Let x be the length of a side, V be the volume, and s be the surface area of the cube.
Then, V = x3 and S = 6x2 where x is a function of time t.
It is given that
\begin{align} \frac{dV}{dt} = 8 cm^3 / s \end{align}
Then, by using the chain rule, we have:
∴ \begin{align} \frac{dV}{dt} = \frac{d}{dt} (x^3) . \frac{dx}{dt} = 3x^2 . \frac{dx}{dt} =8 \end{align}
⇒ \begin{align} \frac{dx}{dt} = \frac{8}{3 x^2} ……… (1) \end{align}
Now \begin{align} \frac{dS}{dt} = \frac{d}{dx}(6x^2).\frac{dx}{dt} [By Chain Rule] \end{align}
\begin{align} =12x .\frac{dx}{dt} =12x.(\frac{8}{3x^2}) = \frac{32}{x} \end{align}
Thus, when x = 12 cm, \begin{align} \frac{dS}{dt} = \frac{32}{12} cm^2 / s = 8 cm^2 / s \end{align}
Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of \begin{align} \frac{8}{3} cm^2 / s \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.
Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by
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).
Let f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.
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
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.
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