How can I solve this question without using calculator and in only 2.5 minuites.

Question

This is a math subject GRE exam question, that I know how to solve but in more than 2.5 minuets and using a calculator, may be there is some intuition for solving this question that I do not know that makes its solution quicker, could anyone help me please?

Answer 1

$\sin x$ is increasing on the interval, starting at $0$ and ending at $1/\sqrt{2}$. $\cos x$ is decreasing on the interval, starting at $1$ and ending at $1/\sqrt{2}$, so $\cos x$ is bigger. Since $\cos x \leq 1, \cos^2x\leq\cos x$, so $\cos x$ is still bigger. Since $\cos 2x$ is decreasing from $1$ to $0$ on the same interval, it must be decreasing faster, so $\cos x$ is bigger than that. Lastly, since $\sin x \leq 1, \sin x \cos x \leq \cos x$, so yet again, $\cos x$ is bigger. Since $\cos x$ is strictly greater than all of the other functions on the given interval (endpoints notwithstanding), it's integral must be greater as well.

Answer 2

If $f(t) > g(t)$ whenever $a < t < b,$ then $$ \int_a^b f(t)\,dt > \int_a^b g(t)\,dt. $$

You can eliminate two of the functions immediately because $\sin t < 1$ and $\cos t < 1$ when $0 < t < \frac\pi4.$ To figure out which of the other three functions is greatest, spend a few seconds graphing all three from $0$ to $\frac\pi4$ on the same pair of axes.

Answer 3

Compare the values of $\cos t, \sin t, \cos^2 t, \sin^2 t, \cos 2t, \cos t*\sin t$. On $0 \le t \le \frac {\pi}4$.

$0 \le \sin t \le \frac {\sqrt{2}}2 \le \cos t \le 1$. So $\cos t > \sin t$.

$\sin t < 1 $ so $sin^2 t < \sin t$.

$\cos t \le 1$ (with equality only holding and $t = 0$) so $\cos^2 t \le \cos t$ with equality only holding for $t= 0$.

$\cos t$ is decreasing for $1$ to $\frac {\sqrt 2}{2}$ and $\cos 2t$ is decreasing faster from $1$ to $0$ so $\cos 2t \le \cos t$ (with equality only holding at $0$.

$\sin t < 1$ so $\sin t*\cos t < \cos t$. (Also $\cos t \le 1$ so $\sin t *\cos t \le \sin t < \cos t$.

For all points other that $t= 1$, $\cos t$ is the strictly largest function so the values over which the integral are taken are stricly larger so the integral of $\cos t$ is the largest value.

Answer 4

$\cos 2x$ is stretched along the $x-$axis, just like $\sin x \cos x=\dfrac{\sin 2x}{2}$. Thus they are smaller than $\cos$ and $\sin$.

Furthermore squaring numbers less than $1$ gives smaller numbers. So options $c$ to $e$ are out.

Recalling the graph of $\sin$ and $\cos$ is quite easy to answer that the larger is $\cos$.