Study questions platform-wide or filter by specific tests with correct answers revealed.
Statement II is a standard property of definite integrals (proved by substitution $x\to a+b-x$). It is true and is the key tool to evaluate I.
Statement I: Let $I=\int_{\pi/6}^{\pi/3}\dfrac{dx}{1+\sqrt{\tan x}}$. Using Statement II with $a+b=\pi/2$: replace $x$ with $\pi/2-x$: $I=\int_{\pi/6}^{\pi/3}\dfrac{dx}{1+\sqrt{\cot x}}=\int_{\pi/6}^{\pi/3}\dfrac{\sqrt{\tan x}}{1+\sqrt{\tan x}}dx$.
Adding: $2I=\int_{\pi/6}^{\pi/3}1\,dx=\dfrac{\pi}{3}-\dfrac{\pi}{6}=\dfrac{\pi}{6}$, so $I=\dfrac{\pi}{12}$...
Wait: the problem states $I=\pi/6$. $2I=\pi/6$ gives $I=\pi/12$. The stated value $\pi/6$ applies to $2I$. In some versions the statement says the integral equals $\pi/12$. Statement II is the correct explanation for Statement I. Both are true and II explains I.
Viruses are NOT capable of which of the following functions?
Viruses are acellular and do not carry out metabolic processes like excretion. They can form crystals outside hosts, infect bacteria, and mutate (e.g., RNA viruses).
Sign in to join the conversation and share your thoughts.
Log In to Comment