Standard +0.8 This question requires finding a tangent that passes through a specific point (the origin), which involves setting up and solving an equation relating the point of tangency to the gradient. Then it requires calculating an area between a curve and a line using integration. While the individual techniques are standard C3 content, the problem-solving aspect of finding the correct point of tangency and the multi-step nature (find point → verify → integrate → subtract areas) makes this moderately harder than average.
6
\includegraphics[max width=\textwidth, alt={}, center]{fc7679bf-a9a1-493d-bf89-35206382787f-3_576_821_258_662}
The diagram shows the curve with equation \(y = \sqrt { 3 x - 5 }\). The tangent to the curve at the point \(P\) passes through the origin. The shaded region is bounded by the curve, the \(x\)-axis and the line \(O P\). Show that the \(x\)-coordinate of \(P\) is \(\frac { 10 } { 3 }\) and hence find the exact area of the shaded region.
Make sound attempt at triangle area and calculate (triangle area) minus (their area under curve)
M1
or equiv
Obtain \(\frac{10}{3}\sqrt{5} - \frac{10}{3}\sqrt{5}\) and hence \(\frac{8}{5}\sqrt{5}\)
A1
9 marks: or exact equiv involving single term
Or:
Answer
Marks
Guidance
Arrange to \(x = \ldots\) and integrate to obtain \(k_1 y^3 + k_2 y\) form
*M1
Obtain \(\frac{1}{9}y^3 + \frac{3}{3}y\)
A1
Apply limits 0 and \(\sqrt{5}\)
M1
dep *M; the right way round
Make sound attempt at triangle area and calculate (their area from integration) minus (triangle area)
M1
Obtain \(\frac{30}{9}\sqrt{5} - \frac{3}{2}\sqrt{5}\) and hence \(\frac{8}{5}\sqrt{5}\)
A1
(9) or exact equiv involving single term
**Method 1:** (Differentiation; assume value $\frac{10}{3}$; eqn of tangent; through origin)
Differentiate to obtain $k(3x - 5)^{-\frac{1}{2}}$ | M1 | any constant $k$
Obtain $\frac{2}{3}(3x - 5)^{-\frac{1}{2}}$ | A1 | or equiv
Attempt to find equation of tangent at P and attempt to show tangent passing through origin | M1 | assuming value $\frac{10}{3}$; or equiv
Obtain $y = \frac{3}{2\sqrt{5}}x$ and confirm that tangent passes through O | A1 | AG; necessary detail needed
**Method 2:** (Differentiation; equate $\frac{y \text{ change}}{x \text{ change}}$ to deriv; solve for x)
Differentiate to obtain $k(3x - 5)^{-\frac{1}{2}}$ | M1 | any constant $k$
Obtain $\frac{2}{3}(3x - 5)^{-\frac{1}{2}}$ | A1 | or equiv
Equate $\frac{y \text{ change}}{x \text{ change}}$ to deriv and attempt solution | M1 |
Obtain $\frac{\sqrt{3x - 5}}{3} = \frac{2}{3}(3x - 5)^{-\frac{1}{2}}$ and solve to obtain $\frac{10}{3}$ only | A1 |
**Method 3:** (Differentiation; find x from $y = f'(x)$ and $y = \sqrt{3x - 5}$)
Differentiate to obtain $k(3x - 5)^{-\frac{1}{2}}$ | M1 | any constant $k$
Obtain $\frac{2}{3}(3x - 5)^{-\frac{1}{2}}$ | A1 | or equiv
State $y = \frac{2}{3}(3x - 5)^{-\frac{1}{2}} x$, $y = \sqrt{3x - 5}$, eliminate y and attempt solution | M1 | condone this attempt at 'eqn of tangent'
Obtain $\frac{10}{3}$ only | A1 |
**Method 4:** (No differentiation; general line through origin to meet curve at one point only)
Eliminate y from equations $y = kx$ and $y = \sqrt{3x - 5}$ and attempt formation of quadratic eqn | M1 |
Obtain $k^2 x^2 - 3x + 5 = 0$ | A1 | or equiv
Equate discriminant to zero to find k | M1 |
Obtain $k = \frac{1}{2\sqrt{5}}$ or equiv and confirm $x = \frac{10}{3}$ | A1 |
**Method 5:** (No differentiation; use coords of P to find eqn of OP; confirm meets curve once)
Use coordinates $(\frac{10}{3}, \sqrt{5})$ to obtain $y = \frac{3\sqrt{5}}{10}x$ or equiv as equation of OP | B1 |
Eliminate y from this eqn and eqn of curve and attempt quadratic eqn | M1 | should be $9x^2 - 60x + 100 = 0$ or equiv
Attempt solution or attempt discriminant | M1 |
Confirm $\frac{10}{3}$ only or discriminant = 0 | A1 |
Either:
Integrate to obtain $k(3x - 5)^{\frac{3}{2}}$ | *M1 | any constant $k$
Obtain correct $\frac{2}{5}(3x - 5)^{\frac{3}{2}}$ | A1 |
Apply limits $\frac{3}{5}$ and $\frac{10}{3}$ | M1 | dep *M; the right way round
Make sound attempt at triangle area and calculate (triangle area) minus (their area under curve) | M1 | or equiv
Obtain $\frac{10}{3}\sqrt{5} - \frac{10}{3}\sqrt{5}$ and hence $\frac{8}{5}\sqrt{5}$ | A1 | 9 marks: or exact equiv involving single term
**Or:**
Arrange to $x = \ldots$ and integrate to obtain $k_1 y^3 + k_2 y$ form | *M1 |
Obtain $\frac{1}{9}y^3 + \frac{3}{3}y$ | A1 |
Apply limits 0 and $\sqrt{5}$ | M1 | dep *M; the right way round
Make sound attempt at triangle area and calculate (their area from integration) minus (triangle area) | M1 |
Obtain $\frac{30}{9}\sqrt{5} - \frac{3}{2}\sqrt{5}$ and hence $\frac{8}{5}\sqrt{5}$ | A1 | (9) or exact equiv involving single term
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\includegraphics[max width=\textwidth, alt={}, center]{fc7679bf-a9a1-493d-bf89-35206382787f-3_576_821_258_662}
The diagram shows the curve with equation $y = \sqrt { 3 x - 5 }$. The tangent to the curve at the point $P$ passes through the origin. The shaded region is bounded by the curve, the $x$-axis and the line $O P$. Show that the $x$-coordinate of $P$ is $\frac { 10 } { 3 }$ and hence find the exact area of the shaded region.
\hfill \mbox{\textit{OCR C3 2011 Q6 [9]}}