Mathematics Week 8 Graded Assignment

Mathematics Week 8 Graded Assignment


Week 8 Graded Assignment


1. Multiple Select Questions (MSQ)

Question 1: Match the given functions in Column A with the equations of their tangents at the origin $(0,0)$ in column B and the plotted graphs and the tangents in Column C, given in Table M2W2G1.

Function (Column A)Tangent at $(0,0)$ (Column B)Graph (Column C)
i)$f(x)=x2^{x}$a) $y=-4x$1)
ii)$f(x)=x(x-2)(x+2)$b) $y=x$2)
iii)$f(x)=-x(x-2)(x+2)$c) $y=4x$3)

Options: Option 1: ii) $\rightarrow$ a) $\rightarrow 1$ Option 2: i) $\rightarrow$ b) $\rightarrow 3$ Option 3: iii) $\rightarrow$ b) $\rightarrow 1$ Option 4: iii) $\rightarrow$ c) $\rightarrow 2$ Option 5: i) $\rightarrow$ a) $\rightarrow 1$

Solution:

  • i): $f(x)=x2^{x}$, $f’(0)=1$ → tangent: $y=x$ → matches with b) and graph 3 i) $\rightarrow$ b $$\rightarrow 3
  • ii): $f(x)=x(x-2)(x+2)$, $f’(0)=-4$ → tangent: $y=-4x$ → matches with a) and graph 1 ii) $\rightarrow$ a) $$\rightarrow 1
  • iii): $f(x)=-x(x-2)(x+2)$, $f’(0)=4$ → tangent: $y=4x$ → matches with c) and graph 2

iii) $\rightarrow$ c) $$\rightarrow 2

Question 2: Consider the following two functions $f(x)$ and $g(x)$:

$$ f(x)= \begin{cases} \frac{x^{3}-9x}{x(x-3)} & \text{if } x \neq 0,3 \\ 3 & \text{if } x=0 \\ 0 & \text{if } x=3 \end{cases} $$$$ g(x)= \begin{cases} |x| & \text{if } x \leq 2 \\ \lfloor x \rfloor & \text{if } x>2 \end{cases} $$

Options: Option 1: $f(x)$ is discontinuous at both $x=0$ and $x=3$ Option 2: $f(x)$ is discontinuous only at $x=0$ Option 3: $f(x)$ is discontinuous only at $x=3$ Option 4: $g(x)$ is discontinuous at $x=2$ Option 5: $g(x)$ is discontinuous at $x=3$

Solution:

  • f(x): Continuous at $x=0$, discontinuous at $x=3$ → Option 3
  • g(x): Continuous at $x=2$, discontinuous at $x=3$ → Option 5

Question 3: Consider the graphs given below (Curve 1, Curve 2, Curve 3, Curve 4).

Options: Option 1: Curve 1 is both continuous and differentiable at the origin. Option 2: Curve 2 is continuous but not differentiable at the origin. Option 3: Curve 2 has derivative 0 at $x=0$. Option 4: Curve 3 is continuous but not differentiable at the origin. Option 5: Curve 4 is not differentiable anywhere. Option 6: Curve 4 has derivative 0 at $x=0$.

Solution:

  • Curve 1: Continuous and differentiable at origin → Option 1
  • Curve 2: Continuous and differentiable at origin, derivative 0 → Option 2, 3 (but note: Option 2 is false as per solution, but solution says Curve 2 is differentiable at origin with derivative 0. Likely only Option 3 is correct, but solution text is ambiguous. Most likely: Option 3.)
  • Curve 3: Continuous but not differentiable at origin → Option 4
  • Curve 4: Differentiable at $x=1$, not everywhere, but not differentiable at all points is false; option 5 is false. Option 6 is also false. Clarification: Based on solution, only Option 1 and Option 4 are correct, but solution text is unclear about Curve 2; likely Option 3 is also correct, but solution says Curve 2 is differentiable. User should verify with figure. However, as per solution: Curve 2 is differentiable at origin with derivative 0, so Option 3 is correct. Option 2 is incorrect (since Curve 2 is differentiable at origin). Option 4 is correct.

Question 4: Consider the function:

$$ f(x)= \begin{cases} \frac{\sin x}{x} & \text{if } x \neq 0 \\ 1 & \text{if } x=0 \end{cases} $$

Options: Option 1: $f(x)$ is not continuous at $x=0$ Option 2: $f(x)$ is continuous at $x=0$ Option 3: $f(x)$ is not differentiable at $x=0$ Option 4: $f(x)$ is differentiable at $x=0$ Option 5: The derivative of $f(x)$ at $x=0$ (if exists) is 0 Option 6: The derivative of $f(x)$ at $x=0$ (if exists) is 1

Solution:

  • f(x): Continuous at $x=0$ → Option 2
  • Differentiable at $x=0$, derivative is 0Option 4, 5

Question 5: Let $f$ be a polynomial of degree 5:

$$ f(x)=a_{5}x^{5}+a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0} $$

Let $f’(b)$ denote the derivative of $f$ at $x=b$.

