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CBSE Board Examination 2024 · Section C · 3 marks

Q26: Is ∣x2−y2∣<8|x^2 - y^2| < 8 on {1,…,5}\{1, \ldots, 5\} reflexive, symmetric, transitive? or f(x)=ax+bf(x) = ax + b one-one and onto

Board 2024Ch 1 · Relations and Functions
The question

(a) A relation R on set A = {1,2,3,4,5}\{1, 2, 3, 4, 5\} is defined as R = {(x,y):∣x2−y2∣<8}\{(x, y) : |x^2 - y^2| < 8\}⁠. Check whether the relation R is reflexive, symmetric and transitive.

OR

(b) A function f is defined from R→RR \to R as f(x)=ax+bf(x) = ax + b⁠, such that f(1)=1f(1) = 1 and f(2)=3f(2) = 3⁠. Find function f(x)f(x)⁠. Hence, check whether function f(x)f(x) is one-one and onto or not.

Exam answer

(a)

Reflexive: For every x∈Ax \in A⁠, ∣x2−x2∣=0<8|x^2 - x^2| = 0 < 8⁠, so (x,x)∈R(x, x) \in R⁠. Thus R is reflexive.

Why?

Reflexive means every element of A is related to itself: (x,x)∈R(x, x) \in R for every xx in A.

For a pair (x,x)(x, x) the difference of squares is x2−x2=0x^2 - x^2 = 0⁠, and 0<80 < 8 is true for all five elements.

Symmetric: Let (x,y)∈R(x, y) \in R for some x,y∈Ax, y \in A⁠. Then ∣x2−y2∣<8|x^2 - y^2| < 8⁠. As ∣y2−x2∣=∣x2−y2∣|y^2 - x^2| = |x^2 - y^2|⁠, we get ∣y2−x2∣<8|y^2 - x^2| < 8⁠, so (y,x)∈R(y, x) \in R⁠. Thus R is symmetric.

Why?

Symmetric means: whenever (x,y)∈R(x, y) \in R⁠, the reversed pair (y,x)(y, x) is in R too.

The key fact is ∣−u∣=∣u∣\lvert -u\rvert = |u|⁠. Since y2−x2=−(x2−y2)y^2 - x^2 = -(x^2 - y^2)⁠, both orders give the same modulus, so the condition holds either way round.

Transitive: (1,2)∈R(1, 2) \in R as ∣12−22∣=3<8|1^2 - 2^2| = 3 < 8⁠, and (2,3)∈R(2, 3) \in R as ∣22−32∣=5<8|2^2 - 3^2| = 5 < 8⁠. But ∣12−32∣=8|1^2 - 3^2| = 8⁠, which is not less than 8, so (1,3)∉R(1, 3) \notin R⁠. Thus R is not transitive.

Why?

Transitive means: (x,y)∈R(x, y) \in R and (y,z)∈R(y, z) \in R always give (x,z)∈R(x, z) \in R⁠. To show that it fails, one counterexample is enough.

To find one, take two elements whose squares are far apart, joined through a middle element: 1 and 3 give ∣1−9∣=8|1 - 9| = 8⁠, while the steps 1 to 2 and 2 to 3 are each less than 8.

The three checks in full:

∣12−22∣=∣1−4∣=∣−3∣=3∣22−32∣=∣4−9∣=∣−5∣=5∣12−32∣=∣1−9∣=∣−8∣=8\begin{aligned} |1^2 - 2^2| &= |1 - 4| = |-3| = 3 \\ |2^2 - 3^2| &= |4 - 9| = |-5| = 5 \\ |1^2 - 3^2| &= |1 - 9| = |-8| = 8 \end{aligned}

Watch out: the condition is a strict inequality, and 8<88 < 8 is false.

Hence R is reflexive and symmetric, but not transitive.

(b)

f(1)=1⇒a+b=1f(1) = 1 \Rightarrow a + b = 1⁠, and f(2)=3⇒2a+b=3f(2) = 3 \Rightarrow 2a + b = 3⁠.

Why?

Put x=1x = 1 and x=2x = 2 into f(x)=ax+bf(x) = ax + b⁠: f(1)=a+bf(1) = a + b and f(2)=2a+bf(2) = 2a + b⁠. This gives two equations for the two unknowns.

Subtracting, a=2a = 2⁠; then b=1−a=−1b = 1 - a = -1⁠.

Why?

