Week 5 - Graded Assignment - 5

Course: Jan 2026 - Mathematics I

Week 5 - Graded Assignment - 5

Last Submitted: You have last submitted on: 2026-03-20, 16:04 IST

Introduction

• There are some questions which have functions with discrete valued domains (such as day, month, year etc).
• For NAT type questions, enter only one right answer even if you get multiple answers for that particular question.
• Notations:
• R= Set of real numbers
• Q= Set of rational numbers
• Z= Set of integers
• N= Set of natural numbers
• The set of natural numbers includes 0.

Question 1

• For all values of $n$, $y$ is not a one-to-one function.
• For all values of $n$, $y$ is an injective function.
• $y$ is not a function.
• If $n$ is an even number, then $y$ is not an injective function.
• If $n$ is an odd number, then $y$ is an injective function.

Solution

Abstract Solution (Strategy)

1. [Injectivity & Parity]: Test the horizontal line test for even and odd power functions.
2. [Decision rule]: If $f(x) = f(-x)$, the function is not injective. Even powers are symmetric, odd powers are strictly monotonic.

Procedure

• Step 1 – Even Case: If $n=2$, $y=x^2$. $f(2)=4, f(-2)=4$. Since different inputs give the same output, it's NOT injective.
• Step 2 – Odd Case: If $n=3$, $y=x^3$. Every $x$ maps to a unique $y$ across $\mathbb{R}$. It passes the horizontal line test, so it IS injective.
• Result: Even $n \implies$ not injective; Odd $n \implies$ injective.

Question 2

• $\mathbb{R}$
• $\big((-\infty, 1] \cap [-2, \infty)\big) \cup (-\infty,-2)$
• $[1,\infty)$
• $\mathbb{R}\setminus (1,\infty)$

Solution

Abstract Solution (Strategy)

1. [Nested Domains]: The domain of $(g \circ f)$ is the set of $x$ such that $x \in Dom(f)$ and $f(x) \in Dom(g)$.
2. [Decision rule]: Identify the square root constraint ($1-x \ge 0$).

Procedure

• Step 1 – Internal f(x): $f(x) = \sqrt{1-x}$. Requires $1-x \ge 0 \implies x \le 1$. Domain is $(-\infty, 1]$.
• Step 2 – External g(x): $1-x^2$ is defined for all Reals. No additional constraints.
• Step 3 – Combine: The total domain is purely $x \le 1$.
• Step 4 – Set Notation: $\mathbb{R}\setminus (1,\infty)$ is equivalent to $(-\infty, 1]$.
• Result: $\mathbb{R}\setminus (1,\infty)$

Question 3

• $\mathbb{R}$
• $\mathbb{R}\setminus \{1\}$
• $[1, \infty)$
• $\mathbb{R}\setminus [1, \infty)$

Solution

Abstract Solution (Strategy)

1. [Inverse Domains]: The domain of $f^{-1}(x)$ is identical to the range of $f(x)$.
2. [Decision rule]: Determine the range of $x^3 + 1$.

Procedure

• Step 1 – Analyze f(x): $x^3+1$ is a cubic function.
• Step 2 – Range: Since $x^3$ goes from $-\infty$ to $+\infty$, adding 1 doesn't change this infinite span. Range is $\mathbb{R}$.
• Step 3 – Map: Domain of inverse $=$ Range of original $=\mathbb{R}$.
• Result: $\mathbb{R}$

Question 4

• (-1,-1)
• (1,-1)
• (0,0)
• (-2,-8)
• (1,1)
• (2,8)

Solution

Abstract Solution (Strategy)

1. [Identity Intersection]: A function and its inverse intersect at points where $f(x) = x$ (they meet on the line of reflection).
2. [Decision rule]: Solve $x^3 = x$.

Procedure

• Step 1 – Set Equation: $x^3 = x \implies x^3 - x = 0$.
• Step 2 – Factor: $x(x^2 - 1) = 0 \implies x(x-1)(x+1) = 0$.
• Step 3 – Values: $x = 0, 1, -1$.
• Step 4 – Points: Since they must lie on $y=x$, points are $(0,0), (1,1), (-1,-1)$.
• Result: (-1,-1), (0,0), (1,1)

Question 5

• $g(x)=f^{-1}(x)$
• $g(x)=f(x)$
• $g(x)=\sqrt[2]{(x-1)}$
• $g(x)=\sqrt[2]{(x+1)}$

Solution

Abstract Solution (Strategy)

1. [Mirror Reflection Mapping]: A graphical reflection bounded physically across the axis diagonal $y=x$ constitutes the absolute definition of an Inverse Function.
2. [Formula]: $y = f(x) \implies x = f^{-1}(y)$.
3. [Decision rule]: Notice the domain restriction $[0,\infty)$ legally forces the parabola into passing the horizontal line test, permitting a native inverse limit to structurally exist. Calculate it.

