Class JEE Mathematics Sets, Relations, and Functions Q #1051
KNOWLEDGE BASED
APPLY
4 Marks 2021 JEE Main 2021 (Online) 16th March Morning Shift MCQ SINGLE
The number of elements in the set {$x \in R : (|x| - 3) |x + 4| = 6$} is equal to :
(A) 4
(B) 2
(C) 3
(D) 1
Prev Next
Correct Answer: B
Explanation
We need to consider three cases based on the value of $x$:

Case 1: $x \le -4$
In this case, $|x| = -x$ and $|x+4| = -(x+4)$. The equation becomes:
$(-x - 3)(-x - 4) = 6$
$(x + 3)(x + 4) = 6$
$x^2 + 7x + 12 = 6$
$x^2 + 7x + 6 = 0$
$(x + 1)(x + 6) = 0$
$x = -1$ or $x = -6$
Since $x \le -4$, we only accept $x = -6$.

Case 2: $-4 < x < 0$
In this case, $|x| = -x$ and $|x+4| = x+4$. The equation becomes:
$(-x - 3)(x + 4) = 6$
$-x^2 - 4x - 3x - 12 = 6$
$-x^2 - 7x - 18 = 0$
$x^2 + 7x + 18 = 0$
The discriminant is $D = 7^2 - 4(1)(18) = 49 - 72 = -23 < 0$. So, there are no real solutions in this case.

Case 3: $x \ge 0$
In this case, $|x| = x$ and $|x+4| = x+4$. The equation becomes:
$(x - 3)(x + 4) = 6$
$x^2 + 4x - 3x - 12 = 6$
$x^2 + x - 18 = 0$
Using the quadratic formula, $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-18)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 72}}{2} = \frac{-1 \pm \sqrt{73}}{2}$
Since $x \ge 0$, we take the positive root: $x = \frac{-1 + \sqrt{73}}{2} \approx \frac{-1 + 8.54}{2} \approx 3.77$.
So, $x = \frac{-1 + \sqrt{73}}{2}$ is a valid solution.

Therefore, the solutions are $x = -6$ and $x = \frac{-1 + \sqrt{73}}{2}$. The number of elements in the set is 2.

More from this Chapter

NUMERICAL
Let $S=\left\{p_1, p_2 \ldots, p_{10}\right\}$ be the set of first ten prime numbers. Let $A=S \cup P$, where $P$ is the set of all possible products of distinct elements of $S$. Then the number of all ordered pairs $(x, y), x \in S$, $y \in A$, such that $x$ divides $y$, is ________ .
NUMERICAL
Let $A=\{1,2,3, \ldots \ldots \ldots \ldots, 100\}$. Let $R$ be a relation on $\mathrm{A}$ defined by $(x, y) \in R$ if and only if $2 x=3 y$. Let $R_1$ be a symmetric relation on $A$ such that $R \subset R_1$ and the number of elements in $R_1$ is $\mathrm{n}$. Then, the minimum value of $\mathrm{n}$ is _________.
NUMERICAL
Let $A=\{1,2,3, \ldots, 20\}$. Let $R_1$ and $R_2$ two relation on $A$ such that $R_1=\{(a, b): b$ is divisible by $a\}$ $R_2=\{(a, b): a$ is an integral multiple of $b\}$. Then, number of elements in $R_1-R_2$ is equal to _____________.
MCQ_SINGLE
Let the relations $R_1$ and $R_2$ on the set $X = \{1, 2, 3, ..., 20\}$ be given by $R_1 = \{(x, y) : 2x - 3y = 2\}$ and $R_2 = \{(x, y) : -5x + 4y = 0\}$. If $M$ and $N$ be the minimum number of elements required to be added in $R_1$ and $R_2$, respectively, in order to make the relations symmetric, then $M + N$ equals
MCQ_SINGLE
Let $X = R \times R$. Define a relation R on X as: $(a_1, b_1) R (a_2, b_2) \Leftrightarrow b_1 = b_2$ Statement I: $R$ is an equivalence relation. Statement II: For some $(a, b) \in X$, the set $S = \{(x, y) \in X : (x, y)R(a, b)\}$ represents a line parallel to $y = x$. In the light of the above statements, choose the correct answer from the options given below:
View All Questions