19. Find the product of $$ f(x) $$ and $$ g(x) $$ when, $$ f(x)=x^{2}-3 x+5 $$ and $$ g(x)=x^{2}-5 x+6 $$ 20. If $$ f: A \longrightarrow B $$ be defines by $$ f(x)=2 x+3 $$, then find $$ f(0), f(1) $$, $$ f\left(\frac{1}{2} ight), f(-1) $$ Group- $$ D[4 imes 4=16] $$ 21. If a function $$ f $$ is defines as $$ f(x+2)=f(x)+f(2), x \longleftarrow-R $$ Show that i. $$ f(0)=0 $$ ii. $$ f(-2)=-f(2) $$ 22. Find out the co-ordinates of the points which divides internally the line joining the point $$ (2,1) $$ and $$ (6,7) $$ in the ratio( $$ 3: 2) $$. What will be the co-ordinate if it is divided externally? 23. The angles of a triangle are in ratio $$ 2: 3 $$ and the third is $$ \frac{4 \pi^{c}}{9} $$. Find the all angles in degrees. 24. ABC is a triangle with vertices $$ A(1,2), B(2,-1) $$ and $$ C(4,1) $$. Find the image $$ \Delta A^{I} B^{I} C^{I} $$ under enlargement $$ \in[0,2] $$ with centre at origin and scale factor 2 . A/so plot in graph

Question

19. Find the product of $$ f(x) $$ and $$ g(x) $$ when, $$ f(x)=x^{2}-3 x+5 $$ and $$ g(x)=x^{2}-5 x+6 $$ 20. If $$ f: A \longrightarrow B $$ be defines by $$ f(x)=2 x+3 $$, then find $$ f(0), f(1) $$, $$ f\left(\frac{1}{2}ight), f(-1) $$ Group- $$ D[4 	imes 4=16] $$ 21. If a function $$ f $$ is defines as $$ f(x+2)=f(x)+f(2), x \longleftarrow-R $$ Show that i. $$ f(0)=0 $$ ii. $$ f(-2)=-f(2) $$ 22. Find out the co-ordinates of the points which divides internally the line joining the point $$ (2,1) $$ and $$ (6,7) $$ in the ratio( $$ 3: 2) $$. What will be the co-ordinate if it is divided externally? 23. The angles of a triangle are in ratio $$ 2: 3 $$ and the third is $$ \frac{4 \pi^{c}}{9} $$. Find the all angles in degrees. 24. ABC is a triangle with vertices $$ A(1,2), B(2,-1) $$ and $$ C(4,1) $$. Find the image $$ \Delta A^{I} B^{I} C^{I} $$ under enlargement $$ \in[0,2] $$ with centre at origin and scale factor 2 . A/so plot in graph

UpStudy AI · Accepted Answer

1. **Product of $$ f(x) $$ and $$ g(x) $$**  
   Given  
   $$
   f(x)=x^2-3x+5,\quad g(x)=x^2-5x+6
   $$  
   The product is  
   $$
   f(x)\cdot g(x) = (x^2-3x+5)(x^2-5x+6)
   $$  
   Expand the product term‐by‐term:  
   $$
   \begin{aligned}
   (x^2-3x+5)(x^2-5x+6) &= x^2(x^2-5x+6) - 3x(x^2-5x+6) + 5(x^2-5x+6)\[1mm]
   &= \, x^4 - 5x^3 + 6x^2 - 3x^3 + 15x^2 -18x + 5x^2 -25x+30\[1mm]
   &= \, x^4 - (5x^3+3x^3) + (6x^2+15x^2+5x^2) - (18x+25x) +30\[1mm]
   &= \, x^4 - 8x^3 + 26x^2 - 43x + 30.
   \end{aligned}
   $$

2. **Evaluating $$ f(x)=2x+3 $$ at specific values**  
   Calculate:
   - For $$ x=0 $$:  
     $$
     f(0)=2(0)+3=3.
     $$
   - For $$ x=1 $$:  
     $$
     f(1)=2(1)+3=5.
     $$
   - For $$ x=\frac{1}{2} $$:  
     $$
     f\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)+3=1+3=4.
     $$
   - For $$ x=-1 $$:  
     $$
     f(-1)=2(-1)+3=-2+3=1.
     $$

