Linear Inequalities - Class 11 Mathematics - Chapter 5 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Linear Inequalities | NCERT | Mathematics | Class 11
If $a>b$ and $b>c$, then,
A) $a>c$
B) $a < c$ C) $\quad \mathrm{a} = \mathrm{c}$
D) None of these
The correct option is A) $a > c$.
Given that:
- $a > b$
- $b > c$
From these inequalities, it follows transitively that if $a$ is greater than $b$, and $b$ is greater than $c$, then $a$ must be greater than $c$. Hence, A) $a > c$ is the correct relation.
Find the common interval among $x<0, x<-2$ and $x \leq \frac{-5}{2}$.
A) $x \in (-\infty,\frac{-5}{2})$ B) $x \in (-\infty,\frac{-5}{2}]$
C) $x \in (0,\infty)$
D) $x \in (-\infty,-2]$
The correct answer is Option B: $x \in (-\infty,\frac{-5}{2}]$.
Step 1: Analyze the given intervals:
- The interval $x < 0$ represents all values less than 0.
- The interval $x < -2$ represents all values less than -2.
- The interval $x \leq \frac{-5}{2}$ represents all values less than or equal to $-\frac{5}{2}$.
Step 2: Find the intersection (common region) of these intervals:
- The interval $x < -2$ is a subset of the interval $x < 0$ since if $x$ is less than -2, it is also less than 0.
- To find the common region among the intervals, we consider the most restrictive condition, which here is $x \leq -\frac{5}{2}$. Note that $-\frac{5}{2}$ (or -2.5) is less restrictive than -2. Therefore, the interval $x \leq -\frac{5}{2}$ already ensures that $x$ is less than -2.
Step 3: Concluding the intersection: The values that are in all three intervals are those where $x \leq -\frac{5}{2}$. This region is represented by the interval $(-\infty, -\frac{5}{2}]$.
Conclusion: The common interval satisfying all three conditions is $x \leq -\frac{5}{2}$, and hence Option B is the correct choice.
Solve the following systems of inequalities graphically: $$ x + y \leq 6, x + y \geq 4 $$
Solution
We need to solve the system of inequalities graphically:
- First Inequality: $x + y \leq 6$
- Second Inequality: $x + y \geq 4$
Graphing Steps:
1. Graph the line $x + y = 6$:
We plot some points which satisfy $x + y = 6$:
- If $x = 3$, then $y = 6 - 3 = 3$.
- If $x = 4$, then $y = 6 - 4 = 2$.
Plot these points, (3, 3) and (4, 2), and draw a straight line through them.
Test a point to determine the region to shade:
- Use the origin (0, 0), we have $0 + 0 \leq 6$ which is true.
- Hence, shade the area including and below this line.
2. Graph the line $x + y = 4$:
We plot some points which satisfy $x + y = 4$:
- If $x = 2$, then $y = 4 - 2 = 2$.
- If $x = 1$, then $y = 4 - 1 = 3$.
Plot these points, (2, 2) and (1, 3), and draw a straight line through them.
Test a point to determine the region to shade:
- Use the origin (0, 0) again, we have $0 + 0 \geq 4$ which is false.
- Hence, shade the area above and including this line.
Result:
The solution to the system is the region where the two shaded areas overlap. This region includes points that satisfy both inequalities simultaneously. The overlapping shaded area represents all combinations of $x$ and $y$ that solve the system of inequalities $x + y \leq 6$ and $x + y \geq 4$.
In these questions, the relationship between different elements is shown in the statements. The statements are followed by two conclusions. Choose the correct option.
Statement: $$ A \geq B = C; B < D \leq E $$
Conclusions: $$ \begin{array}{l} \text{I. D > A} \ \text{II. E > C} \end{array} $$
A) Only conclusion I is true
B) Only conclusion II is true
C) Either conclusion I or II is true
D) Neither conclusion I nor II is true
E) Both conclusions I and II are true
Solution
The correct option is B) Only conclusion II is true.
Given statements: $$ A \geq B = C; \quad B < D \leq E $$
From the above statements, we establish:
- $A \geq B$
- $B = C$
- $B < D \leq E$
Combining these, we deduce: $$ A \geq C < D \leq E $$
Let's analyze the conclusions:
-
$D > A$: With $D$ being greater than $B$ and $B$ being at most equal to $A$, there isn't enough information to definitively state that $D > A$. Therefore, Conclusion I is false.
-
$E > C$: Since $C = B < D \leq E$, it follows directly that $C < E$. Thus, Conclusion II is true.
