Mathematics Week 4 Graded Assignment

Mathematics Week 4 Graded Assignment


Graded Assignment


Multiple Select Question (MSQ)

1. Consider the line $\left(L_{x}\right)$ and parabola $\left(P_{x}\right)$ as shown in below figure.

Which among the following represents the graph of $\frac{P_{x}}{L_{x}}$ ?

Answer: Option b

Solution:

The equation of $L_{x}$ from the graph is $f(x)=3x-9$.

The equation of $P_{x}$ from the graph is $f(x)=3x^{2}-18x+27$.

$$ \frac{P_{x}}{L_{x}}=\frac{3x^{2}-18x+27}{3x-9}=\frac{x^{2}-6x+9}{x-3}=\frac{(x-3)^{2}}{x-3}=x-3 $$

Therefore, the graph of $y=x-3$ is option 2.

2. Consider $f(x)=x^{3}-4x^{2}-17x+60$ and $g(x)=x^{3}+5x^{2}-8x-12$, whose one of the roots are given in the set ${3,2,-3,-2}$. Choose the set of correct options regarding $f(x)$ and $g(x)$.

Answer: Option b

Solution:

From the given set, $x=3$ is one of the root of $f(x)$, since $f(3)=0$. To find the other roots:

$$ \frac{x^{3}-4x^{2}-17x+60}{x-3}=x^{2}-x-20 \Rightarrow x=-4,5 $$

So, option a is incorrect.

From the given set, $x=2$ is one of the root of $g(x)$, since $g(2)=0$. To find the other roots:

$$ \frac{x^{3}+5x^{2}-8x-12}{x-2}=x^{2}+7x+6 \Rightarrow x=-6,-1 $$

So, option b is correct.

$$ f(x)=(x-3)(x+4)(x-5) $$$$ f(x)\text{ is positive when }x\in(-4,3)\cup(5,\infty) $$

Thus, option d is incorrect.

$$ g(x)\text{ is negative when }x\in(-\infty,-6)\cup(-1,-2) $$

So, option c and e are incorrect.

Numerical Answer Type (NAT)

3. Consider a rectangular box whose length is $x$, height is 3 units less than length, and breadth is 5 units more than height. If the volume of the rectangular box is 24 unit$^{3}$, then choose the correct value of $x$ from the given options.

Answer: Option b (which is 4, as seen from the solution)

Solution:

Let the length of rectangular box be $x$, then the height is $h=x-3$ and breadth is $b=h+5=x-3+5=x+2$.

The volume of rectangular box is:

$$ V=x(x-3)(x+2)=x^{3}-x^{2}-6x $$

Given that volume is 24:

$$ x^{3}-x^{2}-6x=24 $$

From the given options, option b (which is 4) satisfies the equation.

Multiple Select Question (MSQ)

4. Consider a polynomial function $f(x)$ of degree 4 which intersects the X-axis at $x=2, x=-3$ and $x=-4$. Moreover, $f(x)<0$ when $x\in(1,2)$, and $f(x)>0$ when $x\in(-1,1)$. Find out the equation of the polynomial.

Answer: Option b

$$ a(x^{4}+4x^{3}-7x^{2}-22x+24), a>0 $$

Solution:

Given that $f(x)$ has degree 4 and intersects the X-axis at $x=2, x=-3, x=-4$. Also, $f(x)<0$ when $x\in(1,2)$, and $f(x)>0$ when $x\in(-1,1)$. This means when $x=1, f(x)=0$.

Hence, the equation of $f(x)$ is:

$$ f(x)=a(x-2)(x+3)(x+4)(x-1), a>0 $$$$ =a(x^{4}+4x^{3}-7x^{2}-22x+24) $$

Hence, option b is correct.

5. Consider a polynomial function $f(x)=\frac{-1}{300}(x-2)^{2}(x-3)(x+1)^{2}(x+4)(x-5)$. Choose the correct set of options.

Answer: Option b, c, d and e

Solution:

From the above graph,

  • $f(x)$ has exactly 6 turning points. So, option b is correct.
  • In the interval $x\in(3,5), f(x)$ is increasing first and then decreasing. So, option d is correct.
  • The function $f(x)$ is negative when $x\in(-1,2)$. So, option e is correct.

Also, if we replace $x$ by $-x$ in $f(x)$, then $f(-x) \neq -f(x)$ or $f(-x) \neq f(x)$. So, $f(x)$ is neither even nor odd function.