(2a+b)−(a+b)=3−1(2a + b) - (a + b) = 3 - 1 removes b and leaves a=2a = 2⁠. Then the first equation gives b=1−2=−1b = 1 - 2 = -1⁠.

(2a+b)−(a+b)=3−12a+b−a−b=2a=2a+b=12+b=1b=1−2=−1\begin{aligned} (2a + b) - (a + b) &= 3 - 1 \\ 2a + b - a - b &= 2 \\ a &= 2 \\[4pt] a + b &= 1 \\ 2 + b &= 1 \\ b &= 1 - 2 = -1 \end{aligned}

Thus a=2a = 2 and b=−1b = -1⁠. So f(x)=2x−1f(x) = 2x - 1⁠.

One-one: Let f(x1)=f(x2)f(x_1) = f(x_2) for some x1,x2∈Rx_1, x_2 \in R⁠.

2x1−1=2x2−1⇒x1=x22x_1 - 1 = 2x_2 - 1 \Rightarrow x_1 = x_2
Why?

One-one means equal outputs can come only from equal inputs. So we assume the outputs are equal and show the inputs must be equal: add 1 to both sides, then divide by 2.

2x1−1=2x2−12x1=2x2(add 1)x1=x2(divide by 2)\begin{aligned} 2x_1 - 1 &= 2x_2 - 1 \\ 2x_1 &= 2x_2 \quad (\text{add } 1) \\ x_1 &= x_2 \quad (\text{divide by } 2) \end{aligned}

Hence f is one-one.

Onto: Let y∈Ry \in R (codomain) and y=2x−1y = 2x - 1⁠. Then x=y+12∈Rx = \frac{y + 1}{2} \in R (domain), and

Why?

Onto means every element of the codomain is the image of some element of the domain. So we take any yy and solve y=2x−1y = 2x - 1 for xx⁠.

y=2x−1y+1=2x(add 1)y+12=x(divide by 2)\begin{aligned} y &= 2x - 1 \\ y + 1 &= 2x \quad (\text{add } 1) \\ \frac{y + 1}{2} &= x \quad (\text{divide by } 2) \end{aligned}

Watch out: we must check that this xx is in the domain. Here y+12\frac{y + 1}{2} is a real number for every real yy⁠, so it is.

f(y+12)=2⋅y+12−1=yf\left(\frac{y + 1}{2}\right) = 2\cdot\frac{y + 1}{2} - 1 = y
Why?

Put x=y+12x = \frac{y + 1}{2} into f(x)=2x−1f(x) = 2x - 1⁠. The 2 on top cancels the 2 below:

2⋅y+12−1=(y+1)−1=y\begin{aligned} 2\cdot\frac{y + 1}{2} - 1 &= (y + 1) - 1 \\ &= y \end{aligned}

So every yy in the codomain is an image. Hence f is onto.

∴ f(x)=2x−1\therefore\ f(x) = 2x - 1⁠, and f is one-one and onto.

Where this comes from

Chapter 1, Relations and Functions: Relation properties; one-one and onto function.

Question type: (a) A relation given by a rule; (b) Prove one-one and onto

How to spot it: click to see

(a) Type 2. A relation given by a rule: letters to prove, numbers to disprove

Spot it: RR is given by a condition, such as a≤ba \le b on R\R⁠, and the set is too big to list every pair.

Here: RR is given by a rule on ∣x2−y2∣|x^2 - y^2|⁠. Prove reflexive and symmetric with letters (∣x2−x2∣=0|x^2 - x^2| = 0⁠, and ∣y2−x2∣=∣x2−y2∣|y^2 - x^2| = |x^2 - y^2|⁠). Disprove transitive with numbers: (1,2),(2,3)∈R(1, 2), (2, 3) \in R⁠, but ∣12−32∣=8|1^2 - 3^2| = 8 is not less than 8.

(b) Type 5. Prove one-one and onto: the bijective function

Spot it: "show that ff is one-one and onto", or "is ff bijective?", with a formula like f(x)=3−4xf(x) = 3 - 4x⁠.

Here: First find f(x)=2x−1f(x) = 2x - 1 from f(1)=1f(1) = 1 and f(2)=3f(2) = 3⁠. Then 2x1−1=2x2−12x_1 - 1 = 2x_2 - 1 gives x1=x2x_1 = x_2⁠, and x=y+12x = \frac{y + 1}{2} is real for every yy⁠.

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