The number of elements in the set S=\ x ∈ ℤ : x^2-7 x+6 ≤ 0 and .x^2-3 x>0\ is

Correct answer: C. 3 Given S=\ x ∈ Z: x²-7 x+6 ≤ 0 and x^2-3 x>0\ ...(i) Now array l|l x²-7 x+6 ≤ 0 and and x^2-3 x>0 (x-6)(x-1) ≤ 0 and (x-0)(x-3)>0 x ∈[1,6] and x ∈(-∞,0) (3, ∞) array Hence equation (i) can be reduced to aligned and S=\ x ∈ Z: x ∈[1,6] ((-∞,0) (3, ∞)) and S=\ x ∈ Z: x ∈(3,6]\ and S=\ 4,5,6\ aligned Hence number of elements in set ' S ' are ' 3 '

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Sets and Relations

JEE Main · Mathematics · Sets and Relations

The number of elements in the set S=\ x ∈ ℤ : x^2-7 x+6 ≤ 0 and...

The number of elements in the set

S=\{x \in \mathbb{Z} : x^2-7 x+6 \leq 0\text{ and } \left.x^2-3 x>0\right\}

Options

A $\infty$
B $2$
C $3$
D $4$

Correct answer

C. $3$

Step-by-step solution

Given S=\left\{x \in Z: x²-7 x+6 \leq 0

\text { and }

x^2-3 x>0\right\} ...(i) Now

\begin{array}{l|l}x²-7 x+6 \leq 0 \text{ and } \text { and }x^2-3 x>0 \\ \Rightarrow(x-6)(x-1) \leq 0 \text{ and } \Rightarrow(x-0)(x-3)>0 \\ \Rightarrow x \in[1,6] \text{ and } \Rightarrow x \in(-\infty,0) \cup(3, \infty) \end{array}

Hence equation (i) can be reduced to

\begin{aligned} \text{ and } S=\{x \in Z: x \in[1,6] \cap((-\infty,0) \cup(3, \infty)) \\ \text{ and } \Rightarrow S=\{x \in Z: x \in(3,6]\} \\ \text{ and } \Rightarrow S=\{4,5,6\} \end{aligned}

Hence number of elements in set '

S

' are ' 3 '

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