Options: Option 1: $a_{1}=f’(0)$ Option 2: $5a_{5}+3a_{3}=\frac{1}{2}\left(f’(1)+f’(-1)-2f’(0)\right)$ Option 3: $4a_{4}+2a_{2}=\frac{1}{2}\left(f’(1)-f’(-1)\right)$ Option 4: None of the above

Solution:

  • Option 1: $a_{1}=f’(0)$ → True
  • Option 2: $5a_{5}+3a_{3}=\frac{1}{2}\left(f’(1)+f’(-1)-2f’(0)\right)$ → True
  • Option 3: $4a_{4}+2a_{2}=\frac{1}{2}\left(f’(1)-f’(-1)\right)$ → True
  • Option 4: False

2. Numerical Answer Type (NAT)

Question 6: Let $f$ be a differentiable function at $x=3$. The tangent line to the graph of the function $f$ at the point $(3,0)$, passes through the point $(5,4)$. What will be the value of $f’(3)$?
Solution: Slope of tangent = $\frac{4-0}{5-3}=2$. Answer: $f’(3)=2$
Question 7: Let $f$ and $g$ be two functions which are differentiable at each $x \in \mathbb{R}$. Suppose that, $f(x)=g(x^{2}+5x)$, and $f’(0)=10$. Find the value of $g’(0)$.
Solution: $f’(x)=g’(x^{2}+5x)(2x+5)$ At $x=0$, $f’(0)=g’(0)\times5=10$ Answer: $g’(0)=2$

3. Comprehension Type Questions

Question 8: Consider the following statements:

  • Statement P: Both the functions $p(t)$ and $q(t)$ are continuous.
  • Statement Q: $p(t)$ is continuous, but $q(t)$ is not.
  • Statement R: $q(t)$ is continuous, but $p(t)$ is not.
  • Statement S: Neither $p(t)$ nor $q(t)$ is continuous.

Given the functions:

$$ p(t)= \begin{cases} \frac{t^{3}-27}{t-3} & \text{if } 0 \leq t < 3 \\ 27 & t=3 \\ \frac{1}{e^{81}(t-3)}\left(e^{27t}-e^{81}\right) & \text{if } t>3 \end{cases} $$$$ q(t)= \begin{cases} (5t-9)^{\frac{1}{t-2}} & \text{if } 0 \leq t < 2 \\ e^{4} & t=2 \\ \frac{e^{t+2}-e^{4}}{t-2} & \text{if } t>2 \end{cases} $$

Find the number of the correct statements.

Solution:

  • p(t): Continuous at $t=3$
  • q(t): Not continuous at $t=2$ (left limit $e^5$, right limit $e^4$) Only Statement Q is correct. Answer: 1

Question 9: If $L_p(t)=At+B$ denotes the best linear approximation of the function $p(t)$ at the point $t=1$, then find the value of $2A+B$.

Solution: $p(t)=\frac{t^{3}-27}{t-3}$ for $0 \leq t < 3$ $p(1)=13$, $p’(1)=5$ Linear approximation: $L_p(t)=5t+8$ So, $A=5$, $B=8$ Answer: $2A+B=18$

Question 10: If $L_q(t)=e^{4}(At+B)+Ce^{5}$ denotes the best linear approximation of the function $q(t)$ at the point $t=3$, then find the value of $A+B+C$.
Solution: $q(3)=e^{5}-e^{4}$, $q’(3)=e^{4}$ Linear approximation: $L_q(t)=e^{4}(t-4)+e^{5}$ So, $A=1$, $B=-4$, $C=1$ Answer: $A+B+C=-2$

Question 11: Consider a function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined as:

$$ f(x)= \begin{cases} \frac{\sin 14x + A \sin x}{19x^{3}} & \text{if } x \neq 0 \\ B & \text{if } x=0 \end{cases} $$

If $f(x)$ is continuous at $x=0$, then find the value of $114B-A$.

Solution: For continuity, $A=-14$ $B=\frac{-2730}{114}$ Answer: $114B-A=-2716$

Question 12: The distance (in meters) traveled by a car after $t$ minutes is given by the function $d(t)=g(4t^{3}+2t^{2}+5t+2)$, where $g$ is a differentiable function with domain $\mathbb{R}$. Find the instantaneous speed of the car after 5 min, where $g’(577)=2$.

Solution: $d’(t)=g’(4t^{3}+2t^{2}+5t+2)(12t^{2}+4t+5)$ At $t=5$, $d’(5)=g’(577)\times325=2\times325=650$ Answer: 650

Question 13: Consider the functions:

$$ p(t)= \begin{cases} \frac{2e^{(t-2)}-2}{t-2} & \text{if } 0 \leq t < 2 \\ 2 & t=2 \\ 2(t^{2}-4)^{\frac{1}{\ln(t-2)}} & \text{if } t>2 \end{cases} $$$$ q(t)=|t(t-7)(t-8)| $$

Statements: P: Both $p(t)$ and $q(t)$ are continuous Q: Both are not differentiable R: $p(t)$ is continuous, $q(t)$ is differentiable S: $q(t)$ is continuous, $p(t)$ is not differentiable T: Neither is continuous

Find the number of correct statements.

Solution:

  • p(t): Not continuous at $t=2$
  • q(t): Continuous, but not differentiable at $x=0,7,8$ Only S is correct. Answer: 1
Question 14: If linear function $L_p(t)=At+B$ denotes the best linear approximation of the function $p(t)$ at the point $t=1$, find the value of $\frac{-2}{e^{-1}-1}(A+B)$.
Solution: $A+B=p(1)$ $p(1)=\frac{2e^{-1}-2}{-1}=2(1-e^{-1})$ $\frac{-2}{e^{-1}-1}(A+B)=4$ Answer: 4

Question 15: Consider the function:

$$ q(t)=|t(t-7)(t-8)| $$

If $m$ is slope of the tangent of the function $q(t)$ at point $t=\frac{3}{2}$, find the value $m-\frac{27}{4}$.

Solution: For $0 \leq t < 7$, $q(t)=t(t-7)(t-8)$ $q’(t)=3t^{2}-30t+56$ $q’\left(\frac{3}{2}\right)=\frac{27}{4}-45+56=\frac{27}{4}+11$ But solution says $m-\frac{27}{4}=11$ Answer: 11

This covers all questions and solutions from the provided PDF1.


  1. Week-8.pdf ↩︎