Procedure

• Step 1 – Inverse Identification: Reflecting the ant along $y=x$ signifies explicitly that the tracking curve $g(x)$ functionally equals $f^{-1}(x)$. (True)
• Step 2 – Algebraic inversion: Start with $y = x^2 + 1$.
• Step 3 – Isolate parameter: Substitute $x^2 = y - 1$.
• Step 4 – Extract limits: Given initial bounds strictly demanded $x \ge 0$, pulling the square root yields exclusively the positive limit string: $x = \sqrt{y-1}$.
• Step 5 – Remap to standard variables: $f^{-1}(x) = \sqrt{x-1}$. (True)
• Result: $g(x) = f^{-1}(x)$ and formally equates directly to $g(x) = \sqrt{x-1}$.

Question 6

• The minimum amount she should pay cannot be determined.
• The minimum amount she should pay is approximately ₹6,999.
• The amount after applying $D_2$ only is approximately ₹11,199.
• The amount after applying $D_1$ only is approximately ₹9,999.
• To pay minimum, Shalini should avail $D_1$ first then $D_2$.
• If Shalini avails $D_2$ first, she might not qualify for $D_1$.
• In order to pay minimum, Shalini should avail $D_2$ first then $D_1$.

Solution

Abstract Solution (Strategy)

1. [Composite Functions]: Compare $f(g(x))$ and $g(f(x))$.
2. [Decision rule]: Evaluate which sequence leads to the lower price. Note that $D_1$ only applies if the total is $>14999$.

Procedure

• Step 1 – Cart: Cart $X \ge 16000$.
• Step 2 – D1 only: Final $= 9999$.
• Step 3 – D2 only: Final $= 0.7 \times 16000 = 11200$. (Correction: at 17999, $0.7 \times 17999 = 12599$).
• Step 4 – D1 then D2: $(17999 \to 9999) \to 9999 \times 0.7 = 6999$.
• Step 5 – D2 then D1: $17999 \times 0.7 = 12599$. Since $12599 < 14999$, $D_1$ no longer applies! Final is $12599$.
• Result: $D_1$ then $D_2$ is optimal.

Question 7

• $f\circ h$ is not an injective function.
• ${f(f(h(x)))}\times h(x)=(x-1)^5$
• $h\circ f$ is not an injective function.
• There are two distinct solution for ${h(h(f(x)))}=0$.
• $f \circ h$ is an injective function.
• ${f(f(h(x)))}\times h(x)=(x-1)^4$.
• $h\circ f$ is an injective function.

Solution

Abstract Solution (Strategy)

1. [Injectivity Analysis]: Parabolas are not injective on $\mathbb{R}$.
2. [Algebraic Expansion]: $f(f(h(x))) \times h(x) = ((x-1)^2)^2 \times (x-1) = (x-1)^5$.
3. [Decision rule]: Identify the mathematical falsehoods.

Procedure

• Step 1 – Composition Check: $f(h(x)) = (x-1)^2$ and $h(f(x)) = x^2-1$. Both are quadratic parabolas. They are NOT injective. Therefore, saying they are injective is incorrect.
• Step 2 – Power Check: $((x-1)^2)^2 \times (x-1) = (x-1)^5$. Therefore, saying it equals $(x-1)^4$ is incorrect.
• Result: Select the injective claims and the incorrect power claim.

Question 8

• $g(x)$ may be the inverse of $f(x)$.
• $p(x)$ and $q(x)$ are even functions but $f(x)$ and $g(x)$ are neither.
• $q(x)$ could not be the inverse function of $p(x)$.
• $p(x), q(x)$ can be even degree and $f(x)$ can be odd degree.

Solution

Abstract Solution (Strategy)

1. [Geometric Parity]: Visual symmetry across Y-axis implies Even function.
2. [Inverses]: Reflection across $y=x$ implies Inverse. Function must be one-to-one to have an inverse.
3. [Decision rule]: $p, q$ are parabolic (even degree, even parity). $f, g$ are monotonic (odd degree).

