SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Linear Inequalities

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Miscellaneous Exercise 1–10 (part 4 of 5)

  1. Solve the inequalities in Exercises $\displaystyle 1$ to $\displaystyle 6$ .

    Exercise 1

    2≤3x−4≤5\displaystyle 2 \leq 3 x-4 \leq 5

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    NCERT’s answer
    [$\displaystyle 2$, $\displaystyle 3$]
    Double inequality — operate on all three parts. Whatever is added to, or multiplied into, one part must be applied to all three.\[2 \leq 3x - 4 \leq 5\]Add \(\displaystyle 4\) throughout:\[6 \leq 3x \leq 9\]Divide throughout by \(\displaystyle 3\) (positive, so the senses are unchanged):\[2 \leq x \leq 3\]Solution set: \(\displaystyle \{x : 2 \leq x \leq 3\} = [2,\ 3]\).

  2. Exercise 2

    6≤−3(2x−4)<12\displaystyle 6 \leq-3(2 x-4)<12

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    NCERT’s answer
    ($\displaystyle 0$, $\displaystyle 1$]
    Double inequality — dividing by a negative number reverses the signs.\[6 \leq -3(2x-4) < 12\]Divide all three parts by \(\displaystyle -3\). Because \(\displaystyle -3\) is negative, both inequality signs must be reversed:\[-2 \geq 2x - 4 > -4\]Add \(\displaystyle 4\) throughout:\[2 \geq 2x > 0\]Divide throughout by \(\displaystyle 2\) (positive):\[1 \geq x > 0, \qquad\text{that is}\qquad 0 < x \leq 1\]Check the ends: \(\displaystyle x=1\) gives \(\displaystyle -3(-2)=6\), and \(\displaystyle 6 \leq 6\) holds; \(\displaystyle x=0\) gives \(\displaystyle 12\), which fails \(\displaystyle <12\).Solution set: \(\displaystyle \{x : 0 < x \leq 1\} = (0,\ 1]\).

  3. Exercise 3

    −3≤4−7x2≤18\displaystyle -3 \leq 4-\frac{7 x}{2} \leq 18

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    NCERT’s answer
    [- $\displaystyle 4$, $\displaystyle 2$]
    Double inequality — isolate \(\displaystyle x\) in all three parts.\[-3 \leq 4 - \frac{7x}{2} \leq 18\]Subtract \(\displaystyle 4\) throughout:\[-7 \leq -\frac{7x}{2} \leq 14\]Multiply throughout by \(\displaystyle -\dfrac{2}{7}\). The multiplier is negative, so both signs reverse:\[2 \geq x \geq -4, \qquad\text{that is}\qquad -4 \leq x \leq 2\]Check the ends: \(\displaystyle x=-4\) gives \(\displaystyle 4+14 = 18\) and \(\displaystyle x=2\) gives \(\displaystyle 4-7=-3\); both endpoints satisfy the inequality, so both are included.Solution set: \(\displaystyle \{x : -4 \leq x \leq 2\} = [-4,\ 2]\).

  4. Exercise 4

    −15<3(x−2)5≤0\displaystyle -15<\frac{3(x-2)}{5} \leq 0

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    NCERT’s answer
    (-$\displaystyle 23$, $\displaystyle 2$]
    Double inequality — clear the fraction first.\[-15 < \frac{3(x-2)}{5} \leq 0\]Multiply throughout by \(\displaystyle \dfrac{5}{3}\) (positive, so the senses are unchanged):\[-25 < x - 2 \leq 0\]Add \(\displaystyle 2\) throughout:\[-23 < x \leq 2\]Check the ends: \(\displaystyle x=2\) gives \(\displaystyle 0\), and \(\displaystyle 0 \leq 0\) holds; \(\displaystyle x=-23\) gives exactly \(\displaystyle -15\), which fails the strict \(\displaystyle -15 <\).Solution set: \(\displaystyle \{x : -23 < x \leq 2\} = (-23,\ 2]\).

  5. Exercise 5

    −12<4−3x−5≤2\displaystyle -12<4-\frac{3 x}{-5} \leq 2

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    NCERT’s answer
    $\displaystyle \left(\frac{-80}{3}, \frac{-10}{3}\right]$
    Simplify the sign first. Note that \(\displaystyle -\dfrac{3x}{-5} = +\dfrac{3x}{5}\), so the middle part is \(\displaystyle 4 + \dfrac{3x}{5}\):\[-12 < 4 + \frac{3x}{5} \leq 2\]Subtract \(\displaystyle 4\) throughout:\[-16 < \frac{3x}{5} \leq -2\]Multiply throughout by \(\displaystyle \dfrac{5}{3}\) (positive, so the senses are unchanged):\[-\frac{80}{3} < x \leq -\frac{10}{3}\]Check the ends: \(\displaystyle x=-\tfrac{10}{3}\) gives \(\displaystyle 4-2 = 2\), and \(\displaystyle 2 \leq 2\) holds; \(\displaystyle x=-\tfrac{80}{3}\) gives exactly \(\displaystyle -12\), which fails the strict \(\displaystyle -12<\).Solution set: \(\displaystyle \left\{x : -\dfrac{80}{3} < x \leq -\dfrac{10}{3}\right\} = \left(-\dfrac{80}{3},\ -\dfrac{10}{3}\right]\).

  6. Exercise 6

    7≤(3x+11)2≤11\displaystyle 7 \leq \frac{(3 x+11)}{2} \leq 11.

