of the line is
Solution of $|3\text{x}+2| <1$ is:
A straight line $L$ at a distance of $4$ units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of $60^o$ with the line $x + y = 0$. Then an equation of the line $L$ is
Let $b, d>0$. The locus of all points $P(r, \theta)$ for which the line $P$ (where, $O$ is the origin) cuts the line $r \sin \theta=b$ in $Q$ such that $P Q=d$ is
If $\left(\frac{3^{6}}{4^{4}}\right) \mathrm{k}$ is the term, independent of $\mathrm{x}$, in the binomial expansion of $\left(\frac{\mathrm{x}}{4}-\frac{12}{\mathrm{x}^{2}}\right)^{12}$, then $\mathrm{k}$ is equal to ...... .
$\frac{{{C_0}}}{1} + \frac{{{C_2}}}{3} + \frac{{{C_4}}}{5} + \frac{{{C_6}}}{7} + ....$=
If $a_1 , a_2, a_3, . . . . , a_n, ....$ are in $A.P.$ such that $a_4 - a_7 + a_{10}\, = m$, then the sum of first $13$ terms of this $A.P.$, is .............. $\mathrm{m}$
The angle between the tangents drawn at the end points of the latus rectum of parabola ${y^2} = 4ax$, is
If the roots of equation $\frac{{{x^2} - bx}}{{ax - c}} = \frac{{m - 1}}{{m + 1}}$ are equal but opposite in sign, then the value of $m$ will be