Procedure

• Step 1 – Symmetries: $p, q$ are symmetrical about Y. They are even. $f, g$ are not.
• Step 2 – Degree: $p, q$ tails go same way (even degree). $f$ tails go opposite ways (odd degree).
• Step 3 – Inverse: $p$ is not 1-to-1 (it's even). Thus $q$ cannot be its inverse. Note: the accepted key says "q could not be the inverse" is a correct statement.
• Step 4 – Graph Reflection: $f$ and $g$ look like mirror images across $y=x$, so $g$ could be $f^{-1}$.
• Result: Even/Odd parity, degree types, and inverse feasibility are all correctly described.

Question 9

Solution

Abstract Solution (Strategy)

1. [Composite Inequality]: The input to $f$ must lie in its domain.
2. [Formula]: $-5 \le |2x + 5| \le 7$.
3. [Decision rule]: Solve the absolute value inequality.

Procedure

• Step 1 – Setup: $-5 \le |2x + 5| \le 7$.
• Step 2 – Abs Logic: $|2x+5| \ge -5$ is always true for all Reals.
• Step 3 – Upper Bound: $|2x+5| \le 7 \implies -7 \le 2x+5 \le 7$.
• Step 4 – Solve: Subtracting 5: $-12 \le 2x \le 2 \implies -6 \le x \le 1$.
• Step 5 – Sum: $a = -6, b = 1 \implies a+b = -5$.
• Result: -5

Question 10

Solution

Abstract Solution (Strategy)

1. [Exponential Bases]: Convert all bases to common primes (2 and 3).
2. [Decision rule]: Substitute $n = 4m$ into the expression.

Procedure

• Step 1 – Term 1: $\frac{16^m}{2^n} = \frac{(2^4)^m}{2^{4m}} = \frac{2^{4m}}{2^{4m}} = 1$.
• Step 2 – Term 2: $\frac{27^n}{9^{6m}} = \frac{(3^3)^{4m}}{(3^2)^{6m}} = \frac{3^{12m}}{3^{12m}} = 1$.
• Step 3 – Sum: $1 + 1 = 2$.
• Result: 2

Question 11

Solution

Abstract Solution (Strategy)

1. [Substitution]: Let $3^x = u$. This converts the exponential equation into a quadratic.
2. [Constraint]: Exponential outputs $3^x$ are always strictly positive.

Procedure

• Step 1 – Eq: $u^2 + u - 6 = 0$.
• Step 2 – Roots: $(u+3)(u-2) = 0 \implies u = -3$ or $u = 2$.
• Step 3 – Check: $3^x = -3$ (Impossible) or $3^x = 2$ (Possible).
• Result: Exactly 1 solution.

Question 12

• $f(x)$ is invertible.
• $g(x)$ is invertible.
• Domain of $(f \circ g)(x)$ is $(0, \infty)$.
• Domain of $(g \circ f)(x)$ is $(2, \infty)$.

Solution

Abstract Solution (Strategy)

1. [Injectivity]: Invertibility requires a 1-to-1 mapping.
2. [Decision rule]: Check for symmetry or absolute values that break injectivity.

Procedure

• Step 1 – f(x): $|x-1|$ is symmetric about $1$. E.g., $x=0$ and $x=2$ both give $\log(1)=0$. NOT injective.
• Step 2 – g(x): $g(x) = 3\log(x)$ for $x>0$. It is strictly increasing across its entire domain. IS injective.
• Step 3 – Result: $g(x)$ is invertible.

Question 13

• $f(x) \ge g(x)$ for all $x \ge x_0$.
• There exists $x_1 > x_0$ such that $f(x_1) = g(x_1)$.
• $g(x) \ge f(x)$ for all $x \ge x_0$.
• $g(x) \ge f(x)$ for all $x \le x_0$.
• $f(x) \ge g(x)$ for all $x \le x_0$.

Solution

Abstract Solution (Strategy)

1. [Monotonicity]: If $f$ is increasing, $f(x) > f(x_0)$ for $x > x_0$. If $g$ is decreasing, $g(x) < g(x_0)$ for $x > x_0$.
2. [Decision rule]: Compare the positions of $f$ and $g$ relative to their intersection point $x_0$.

Procedure

• Step 1 – Forward $(x > x_0)$: $f(x)$ is climbing above the meeting point, $g(x)$ is falling below it. Clear $f(x) > g(x)$.
• Step 2 – Backward $(x < x_0)$: $f(x)$ was lower before the meeting point, $g(x)$ was higher. Clear $g(x) > f(x)$.
• Result: $f \ge g$ for $x \ge x_0$ and $g \ge f$ for $x \le x_0$.