Hence, the correct answer is B) Only conclusion II is true.
$$ \begin{array}{l} y \leq 3x+1 \ x-y>1 \end{array} $$
Which of the following ordered pairs $(x, y)$ satisfies the system of inequalities above?
A. $(-2, -1)$
B. $(-1, 3)$
C. $(1, 5)$
D. $(2, -1)$
To determine which ordered pair $(x, y)$ satisfies the given system of inequalities: $$ y \leq 3x+1 \ x-y > 1 $$ We first test each of the given options by substituting the values of $x$ and $y$ into both inequalities.
-
Option A: $(-2, -1)$
- For $y \leq 3x+1$: $$ -1 \leq 3(-2) + 1 = -5 $$ This is true since $-1 \geq -5$.
- For $x-y>1$: $$ -2 - (-1) = -1 $$ This is false since $-1 > 1$ is not true.
Thus, option A does not satisfy both inequalities.
-
Option B: $(-1, 3)$
- For $y \leq 3x+1$: $$ 3 \leq 3(-1) + 1 = -2 $$ This is false.
- For $x-y>1$: $$ -1 - 3 = -4 $$ This is also false.
Option B fails to satisfy both inequalities.
-
Option C: $(1, 5)$
- For $y \leq 3x+1$: $$ 5 \leq 3(1) + 1 = 4 $$ This is false.
- For $x-y>1$: $$ 1 - 5 = -4 $$ This is false as well.
Option C does not satisfy either inequality.
-
Option D: $(2, -1)$
- For $y \leq 3x+1$: $$ -1 \leq 3(2) + 1 = 7 $$ This is true since $-1 \leq 7$.
- For $x-y>1$: $$ 2 - (-1) = 3 $$ This is true because $3 > 1$.
Option D satisfies both inequalities.
Conclusion: The correct choice is Option D: $(2, -1)$. Testing shows that this ordered pair is the only one that satisfies both inequalities in the system.
In these questions, the relationship between different elements is shown in the statements. The statements are followed by conclusions. Study the conclusions based on the given statements and select the appropriate answer.
Statements: $$ M>A \geq B=Q \leq P<J \leq Y ; Z \geq A>X $$
Conclusions: I. $\mathrm{B} < \mathrm{Y}$
II. $X \geq J$
A. Only conclusion I is true.
B. Neither conclusion I nor II is true.
C. Only conclusion II is true.
D. Both conclusions are true.
E. Either conclusion I or II is true.
The correct option is A. Only conclusion I is true.
Analyzing the statement: $$ M > A \geq B = Q \leq P < J \leq Y ; Z \geq A > X $$
Let's check the conclusions step-by-step,
Conclusion I: $B < Y$
From the statement, we know that since $B = Q \leq P < J \leq Y$, we can infer that: $$ Y \geq J > P \geq Q = B \quad \text{or simply} \quad Y > B $$ Therefore, Conclusion I is true.
Conclusion II: $X \geq J$
From the relationships in the statement, we see: $$ X < A \text{ and } A \geq B = Q \leq P < J; $$ which implies: $$ X < J $$ Thus, Conclusion II is false.
Therefore, the correct answer is A. Only conclusion I is true.
Solve the following inequality: $2y - 3 < y + 1 \leq 4y + 7$, where $y \in \mathbb{R}$.
A) $-2 \leq y < 4$ B) $-2 \leq y < 4$ C) $-2 < y < 4$ D) $-2 < y \leq 4$
The correct answer is option B: $-2 \leq y < 4$.
To solve the inequality $2y - 3 < y + 1 \leq 4y + 7$, we divide it into two parts for simplicity:
- $2y - 3 < y + 1$
- $y + 1 \leq 4y + 7$
For the first inequality: $$ \begin{align*} 2y - 3 &< y + 1 \ 2y - y &< 1 + 3 \ y &< 4 \end{align*} $$
For the second inequality: $$ \begin{align*} y + 1 &\leq 4y + 7 \ y - 4y &\leq 7 - 1 \ -3y &\leq 6 \ y &\geq -2 \end{align*} $$
Combining these results gives: $$ -2 \leq y < 4 $$
Thus, the interval of solutions for $y$ where $y$ is greater than or equal to $-2$ and less than $4$ matches option B: $-2 \leq y < 4$.
