\begin{align} y = xsinx:xy{'}=y +x\sqrt{x^2 -y^2}(x\neq0\; and\; x>y\; or\; x<-y)\end{align}
y= x.sinx
Differentiating both sides of this equation with respect to x, we get:
\begin{align} y^{'} =\frac{d}{dx}\left(x.sinx\right)\end{align}
\begin{align}\Rightarrow y^{'} =sinx. \frac{d}{dx}\left(x\right)+ x. \frac{d}{dx}\left(sinx\right)\end{align}
\begin{align} \Rightarrow y^{'} =sinx + x.cosx\end{align}
Differentiating both sides of this equation with respect to x, we get:
L.H.S. =xy' = x(sinx + xcosx)
\begin{align} =x.sinx + x^2.cosx\end{align}
\begin{align} =y + x^2.\sqrt{1-sin^2x}\end{align}
\begin{align} =y + x^2.\sqrt{1-\left(\frac{y}{x}\right)^2}\end{align}
\begin{align} =y + x^2.\sqrt{\frac{x^2-y^2}{x^2}}\end{align}
\begin{align} =y + x.\sqrt{x^2-y^2}\end{align}
R.H.S.
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.
The total revenue in Rupees received from the sale of x units of a product is given by
R (x) = 13x2 + 26x + 15
Find the marginal revenue when x = 7.
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
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
The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by
Maximise Z = 3x + 4y
Subject to the constraints:x + y ≤ 4, x ≥ 0, y ≥ 0
Let f: X → Y be an invertible function. Show that the inverse of f –1 is f, i.e., (f–1)–1 = f.