ECTION B: Attempt ANY TWO Questions QUESTION TWO (15 Marks) a) A particle moves along the x-axis in such a way that it's distance x metres from the origin after t seconds is given by x(t)=2t^3+6t^2-6t+1 Find the time when the particle's velocity and acceleration are equal. (4mks) b) If f(t)=2t^2,g(t)=sqrt (t) ,and h(t)=3t-1 .Find (i) g(h(f(t))) (2mks) (ii) f(g(h(t))) (2mks) c) Find (dy)/(dx) given y=3x^2cosx+tan3x (4mks) d) Show that the lim _(xarrow -2)(-x)/(x+2) does not exist UESTION THREE (15 Marks) a) Evaluate lim _(xarrow 0)u(x) given that (1-(x^4)/(3))leqslant u(x)leqslant (1+(x^2)/(2)) (4mks) b) Find the derivative of f(x)=(x^4-3x)/(x^4)+2 x=0 c) Find (dy)/(dx) given y=ln[(5x+9)(x^2+1)] d) A company sells xitems at a cost of Ksh. 9x. If the cost of producing x items is give by x^3-6x^2+15x . What production unit maximizes the profit? (4mks) (3mks) (0 (4mks)

QUESTION TWO:

a) To find the time when the particle's velocity and acceleration are equal, we need to find the velocity and acceleration functions first.

The velocity function is the first derivative of the position function with respect to time:
$$v(t) = \frac{dx}{dt} = \frac{d}{dt}(2t^3 + 6t^2 - 6t + 1) = 6t^2 + 12t - 6$$

The acceleration function is the first derivative of the velocity function with respect to time:
$$a(t) = \frac{dv}{dt} = \frac{d}{dt}(6t^2 + 12t - 6) = 12t + 12$$

To find the time when the velocity and acceleration are equal, we set $$v(t) = a(t)$$ and solve for $$t$$ :
$$6t^2 + 12t - 6 = 12t + 12$$
$$6t^2 - 18 = 0$$
$$t^2 - 3 = 0$$
$$t = \sqrt{3}$$ or $$t = -\sqrt{3}$$

Therefore, the particle's velocity and acceleration are equal at $$t = \sqrt{3}$$ seconds and $$t = -\sqrt{3}$$ seconds.

b) (i) To find $$g(h(f(t)))$$ , we first substitute $$f(t)$$ into $$h(t)$$ and then substitute the result into $$g(t)$$ :
$$g(h(f(t))) = g(h(2t^2)) = g(3(2t^2) - 1) = g(6t^2 - 1) = \sqrt{6t^2 - 1}$$

(ii) To find $$f(g(h(t)))$$ , we first substitute $$h(t)$$ into $$g(t)$$ and then substitute the result into $$f(t)$$ :
$$f(g(h(t))) = f(g(3t - 1)) = f(\sqrt{3t - 1}) = 2(\sqrt{3t - 1})^2 = 2(3t - 1) = 6t - 2$$

c) To find $$\frac{dy}{dx}$$ given $$y = 3x^2 \cos(x) + \tan(3x)$$ , we differentiate each term with respect to $$x$$ :
$$\frac{dy}{dx} = \frac{d}{dx}(3x^2 \cos(x)) + \frac{d}{dx}(\tan(3x))$$
$$\frac{dy}{dx} = 3(2x \cos(x) - x^2 \sin(x)) + 3 \sec^2(3x)$$

d) To show that $$\lim_{x \to -2} \frac{-x}{x+2}$$ does not exist, we need to analyze the behavior of the function as $$x$$ approaches $$-2$$ from both the left and right sides.

As $$x$$ approaches $$-2$$ from the left side, the denominator $$(x+2)$$ approaches $$0$$ from the negative side, making the fraction $$\frac{-x}{x+2}$$ approach positive infinity.

As $$x$$ approaches $$-2$$ from the right side, the denominator $$(x+2)$$ approaches $$0$$ from the positive side, making the fraction $$\frac{-x}{x+2}$$ approach negative infinity.

Since the function approaches different values from the left and right sides as $$x$$ approaches $$-2$$ , the limit does not exist.

QUESTION THREE:

a) To evaluate $$\lim_{x \to 0} u(x)$$ given that $$(1 - \frac{x^4}{3}) \leq u(x) \leq (1 + \frac{x^2}{2})$$ , we can use the Squeeze Theorem.

As $$x$$ approaches $$0$$ , $$(1 - \frac{x^4}{3})$$ approaches $$1$$ and $$(1 + \frac{x^2}{2})$$ approaches $$1$$ . Therefore, by the Squeeze Theorem, $$\lim_{x \to 0} u(x) = 1$$ .

b) To find the derivative of $$f(x) = \frac{x^4 - 3x}{x^4 + 2}$$ at $$x = 0$$ , we can use the quotient rule.

The derivative of $$f(x)$$ is given by:
$$f'(x) = \frac{(x^4 + 2)(4x^3 - 3) - (x^4 - 3x)(4x^3)}{(x^4 + 2)^2}$$

Substituting $$x = 0$$ into the derivative, we get:
$f'(0) = \frac{(0^4 + 2)(4 \cdot 0^3 - 3) - (0^4