Question

A curve is such that $$\frac{\mathrm{d} y}{\mathrm{d} x}=x^{\frac{1}{2}}-x^{-\frac{1}{2}}$$ .The curve passes through the point $$\left ( 4,\frac{2}{3} \right )$$

(i)Find the equation of the curve .

(ii)Find $$\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}$$

(iii)Find the coordinates of the stationary point and determine its nature.

(i)$$y=\frac{2}{3}x^{\frac{3}{2}}-2x^{\frac{1}{2}}+c$$

$$\frac{2}{3}=\frac{16}{3}-4+c$$

$$c=-\frac{2}{3}$$

(ii)$$\frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{2}^{-\frac{3}{2}}$$

(iii)$$x^{\frac{1}{2}}-x^{-\frac{1}{2}}=0$$$$\rightarrow \frac{x-1}{\sqrt{x}}=0$$

x=1

When x=1,$$y=\frac{2}{3}-2-\frac{2}{3}=-2$$

When x=1 ,$$\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=1> 0$$ Hence minimum

Question

A curve has equation $$y=\frac{8}{x}+2x$$
(i) Find $$\frac{\mathrm{d} y}{\mathrm{d} x}$$ and $$\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}$$.

(ii) Find the coordinates of the stationary points and state, with a reason, the nature of each stationary point.

(i)$$\frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{8}{x^{2}}+2$$

$$\frac{\mathrm{d}^{2} y}{\mathrm{d} x^{2}}=\frac{16}{x^{3}}$$

(ii)$$-\frac{8}{x^{2}}+2=0\rightarrow 2x^{2}-8=0$$

$$x=\pm 2$$

$$y=\pm 8$$

$$\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}> 0$$  when x=2 hence MiNIMUM

$$\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}<0$$ when x=-2 MAXIMUM

#### Question.

The function f is defined for x≥ 0 by $$f(x) =(4x + 1)^\frac{3}{2}$$.

(i) Find f′(x) and f′′(x).

The first, second and third terms of a geometric progression are respectively f(2), f′(2) and kf′′(2).
(ii) Find the value of the constant k.

(i) $$f{}’\left ( x \right )=\left [ \frac{3}{2}\left ( 4x+1 \right )^{\frac{1}{2}} \right ]\left [ 4 \right ]$$

$$f{}”\left ( x \right )=6\times \frac{1}{2}\times \left ( 4x+1 \right )^{-\frac{1}{2}}\times 4$$

(ii) f(2),$$f{}’\left ( 2 \right )$$,$$kf{}”\left ( 2 \right )=27$$,4k OR 12

$$\frac{27}{18}=\frac{18}{4k}$$ oe OR $$kf{}”\left ( 2 \right )=12\Rightarrow k=3$$

Question

The function f is defined by f(x) = 2 − 3 cos x for 0 ≤ x ≤ 20.

(i) State the range of f.

(ii) Sketch the graph of y=f(x).

The function g is defined by g(x)= 2 − 3 cos x for 0 ≤ x ≤ p, where p is a constant.

(iii) State the largest value of p for which g has an inverse.

(iv) For this value of p, find an expression for $$g^{-1}(x)$$

(i)
−1 ⩽ f(x) ⩽ 5 or [–1, 5] (may use y or f instead of f(x)) (iii) (greatest value of p =)π

9(iv)
$$x = 2− 3cosx → cosx =1⁄3(2-x)$$

$$g^{-1}=cos^{-1}\frac{2-x}{3}$$

Question

A curve has equation $$y =(2x-1)^{-1}+2x$$

(i).Find $$\frac{\partial y}{\partial x}$$ and  $$\frac{\partial^2 y}{\partial x^2}$$

(ii) Find the x-coordinates of the stationary points and, showing all necessary working, determine
the nature of each stationary point.