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    NCERT’s answer
    $\displaystyle \left[1, \frac{11}{3}\right]$
    Double inequality — clear the denominator first.\[7 \leq \frac{3x + 11}{2} \leq 11\]Multiply throughout by \(\displaystyle 2\) (positive):\[14 \leq 3x + 11 \leq 22\]Subtract \(\displaystyle 11\) throughout:\[3 \leq 3x \leq 11\]Divide throughout by \(\displaystyle 3\):\[1 \leq x \leq \frac{11}{3}\]Solution set: \(\displaystyle \left\{x : 1 \leq x \leq \dfrac{11}{3}\right\} = \left[1,\ \dfrac{11}{3}\right]\).
  7. Solve the inequalities in Exercises $\displaystyle 7$ to $\displaystyle 10$ and represent the solution graphically on number line.

    Exercise 7

    5x+1>−24,5x−1<24\displaystyle 5 x+1>-24,5 x-1<24

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    NCERT’s answer
    $\displaystyle (-5, 5)$ ![](https://cdn.mathpix.com/cropped/952f948a-c38e-4ca5-a31f-e9bbea10a1ba-08.jpg?height=$\displaystyle 96$&width=$\displaystyle 852$&top_left_y=$\displaystyle 645$&top_left_x=$\displaystyle 539$)
    Solve each inequality, then take the intersection. A value of \(\displaystyle x\) must satisfy both, so the solution set is the overlap of the two.First inequality:\[5x + 1 > -24 \quad\Longrightarrow\quad 5x > -25 \quad\Longrightarrow\quad x > -5\]Second inequality:\[5x - 1 < 24 \quad\Longrightarrow\quad 5x < 25 \quad\Longrightarrow\quad x < 5\]The common part of \(\displaystyle (-5,\infty)\) and \(\displaystyle (-\infty,5)\) is\[-5 < x < 5\]On the number line, put hollow circles at \(\displaystyle -5\) and at \(\displaystyle 5\) and shade the segment between them.Solution set: \(\displaystyle \{x : -5 < x < 5\} = (-5,\ 5)\).

  8. Exercise 8

    2(x−1)<x+5,3(x+2)>2−x\displaystyle 2(x-1)<x+5,3(x+2)>2-x

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    Solve each inequality, then take the intersection.First inequality:\[2(x-1) < x + 5\] \[2x - 2 < x + 5 \quad\Longrightarrow\quad x < 7\]Second inequality:\[3(x+2) > 2 - x\] \[3x + 6 > 2 - x \quad\Longrightarrow\quad 4x > -4 \quad\Longrightarrow\quad x > -1\]Both must hold, so we intersect \(\displaystyle (-\infty,7)\) with \(\displaystyle (-1,\infty)\):\[-1 < x < 7\]On the number line, put hollow circles at \(\displaystyle -1\) and at \(\displaystyle 7\) and shade the segment between them.Solution set: \(\displaystyle \{x : -1 < x < 7\} = (-1,\ 7)\).

  9. Exercise 9

    3x−7>2(x−6),6−x>11−2x\displaystyle 3 x-7>2(x-6), 6-x>11-2 x

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    Solve each inequality, then take the intersection.First inequality:\[3x - 7 > 2(x - 6)\] \[3x - 7 > 2x - 12 \quad\Longrightarrow\quad x > -5\]Second inequality:\[6 - x > 11 - 2x\] \[6 - x + 2x > 11 \quad\Longrightarrow\quad x > 5\]Both must hold. Since every \(\displaystyle x > 5\) already exceeds \(\displaystyle -5\), the intersection of \(\displaystyle (-5,\infty)\) and \(\displaystyle (5,\infty)\) is the smaller set:\[x > 5\]On the number line, put a hollow circle at \(\displaystyle 5\) and shade the ray to its right.Solution set: \(\displaystyle \{x : x > 5\} = (5,\ \infty)\).

  10. Exercise 10

    5(2x−7)−3(2x+3)≤0,2x+19≤6x+47\displaystyle 5(2 x-7)-3(2 x+3) \leq 0, \quad 2 x+19 \leq 6 x+47.

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    NCERT’s answer
    [-$\displaystyle 7$, $\displaystyle 11$] ![](https://cdn.mathpix.com/cropped/952f948a-c38e-4ca5-a31f-e9bbea10a1ba-08.jpg?height=$\displaystyle 94$&width=$\displaystyle 1024$&top_left_y=$\displaystyle 1043$&top_left_x=$\displaystyle 301$)
    Solve each inequality, then take the intersection.First inequality:\[5(2x-7) - 3(2x+3) \leq 0\] \[10x - 35 - 6x - 9 \leq 0\] \[4x - 44 \leq 0 \quad\Longrightarrow\quad x \leq 11\]Second inequality:\[2x + 19 \leq 6x + 47\] \[19 - 47 \leq 6x - 2x \quad\Longrightarrow\quad -28 \leq 4x \quad\Longrightarrow\quad x \geq -7\]Both must hold, so we intersect \(\displaystyle (-\infty,11]\) with \(\displaystyle [-7,\infty)\):\[-7 \leq x \leq 11\]On the number line, put solid dots at \(\displaystyle -7\) and at \(\displaystyle 11\) — both endpoints belong to the solution — and shade the segment between them.Solution set: \(\displaystyle \{x : -7 \leq x \leq 11\} = [-7,\ 11]\).