CBSE Class 11 Maths Chapter 6 Linear Inequalities NCERT Solutions

Solution:

We are asked to solve the compound inequality $1\; \backslash frac\{4\}\{x-1\}\; \backslash le\; 3\; \backslash le\; \backslash frac\{6\}\{x+1\}$ with the condition $x\; >\; 0$. This can be broken down into two separate inequalities:

1. $\backslash frac\{4\}\{x-1\}\; \backslash le\; 3$
2. $3\; \backslash le\; \backslash frac\{6\}\{x+1\}$

Let's solve the first inequality:

$\backslash frac\{4\}\{x-1\}\; \backslash le\; 3$

Subtract 3 from both sides:

$\backslash frac\{4\}\{x-1\}\; -\; 3\; \backslash le\; 0$

$\backslash frac\{4\; -\; 3(x-1)\}\{x-1\}\; \backslash le\; 0$

$\backslash frac\{4\; -\; 3x\; +\; 3\}\{x-1\}\; \backslash le\; 0$

$\backslash frac\{7\; -\; 3x\}\{x-1\}\; \backslash le\; 0$

For this inequality to hold, the numerator and denominator must have opposite signs, or the numerator is zero. The critical points are $x\; =\; 7/3$ (where numerator is 0) and $x\; =\; 1$ (where denominator is 0).

Considering the condition $x\; >\; 0$ and the critical points, we test intervals:

• If : $\backslash frac\{7\; -\; 3x\}\{x-1\}$ is $\backslash frac\{+\}\{-\}$ = negative. This satisfies the inequality.
• If : $\backslash frac\{7\; -\; 3x\}\{x-1\}$ is $\backslash frac\{+\}\{+\}$ = positive. This does not satisfy the inequality.
• If $x\; >\; 7/3$: $\backslash frac\{7\; -\; 3x\}\{x-1\}$ is $\backslash frac\{-\}\{+\}$ = negative. This satisfies the inequality.

So, from the first inequality, we get $x\; \backslash in\; (0,\; 1)\; \backslash cup\; [7/3,\; \backslash infty)$.

Now let's solve the second inequality:

$3\; \backslash le\; \backslash frac\{6\}\{x+1\}$

Subtract $6/(x+1)$ from both sides:

$3\; -\; \backslash frac\{6\}\{x+1\}\; \backslash le\; 0$

$\backslash frac\{3(x+1)\; -\; 6\}\{x+1\}\; \backslash le\; 0$

$\backslash frac\{3x\; +\; 3\; -\; 6\}\{x+1\}\; \backslash le\; 0$

$\backslash frac\{3x\; -\; 3\}\{x+1\}\; \backslash le\; 0$

The critical points are $x\; =\; 1$ (where numerator is 0) and $x\; =\; -1$ (where denominator is 0).

Considering the condition $x\; >\; 0$ and the critical points, we test intervals:

• If : $\backslash frac\{3x\; -\; 3\}\{x+1\}$ is $\backslash frac\{-\}\{+\}$ = negative. This satisfies the inequality.
• If $x\; >\; 1$: $\backslash frac\{3x\; -\; 3\}\{x+1\}$ is $\backslash frac\{+\}\{+\}$ = positive. This does not satisfy the inequality.

So, from the second inequality, we get $x\; \backslash in\; (0,\; 1]$.

To satisfy the original compound inequality, x must satisfy both conditions. We need to find the intersection of $(0,\; 1)\; \backslash cup\; [7/3,\; \backslash infty)$ and $(0,\; 1]$.

The intersection is $(0,\; 1)$.

Final Answer: The solution set is $x\; \backslash in\; (0,\; 1)$.

Solution:

Let $y\; =\; |x-2|$. Substituting this into the inequality, we get:

$$\backslash frac\{y-1\}\{y-2\}\; \backslash le\; 0$$

This inequality holds when the numerator and denominator have opposite signs. The critical points are $y=1$ (where the numerator is zero) and $y=2$ (where the denominator is zero).

We analyze the sign of $\backslash frac\{y-1\}\{y-2\}$ in different intervals for y:

• If : $y-1$ is negative, $y-2$ is negative. The fraction is positive.
• If : $y-1$ is non-negative, $y-2$ is negative. The fraction is non-positive. This interval satisfies the inequality.
• If $y\; >\; 2$: $y-1$ is positive, $y-2$ is positive. The fraction is positive.

Thus, the solution for y is .

Now, substitute back $y\; =\; |x-2|$:

This compound inequality can be split into two inequalities:

1. $|x-2|\; \backslash ge\; 1$

Solving the first inequality, $|x-2|\; \backslash ge\; 1$:

This means $x-2\; \backslash ge\; 1$ or $x-2\; \backslash le\; -1$.