(i).$$\frac{\partial y}{\partial x}= -2(2x-1)^{-2}+2$$

$$\frac{\partial^2 y}{\partial x^2}=8(2x-1)^{-3}$$

(ii)  Set $$\frac{\partial y}{\partial x}$$ to zero and attempt to solve  at least one correct step x = 0, 1

when x=0,$$\frac{\partial^2y}{\partial x^2}$$=-8(or<0),Hence MAX

when x=1,$$\frac{\partial^2y}{\partial x^2}$$=-8(or>0),Hence MIN

### Question.

The equation of a curve is $$y =(3 − 2x)^{3} + 24x$$.

(a) Find expressions for $$\frac{\partial y}{\partial x}$$ and  $$\frac{\partial^2 y}{\partial x^2}$$.

(b) Find the coordinates of each of the stationary points on the curve.

(a) $$\frac{dx}{dy}=3(3-2x)^{2}x-2+24=-6(3-2x)^{2}+24$$

(B1 without ×−2. B1 for ×−2)

$$\frac{d^{2}y}{dx^{2}}=-12(3-2x)\times -2=24(3-2x)$$

(B1FT from  $$\frac{dy}{dx}$$ without – 2)

(b) $$\frac{dy}{dx}=0 \ when \ 6(3-2x)^{2}=24\rightarrow 3-2x=\pm 2$$

x = ½,     y = 20 or     x = 2½,          y = 52
(A1 for both x values or a correct pair)

9 (c) If x = ½, $$\frac{d^{2}y}{dx^{2}}=48 \ Minimum$$

If x = 2½, $$\frac{d^{2}y}{dx^{2}}=-48 \ Minimum$$

### Question.

The equation of a curve is $$y =(3 − 2x)^{3} + 24x$$.

(a) Find expressions for $$\frac{\partial y}{\partial x}$$ and  $$\frac{\partial^2 y}{\partial x^2}$$.

(b) Find the coordinates of each of the stationary points on the curve.

(a) $$\frac{dx}{dy}=3(3-2x)^{2}x-2+24=-6(3-2x)^{2}+24$$

(B1 without ×−2. B1 for ×−2)

$$\frac{d^{2}y}{dx^{2}}=-12(3-2x)\times -2=24(3-2x)$$

(B1FT from  $$\frac{dy}{dx}$$ without – 2)

(b) $$\frac{dy}{dx}=0 \ when \ 6(3-2x)^{2}=24\rightarrow 3-2x=\pm 2$$

x = ½,     y = 20 or     x = 2½,          y = 52
(A1 for both x values or a correct pair)

9 (c) If x = ½, $$\frac{d^{2}y}{dx^{2}}=48 \ Minimum$$

If x = 2½, $$\frac{d^{2}y}{dx^{2}}=-48 \ Minimum$$

### Question

A curve has equation y = f(x), and it is given that f′ (x) $$= 2x^{2} – 7 – \frac{4}{x^{2}}$$

(a) Given that f(1) = $$-\frac{1}{3}$$, find f(x).

(b) Find the coordinates of the stationary points on the curve.

(c) Find f′′ (x).

(d) Hence, or otherwise, determine the nature of each of the stationary points.

(a)  $$f(x)=\frac{2}{3}x^{3} – 7x + 4x^{-1}[+c]$$

$$-\frac{1}{3} = \frac{2}{3} – 7 + 4 + c$$  leading to c = 

$$f(x)=\frac{2}{3}x^{3} – 7x + 4x^{-1} + 2$$

(b) 2x4 – 7x2 – 4  = 0

(2x2 + 1) (x2-4) [=0]

x = [±]

$$\left [ \frac{2}{3}\left ( 2 \right )^{3} -7(2)+\frac{4}{2}+2 leading to\right ] \left ( 2, -\frac{14}{2} \right )$$

$$\left [ \frac{2}{3}\left (-2 \right )^{3} -7(-2)+\frac{4}{-2}+2 leading to\right ] \left ( -2, \frac{26}{3} \right )$$

(c) f” (x) = 4x + 8x-3

(d) f” (2) = 9 > 0 MINIMUM at x = their 2

f”(-2) = -9< 0 MAXIMUM at x = their – 2

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