\begin{align} Let \;\; tan^{-1}(1)=x. \;\;Then\;\; tan x = 1 = tan\left(\frac{\pi}{4}\right).\end{align}
\begin{align} \therefore tan^{-1}(1)=tan\left(\frac{\pi}{4}\right)\end{align}
\begin{align} Let \;\;cos^{-1}\left(-\frac{1}{2}\right)=y. \;\;Then,\;\; cos y = -\frac{1}{2} = -cos\left(\frac{\pi}{3}\right)= cos\left(\pi - \frac{\pi}{3}\right) = cos\left(\frac{2\pi}{3}\right)\end{align}
\begin{align} \therefore cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}\end{align}
\begin{align} Let \;\; sin^{-1}\left(-\frac{1}{2}\right)=z. \;\;Then,\;\; sin z = -\frac{1}{2} = -sin\left(\frac{\pi}{6}\right)= sin\left(-\frac{\pi}{6}\right)\end{align}
\begin{align} \therefore sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\end{align}
\begin{align} \therefore tan^{-1} (1) + cos^{-1}\left(-\frac{1}{2}\right) + sin^{-1}\left(-\frac{1}{2}\right)\end{align}
\begin{align} =\frac{\pi}{4}+\frac{2\pi}{3}-\frac{\pi}{6}\end{align}
\begin{align} =\frac{3\pi + 8\pi -2\pi}{12}=\frac{9\pi}{12}=\frac{3\pi}{4}\end{align}
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f : R → R defined by f(x) = 3 – 4x
(ii) f : R → R defined by f(x) = 1 + x2
Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither oneone nor onto, where | x | is x, if x is positive or 0 and |x| is – x, if x is negative.
Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Check the injectivity and surjectivity of the following functions:
(i) f : N → N given by f(x) = x2
(ii) f : Z → Z given by f(x) = x2
(iii) f : R → R given by f(x) = x2
(iv) f : N → N given by f(x) = x3
(v) f : Z → Z given by f(x) = x3
Determine order and degree(if defined) of differential equation (ym)2 + (yn)3 + (y')4 + y5 =0
y = Ax : xy' = y (x ≠ 0)
The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(A) 10π (B) 12π (C) 8π (D) 11π