6. Consider a polynomial function $P(x)=(x^{4}+4x^{3}+x+10)$ and $Q(x)=(x^{3}+2x^{2}-6)$. If $M(x)$ is the equation of the straight line passing through $(2, Q(2))$ and having slope 3, then find out the equation of $P(x)+M(x)Q(x)$.

Answer: Option A

$$ 4x^{4}+14x^{3}+8x^{2}-17x-14 $$

Solution:

$M(x)$ is the equation of the straight line passing through $(2, Q(2))=(2,10)$ and having slope 3.

$$ M(x)=3x+4 $$$$ P(x)+M(x)Q(x)=(x^{4}+4x^{3}+x+10)+(3x+4)(x^{3}+2x^{2}-6) $$$$ =(x^{4}+4x^{3}+x+10)+(3x^{4}+6x^{3}-18x+4x^{3}+8x^{2}-24) $$$$ =4x^{4}+14x^{3}+8x^{2}-17x-14 $$

Thus, option A is correct.

Numerical Answer Type (NAT)

7. A function $f(x)$ which is the best fit for the data given in the Table-1 recorded by a student, is

$$ f(x)=-(x-1)^{2}(x-3)(x-5)(x-7)+c $$

What will be the value of $c$, so that SSE (Sum Squared Error) will be minimum?

Answer: 3.4

Solution:

$x_i$$y_i$$f(x_i)$$(y_i - f(x_i))^2$
14$c$$(4-c)^2$
218$15+c$$(3-c)^2$
34$c$$(4-c)^2$
4-24$-27+c$$(3-c)^2$
53$c$$(3-c)^2$

SSE is:

$$ \sum_{i=1}^{5}(y_i - f(x_i))^2 = 3(3-c)^2 + 2(4-c)^2 $$$$ =3(9-6c+c^2) + 2(16-8c+c^2) = 27-18c+3c^2 + 32-16c+2c^2 $$$$ =5c^2-34c+59 $$

SSE is minimized at:

$$ c=\frac{34}{10}=3.4 $$

8. An ant named $B$, wants to climb an uneven cliff and reach its anthill (i.e., home of ant). On its way home, $B$ makes sure that it collects some food. A group of ants have reached the food locations which are at $x$-intercepts of the function $f(x)=(x^{2}-19)((x-9)^{3}-1)$. As ants secrete pheromones (a form of signals which other ants can detect and reach the food location), $B$ gets to know the food location. Then the sum of the $x$-coordinates of all the food locations is

Answer: 10

Solution:

$$ f(x)=(x^{2}-19)((x-9)^{3}-1)=0 $$$$ x^{2}-19=0 \Rightarrow x=\sqrt{19}, -\sqrt{19} $$$$ (x-9)^{3}-1=0 \Rightarrow x=10 $$

Sum of roots: $\sqrt{19} - \sqrt{19} + 10 = 10$.

9. The Ministry of Road Transport and Highways wants to connect three aspirational districts with two roads $r_1$ and $r_2$. Two roads are connected if they intersect. The shape of the two roads $r_1$ and $r_2$ follows polynomial curve $f(x)=(x-19)(x-17)^2$ and $g(x)=-(x-19)(x-17)$ respectively. What will be the $x$-coordinate of the third aspirational district, if the first two are at $x$-intercepts of $f(x)$ and $g(x)$.

Answer: 16

Solution:

Find $x$ such that $f(x)=g(x)$:

$$ (x-19)(x-17)^2=-(x-19)(x-17) $$$$ (x-19)(x-17)^2 + (x-19)(x-17) = 0 $$$$ (x-19)(x-17)(x-17+1) = 0 $$$$ (x-19)(x-17)(x-16) = 0 $$$$ x=19,17,16 $$

Hence, the third aspirational district is at $x=16$.

10. Let $r(x)$ be a polynomial function which is obtained as the quotient after dividing the polynomial $p(x)=(x+5)(x-3)(x^2-4)$ by the polynomial $q(x)=(x-2)(2+x)$. Choose the correct option which represents the polynomial $r(x)$ most appropriately.

Answer: Option 1

Solution:

$$ \frac{p(x)}{q(x)}=\frac{(x+5)(x-3)(x^2-4)}{(x-2)(2+x)}=\frac{(x+5)(x-3)(x+2)(x-2)}{(x-2)(2+x)}=(x+5)(x-3) $$

Therefore, $r(x)=(x+5)(x-3)$. Option 1 is correct.

This covers all questions and solutions from the PDF1.


  1. Maths-1-Week-4.pdf ↩︎