Question 3
$$ g'(x)=mx^2-10x-37 \text{ where } m \text{ is an integer} $$
The curve $y=g(x)$ passes through the point with coordinates $(1,20)$
Given that $(x-5)$ is a factor of $g(x)$
(a) show that $g(x)=2x^3-5x^2-37x+60$ (5)
(b) Hence, or otherwise, use algebra to solve the equation $g(x)=0$ (3)
Detailed Solution
We are given:
$$ g'(x)=mx^2-10x-37 \quad \text{where } m \text{ is an integer.} $$The curve $y=g(x)$ passes through the point $(1,20)$, and $(x-5)$ is a factor of $g(x)$.
(a) Show that $g(x)=2x^3-5x^2-37x+60$.
Step 1: Integrate $g'(x)$
Integrating $g'(x)$ with respect to $x$ gives:
$$ g(x) = \int (mx^2-10x-37)\,dx = \frac{m}{3}x^3-5x^2-37x+C, $$where $C$ is the constant of integration.
Step 2: Use the given point
Since $g(x)$ passes through $(1,20)$, substitute $x=1$ and $g(1)=20$:
$$ 20 = \frac{m}{3}(1)^3 - 5(1)^2 - 37(1) + C. $$Simplify:
$$ 20 = \frac{m}{3} - 5 - 37 + C. $$ $$ 20 = \frac{m}{3} - 42 + C. $$ $$ C = 62-\frac{m}{3}. \tag{1} $$Step 3: Use the factor theorem
Since $(x-5)$ is a factor of $g(x)$, substituting $x=5$ into $g(x)$ must yield $0$:
$$ g(5) = \frac{m}{3}(5)^3 - 5(5)^2 - 37(5) + C = 0. $$Simplify:
$$ \frac{m}{3}(125)-5(25)-37(5)+C=0. $$ $$ \frac{125m}{3}-125-185+C=0. $$ $$ \frac{125m}{3}-310+C=0. \tag{2} $$Step 4: Solve for $m$
Substitute $C=62-\frac{m}{3}$ from equation (1) into equation (2):
$$ \frac{125m}{3}-310+62-\frac{m}{3}=0. $$Simplify:
$$ \frac{124m}{3}-248=0. $$ $$ \frac{124m}{3}=248. $$ $$ m=6. $$Step 5: Substitute $m$ back into $g(x)$
Substituting $m=6$ into $g(x)$:
$$ g(x) = \frac{6}{3}x^3-5x^2-37x+C. $$ $$ g(x)=2x^3-5x^2-37x+C. $$Using $C=62-\frac{6}{3}=60$:
$$ g(x)=2x^3-5x^2-37x+60. $$Thus, we have shown that:
$$ g(x)=2x^3-5x^2-37x+60. $$(b) Solve the equation $g(x)=0$
We need to solve:
$$ 2x^3-5x^2-37x+60=0. $$Step 1: Use the factor theorem
Since $(x-5)$ is a factor, divide $g(x)$ by $(x-5)$.
Performing synthetic division: \[\begin{array}{r|rrrr} 5& 2&-5&-37&60\\ & &10&25&-60\\ \hline &2& 5&-12&0 \end{array}\]Thus:
$$ g(x)=(x-5)(2x^2+5x-12). $$Step 2: Solve the quadratic
Solve $2x^2+5x-12=0$ using the quadratic formula:
$$ x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} \quad \text{where } a=2,\; b=5,\; c=-12. $$ $$ x = \frac{-5\pm\sqrt{5^2-4(2)(-12)}}{2(2)}. $$ $$ x = \frac{-5\pm\sqrt{25+96}}{4}. $$ $$ x = \frac{-5\pm\sqrt{121}}{4}. $$ $$ x = \frac{-5\pm11}{4}. $$ $$ x=\frac{6}{4}=\frac{3}{2}, \quad x=\frac{-16}{4}=-4. $$Final Solution
The solutions to $g(x)=0$ are:
$$ x=5, \quad x=\frac{3}{2}, \quad x=-4. $$Question 4
The point $A$ with coordinates $(12,14)$ and the point $B$ with coordinates $(q,2)$ where $q$ is a constant, lie on the straight line with equation $3y-2x-p=0$ where $p$ is a constant.
(a) Find the value of $p$ and the value of $q$ (3)
The line $L$ is perpendicular to $AB$ and passes through the point $X$, which lies on $AB$ such that $AX:XB=1:2$
(b) Find an equation for $L$ in the form $ax+by+c=0$ where $a$, $b$ and $c$ are integers to be found. (6)
Solution:
Part (a): Find the value of $p$ and the value of $q$
We are given the points $A(12,14)$ and $B(q,2)$, and the line equation
$$ 3y-2x-p=0, $$where $p$ is a constant. Both points lie on this line, so we substitute their coordinates into the equation.
Step 1: Substitute the coordinates of point $A(12,14)$ into the equation.
$$ 3(14)-2(12)-p=0 $$ $$ 42-24-p=0 $$ $$ 18=p $$Thus, the value of $p$ is $18$.
Step 2: Substitute the coordinates of point $B(q,2)$ into the equation.
$$ 3(2)-2(q)-p=0 $$Substitute $p=18$:
$$ 6-2q-18=0 $$ $$ -2q-12=0 $$ $$ -2q=12 $$ $$ q=-6 $$Thus, the value of $q$ is $-6$.
Part (b): Find an equation for the line $L$
The line $L$ is perpendicular to the line $AB$ and passes through a point $X$ on $AB$, such that the ratio
$$ AX:XB=1:2. $$We need to find the equation of line $L$.
Step 1: Find the coordinates of point $X$ using the section formula.
The coordinates of point $X$ dividing the line segment $AB$ in the ratio $m:n$ are:
$$ x_X=\frac{mx_B+nx_A}{m+n}, \qquad y_X=\frac{my_B+ny_A}{m+n} $$Here,
$$ A(12,14), \qquad B(-6,2), \qquad m:n=1:2. $$Substituting the values:
$$ x_X = \frac{1(-6)+2(12)}{1+2} = \frac{-6+24}{3} = \frac{18}{3} = 6 $$ $$ y_X = \frac{1(2)+2(14)}{1+2} = \frac{2+28}{3} = \frac{30}{3} = 10 $$Thus, the coordinates of point $X$ are:
$$ (6,10). $$Step 2: Find the slope of line $AB$.
The slope of line $AB$ is:
$$ m_{AB} = \frac{y_B-y_A}{x_B-x_A} = \frac{2-14}{-6-12} = \frac{-12}{-18} = \frac{2}{3} $$Step 3: Find the slope of line $L$.
Since line $L$ is perpendicular to $AB$, its slope is the negative reciprocal of $\frac{2}{3}$:
$$ m_L=-\frac{3}{2} $$Step 4: Use the point-slope form of the line equation.
The point-slope form is:
$$ y-y_1=m(x-x_1) $$Substituting $m_L=-\frac{3}{2}$ and the point $(6,10)$:
$$ y-10=-\frac{3}{2}(x-6) $$Step 5: Simplify the equation.
$$ y-10=-\frac{3}{2}x+9 $$ $$ y=-\frac{3}{2}x+19 $$Multiply through by $2$:
$$ 2y=-3x+38 $$Rearranging into the form $ax+by+c=0$:
$$ 3x+2y-38=0 $$Thus, the equation of line $L$ is:
$$ \boxed{3x+2y-38=0} $$
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