$x\; \backslash ge\; 3$ or $x\; \backslash le\; 1$.

Solving the second inequality, :

This means .

Adding 2 to all parts, we get .

We need to find the values of x that satisfy both conditions: ($x\; \backslash le\; 1$ or $x\; \backslash ge\; 3$) AND ().

Let's find the intersection:

• Intersection of $x\; \backslash le\; 1$ and is $(0,\; 1]$.
• Intersection of $x\; \backslash ge\; 3$ and is $[3,\; 4)$.

Combining these two intervals, the solution is $x\; \backslash in\; (0,\; 1]\; \backslash cup\; [3,\; 4)$.

Final Answer: The solution set is $x\; \backslash in\; (0,\; 1]\; \backslash cup\; [3,\; 4)$.

Solution:

We are given the inequality $\backslash frac\{1\}\{|x|-3\}\; \backslash le\; \backslash frac\{1\}\{2\}$. First, we must note that $|x|-3\; \backslash neq\; 0$, which means $|x|\; \backslash neq\; 3$, so $x\; \backslash neq\; 3$ and $x\; \backslash neq\; -3$.

Subtract $1/2$ from both sides to get zero on one side:

$\backslash frac\{1\}\{|x|-3\}\; -\; \backslash frac\{1\}\{2\}\; \backslash le\; 0$

Find a common denominator:

$\backslash frac\{2\; -\; (|x|-3)\}\{2(|x|-3)\}\; \backslash le\; 0$

$\backslash frac\{2\; -\; |x|\; +\; 3\}\{2(|x|-3)\}\; \backslash le\; 0$

$\backslash frac\{5\; -\; |x|\}\{2(|x|-3)\}\; \backslash le\; 0$

Since the factor of 2 in the denominator is positive, it does not affect the sign of the inequality. So we have:

$\backslash frac\{5\; -\; |x|\}\{|x|-3\}\; \backslash le\; 0$

This inequality holds when the numerator and denominator have opposite signs, or when the numerator is zero.

The critical points are where the numerator $5\; -\; |x|\; =\; 0$ (which means $|x|\; =\; 5$, so $x\; =\; 5$ or $x\; =\; -5$) and where the denominator $|x|-3\; =\; 0$ (which means $|x|\; =\; 3$, so $x\; =\; 3$ or $x\; =\; -3$).

We need to consider the intervals defined by these critical points: $-\backslash infty,\; -5,\; -3,\; 3,\; 5,\; \backslash infty$.

Let's analyze the sign of $\backslash frac\{5\; -\; |x|\}\{|x|-3\}$ in the intervals:

• For $x\; \backslash in\; (-\backslash infty,\; -5]$: Let $x\; =\; -6$. $|x|=6$. $\backslash frac\{5-6\}\{6-3\}\; =\; \backslash frac\{-3\}\{3\}\; =\; -1\; \backslash le\; 0$. This interval is part of the solution.
• For $x\; \backslash in\; (-5,\; -3)$: Let $x\; =\; -4$. $|x|=4$. $\backslash frac\{5-4\}\{4-3\}\; =\; \backslash frac\{1\}\{1\}\; =\; 1\; >\; 0$. This interval is not part of the solution.
• For $x\; \backslash in\; (-3,\; 3)$: Let $x\; =\; 0$. $|x|=0$. . This interval is part of the solution.
• For $x\; \backslash in\; (3,\; 5]$: Let $x\; =\; 4$. $|x|=4$. $\backslash frac\{5-4\}\{4-3\}\; =\; \backslash frac\{1\}\{1\}\; =\; 1\; >\; 0$. This interval is not part of the solution.
• For $x\; \backslash in\; (5,\; \backslash infty)$: Let $x\; =\; 6$. $|x|=6$. $\backslash frac\{5-6\}\{6-3\}\; =\; \backslash frac\{-1\}\{3\}\; \backslash le\; 0$. This interval is part of the solution.

Combining the intervals where the inequality holds true, and remembering that $x\; \backslash neq\; 3$ and $x\; \backslash neq\; -3$, we get:

$x\; \backslash in\; (-\backslash infty,\; -5]\; \backslash cup\; (-3,\; 3)\; \backslash cup\; [5,\; \backslash infty)$.

Final Answer: The solution set is $x\; \backslash in\; (-\backslash infty,\; -5]\; \backslash cup\; (-3,\; 3)\; \backslash cup\; [5,\; \backslash infty)$.