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:
$x_i$ | $y_i$ | $f(x_i)$ | $(y_i - f(x_i))^2$ |
---|---|---|---|
1 | 4 | $c$ | $(4-c)^2$ |
2 | 18 | $15+c$ | $(3-c)^2$ |
3 | 4 | $c$ | $(4-c)^2$ |
4 | -24 | $-27+c$ | $(3-c)^2$ |
5 | 3 | $c$ | $(3-c)^2$ |
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.
Maths-1-Week-4.pdf ↩︎