BITSAT Practice Papers

Question

Given that x is a real number such that $\sec x-\tan x = 2$ . Find $\sec x+\tan x.$

Possible Answers

Selected	Possible Answer
	$\frac {1}{2}$
	3
	7
	2

Question

Let $0^0 < \theta < 45^0$ Arrange

$t_1=(tan \theta)^{tan \theta}, \quad \quad \quad t_2=(tan \theta)^{cot \theta},$
$t_3=(cot \theta)^{tan \theta}, \quad \quad \quad t_4=(cot \theta)^{cot \theta},$

in decreasing order.

Possible Answers

Selected	Possible Answer

Question

Compute
$\sin \frac{ \pi}{12} , \cos \frac {\pi}{12},$ and $\tan \frac {\pi}{12};$

Possible Answers

Selected	Possible Answer
	$3- \sqrt{5}$
	$6- \sqrt{4}$
	$2- \sqrt{3}$
	$4- \sqrt{4}$

Question

Compute
$\cos^4 \frac {\pi}{24} - \sin^4 \frac {\pi}{24}$

Possible Answers

Selected	Possible Answer
	$\frac {\sqrt{2}+ \sqrt{6}}{6}$
	$\frac {\sqrt{4}+ \sqrt{6}}{4}$
	$\frac {\sqrt{7}+ \sqrt{6}}{4}$
	$\frac {\sqrt{2}+ \sqrt{6}}{4}$

Question

Compute
$\cos 36^0 - \cos 72^0$

Possible Answers

Selected	Possible Answer
	$\frac {1}{3}$
	$\frac {1}{5}$
	$\frac {1}{4}$
	$\frac {1}{2}$

Question

Compute

$\sin 10^0 \sin 50^0 \sin 70^0$

Possible Answers

Selected	Possible Answer
	$\frac {6}{9}$
	$\frac {2}{8}$
	$\frac {4}{8}$
	$\frac {1}{8}$

Question

Simplify the expression

$\sqrt{ \sin^4 x + 4 \cos ^2 x } - \sqrt{ \cos^4 x + 4 \sin^2 x }$

Possible Answers

Selected	Possible Answer
	$\cos^2 x -\sin^2 x = \cos 2x$
	$\cos^2 x -\sin^2 x = \cos 5x$
	$\cos^2 x -\sin^2 x = \cos 4x$
	$\cos^2 x -\sin^2 x = \cos 3x$

Question

Region R contains all the points (x, y) such that $x^2 + y^2 \underline < 100$ and $sin(x + y ) \underline >0.$ Find the area of regain of R.

Possible Answers

Selected	Possible Answer
	$70 \pi$
	$10 \pi$
	$30 \pi$
	$50 \pi$

Question

Let I denote the interval $[- \frac {\pi}{4}\frac {\pi}{4}]$ . Determine the funcation f defined on the interval [-1, 1] such that $f(\sin 2x) = \sin x + \cos x$ and simplify $f(\tan^2 x)$ for x in the interval I.

Possible Answers

Selected	Possible Answer
	$f(\tan^2 x) = \sqrt{ 1+ \tan ^2 x }= \cos x$
	$f(\tan^2 x) = \sqrt{ 1+ \tan ^2 x }= \sec x$
	$f(\tan^2 x) = \sqrt{ 1+ \tan ^2 x }= \sin x$
	$f(\tan^2 x) = \sqrt{ 1+ \tan ^2 x }= \tan x$