\begin{align} Let\;\; cos^{-1}\left(-\frac{1}{\sqrt2}\right)=y, \;\;Then,\;\; cos y = -\frac{1}{\sqrt2} = - cos\left(\frac{\pi}{4}\right)=cos\left(\pi - \frac{\pi}{4}\right) = cos\left(\frac{3\pi}{4}\right)\end{align}
We know that the range of the principal value branch of cos−1 is
\begin{align} \left[0,\pi\right] and \;\;cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt2}\end{align}
Therefore, the principal value of
\begin{align} cos^{-1}\left(-\frac{1}{\sqrt2}\right) is \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.
If f: R → R be given by f(x) = , then fof(x) is
(A)
(B) x3
(C) x
(D) (3 – x3).
Determine order and degree(if defined) of differential equation ym + 2yn + y' =0
Let f : X → Y be an invertible function. Show that f has unique inverse.
(Hint: suppose g1 and g2 are two inverses of f. Then for all y ∈ Y, fog1(y) = 1Y(y) = fog2(y). Use one-one ness of f).
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.