y = x2 + 2x + C : y' - 2x - 2 = 0
y = x2 + 2x + C
Differentiating both sides of this equation with respect to x, we get:
\begin{align}y^{'}=\frac{d}{dx}(x^2 + 2x + C)\end{align}
=> y' = 2x + 2
Substituting the value of y' in the given differential equation, we get:
L.H.S. = y' - 2x - 2 = 2x + 2 - 2x - 2 = 0 = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
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.
Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function.
Determine order and degree(if defined) of differential equation \begin{align}\frac{d^2y}{dx^2}=\cos3x + sin3x\end{align}
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.