Show that f : [–1, 1] → R, given by is one-one. Find the inverse of the function f : [–1, 1] → Range f.
(Hint: For y ∈ Range f, y =, for some x in [ - 1, 1], i.e.,
)
f: [ - 1, 1] → R is given as
Let f(x) = f(y).
∴ f is a one-one function.
It is clear that f: [ - 1, 1] → Range f is onto.
∴ f: [ - 1, 1]→ Range f is one-one and onto and therefore, the inverse of the function:
f: [ - 1, 1] → Range f exists.
Let g: Range f → [ - 1, 1] be the inverse of f.
Let y be an arbitrary element of range f.
Since f: [ - 1, 1] → Range f is onto, we have:
Now, let us define g: Range f → [ - 1, 1] as
∴gof =I[-1, 1]and fog = IRange f
∴ f - 1 = g
⇒
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.
Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.
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The order of the differential equation
\begin{align}2x^2\frac{d^2y}{dx^2}\;- \;3\frac{dy}{dx}\;+ y=\;0\end{align}
is (A) 2 (B) 1 (C) 0 (D) not defined
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y = ex +1 : yn -y' = 0
Answer the following as true or false.
\begin{align}(i) \overrightarrow{a}\; and\; \overrightarrow{-a}\; are\; collinear.\end{align}
(ii) Two collinear vectors are always equal in magnitude.
(iii) Two vectors having same magnitude are collinear.
(iv) Two collinear vectors having the same magnitude are equal.
WHERE IS THE RANGE OF THE FUNCTION