3. **Properties of a function satisfying $$ f(x+2)=f(x)+f(2) $$**  
   (i) To show $$ f(0)=0 $$:  
   Let $$ x=0 $$ in the given equation:
   $$
   f(0+2)=f(0)+f(2) \quad \Longrightarrow \quad f(2)= f(0)+f(2).
   $$
   Subtract $$ f(2) $$ from both sides:
   $$
   f(0)=0.
   $$  
     
   (ii) To show $$ f(-2)=-f(2) $$:  
   Let $$ x=-2 $$ in the equation:
   $$
   f(-2+2)=f(-2)+f(2) \quad \Longrightarrow \quad f(0)= f(-2)+f(2).
   $$
   Since $$ f(0)=0 $$, we have:
   $$
   f(-2)+f(2)=0 \quad \Longrightarrow \quad f(-2)=-f(2).
   $$

4. **Dividing a line segment internally and externally**  
   Let the endpoints be $$ P(2,1) $$ and $$ Q(6,7) $$ and the ratio be $$ 3:2 $$.

- *Internal Division:*  
     The coordinates of the point dividing $$ PQ $$ internally in the ratio $$ 3:2 $$ are given by:
     $$
     \left(\frac{3\cdot6+2\cdot2}{3+2}, \frac{3\cdot7+2\cdot1}{3+2}\right) = \left(\frac{18+4}{5}, \frac{21+2}{5}\right)=\left(\frac{22}{5}, \frac{23}{5}\right).
     $$
     
   - *External Division:*  
     The coordinates for external division in the ratio $$ 3:2 $$ are given by:
     $$
     \left(\frac{3\cdot6-2\cdot2}{3-2}, \frac{3\cdot7-2\cdot1}{3-2}\right) = \left(\frac{18-4}{1}, \frac{21-2}{1}\right)=\left(14,19\right).
     $$

5. **Finding the angles of a triangle**  
   Two angles are in the ratio $$ 2:3 $$ and the third angle is given as $$ \frac{4\pi}{9} $$ radians. First, convert the third angle to degrees:
   $$
   \frac{4\pi}{9}\text{ radians} = \frac{4\pi}{9}\times\frac{180}{\pi} = \frac{720}{9}=80^\circ.
   $$
   Let the two angles be $$ 2k $$ and $$ 3k $$ in degrees. Since the sum of interior angles is $$ 180^\circ $$,
   $$
   2k+3k+80=180 \quad \Longrightarrow \quad 5k=100 \quad \Longrightarrow \quad k=20.
   $$
   Therefore, the angles are:
   $$
   2k=40^\circ,\quad 3k=60^\circ,\quad 80^\circ.
   $$

6. **Image of triangle $$ \Delta ABC $$ under enlargement with centre at the origin and scale factor 2**  
   Given vertices:  
   $$
   A(1,2),\quad B(2,-1),\quad C(4,1).
   $$
   The image of a point $$ (x,y) $$ under enlargement with centre at the origin and scale factor 2 is given by:
   $$
   (2x,2y).
   $$

Thus, the images are:
   - $$ A^I(2\cdot1,2\cdot2)=(2,4) $$,
   - $$ B^I(2\cdot2,2\cdot(-1))=(4,-2) $$,
   - $$ C^I(2\cdot4,2\cdot1)=(8,2) $$.

*For graphing:* Plot the original triangle with vertices at $$ (1,2) $$, $$ (2,-1) $$, and $$ (4,1) $$. Then plot the enlarged triangle with vertices at $$ (2,4) $$, $$ (4,-2) $$, and $$ (8,2) $$. Both triangles will share the property that the lines joining corresponding vertices pass through the origin.
1. **Product of $$ f(x) $$ and $$ g(x) $$:** $$ f(x) \cdot g(x) = x^4 - 8x^3 + 26x^2 - 43x + 30 $$ 2. **Evaluations of $$ f(x) $$:** - $$ f(0) = 3 $$ - $$ f(1) = 5 $$ - $$ f\left(\frac{1}{2}\right) = 4 $$ - $$ f(-1) = 1 $$ 3. **Function Properties:** - $$ f(0) = 0 $$ - $$ f(-2) = -f(2) $$ 4. **Division of Line Segment:** - **Internal Division:** $$ \left(\frac{22}{5}, \frac{23}{5}\right) $$ - **External Division:** $$ (14, 19) $$ 5. **Triangle Angles:** - $$ 40^\circ $$ - $$ 60^\circ $$ - $$ 80^\circ $$ 6. **Enlarged Triangle Coordinates:** - $$ A^I(2,4) $$ - $$ B^I(4,-2) $$ - $$ C^I(8,2) $$

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