The solution set of the inequality $|x|-2|<|2-x|, x \in \mathbb{R}$ is
A $\emptyset$
B $(-\infty, 0)$
C $(-\infty, -2) \cup (2, \infty)$
D $(-\infty, -2)$
To solve the inequality $|x|-2|<|2-x|$, we first observe the absolute functions involved:
Let $f(x) = ||x| - 2|$ and $g(x) = |2 - x|$.
Expanding $f(x)$, we obtain: $$ f(x) = \left{ \begin{array}{ll} |2 - x|, & \text{if } x \geq 0 \ |2 + x|, & \text{if } x < 0 \end{array} \right. $$
For $x \geq 0$, we notice that: $$ f(x) = |2 - x| = g(x) $$ Thus, $f(x)$ equals $g(x)$ for all non-negative values of $x$.
For $x < 0$, $f(x)$ simplifies to: $$ f(x) = |2 + x| $$ Given the properties of absolute values, it can be shown that $|2 + x| < |2 - x|$ for $x < 0$.
Therefore,
- For $x \geq 0$, $f(x)$ and $g(x)$ are equal.
- For $x < 0$, $f(x) < g(x)$.
Hence, the solution to the inequality $||x|-2| <|2-x|$ is all $x$ such that $x < 0$. Following the answer choices provided:
Option $\mathbf{B} (-\infty, 0)$ is the correct answer. This interval represents all real numbers less than 0, which is consistent with our findings that $f(x) < g(x)$ for $x < 0$.
Solve the following inequality:
$$ \sqrt{-x^{2} + 6x - 5} > 8 - 2x $$
To solve the inequality
$$ \sqrt{-x^2 + 6x - 5} > 8 - 2x, $$
we need to go through several mathematical steps.
Step 1: Squaring Both Sides
First, we square both sides of the inequality to remove the square root:
$$ \left(\sqrt{-x^2 + 6x - 5}\right)^2 > (8 - 2x)^2 $$
This simplifies to:
$$ -x^2 + 6x - 5 > (8 - 2x)^2 $$
Step 2: Expanding and Simplifying
Next, we expand the right side:
$$ -x^2 + 6x - 5 > 64 - 32x + 4x^2 $$
We then bring all terms to one side of the inequality:
$$ -x^2 + 6x - 5 - 64 + 32x - 4x^2 > 0 $$
Combine like terms:
$$ -5x^2 + 38x - 69 > 0 $$
Multiply the entire inequality by -1 to make the leading coefficient positive (remember to reverse the inequality):
$$ 5x^2 - 38x + 69 < 0 $$
Step 3: Factorization
We factorize the quadratic expression:
$$ 5x^2 - 38x + 69 = (5x - 23)(x - 3) $$
Thus, the inequality becomes:
$$ (5x - 23)(x - 3) < 0 $$
Step 4: Solving the Inequality
We find the roots of the equation $ (5x - 23)(x - 3) = 0 $:
$$ x = 3 \quad \text{or} \quad x = \frac{23}{5} $$
This divides the number line into three intervals:
$ (-\infty, 3) $
$ (3, \frac{23}{5}) $
$ (\frac{23}{5}, \infty) $
We test intervals to determine where the product is negative. The viable interval is:
$$ 3 < x < \frac{23}{5} $$
Step 5: Considering the Original Inequality
We also need to consider the domain of $ \sqrt{-x^2 + 6x - 5} $:
$$ -x^2 + 6x - 5 \geq 0 $$
Solving $ -x^2 + 6x - 5 \geq 0 $, we factorize:
$$ -(x^2 - 6x + 5) \geq 0 \implies -(x-1)(x-5) \geq 0 $$
The quadratic inequality $(x-1)(x-5) \leq 0$ is true for:
$$ 1 \leq x \leq 5 $$
Step 6: Intersection of Intervals
Combining the intervals we get from the domain constraint and the solution to the quadratic inequality:
$$ 3 < x < 5 $$
Therefore, the solution to the inequality is:
$$ \boxed{(3, 5]} $$
A woman has 20m notes and 50n notes. The total amount of money in her possession cannot be less than $800$. Represent this as an inequality.
A $20m + 50n < 800$
B $20m + 50n \leq 800$
C $20m + 50n > 800$
D $20m + 50n \geq 800$
The correct option is $\mathbf{D}$
$$ 20m + 50n \geq 800 $$
The total value of the $m$ $20m$ notes and the $50n$ notes can be represented as:
$$ 20m + 50n $$
Given that the minimum amount of money the woman can have is $800$, we can set up the following inequality:
$$ 20m + 50n \geq 800 $$
Thus, the correct answer is $D$.
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