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NCERT Exemplar · Class 11 Mathematics Linear Inequalities

32 questions · 32 still being checked

EXERCISE 6.3 1–10 (part 1 of 3)

  1. Solve for \(\displaystyle x\), the inequalities in Exercises $\displaystyle 1$ to 12.

    Exercise 1

    4x+1≤3≤6x+1,(x>0)\displaystyle \frac{4}{x+1} \leq 3 \leq \frac{6}{x+1},(x>0)

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    NCERT’s answer
    \(\displaystyle \frac{1}{3} \leq x \leq 1\)
    Since \(\displaystyle x>0\), \(\displaystyle x+1>0\), so multiplying by \(\displaystyle x+1\) keeps the signs. \[4 \le 3(x+1) \Rightarrow x \ge \tfrac{1}{3} \] \[3(x+1) \le 6 \Rightarrow x \le 1 \] Both hold with \(\displaystyle x>0\). Answer: \(\displaystyle \tfrac13 \le x \le 1\), i.e. \(\displaystyle x \in \left[\tfrac13,\,1\right]\).

  2. Exercise 2

    ∣x−2∣−1∣x−2∣−2≤0\displaystyle \frac{|x-2|-1}{|x-2|-2} \leq 0

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    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    Put \(\displaystyle t=|x-2|\); the denominator must not vanish, so \(\displaystyle t \ne 2\). \[\frac{t-1}{t-2} \le 0 \Rightarrow 1 \le t < 2 \] \[1 \le |x-2| < 2 \] \[1 \le x-2 < 2 \quad \text{or} \quad -2 < x-2 \le -1 \] \[3 \le x < 4 \quad \text{or} \quad 0 < x \le 1 \] Answer: \(\displaystyle x \in (0,\,1] \cup [3,\,4)\).

  3. Exercise 3

    1∣x∣−3≤12\displaystyle \frac{1}{|x|-3} \leq \frac{1}{2}

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    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    Put \(\displaystyle t=|x|\); \(\displaystyle t \ne 3\). \[\frac{1}{t-3}-\frac12 \le 0 \] \[\frac{5-t}{2(t-3)} \le 0 \Rightarrow \frac{t-5}{t-3} \ge 0 \] \[t<3 \quad \text{or} \quad t \ge 5 \] \[|x|<3 \quad \text{or} \quad |x| \ge 5 \] \[-3<x<3 \quad \text{or} \quad x \le -5 \quad \text{or} \quad x \ge 5 \] Answer: \(\displaystyle x \in (-\infty,\,-5] \cup (-3,\,3) \cup [5,\,\infty)\).

  4. Exercise 4

    ∣x−1∣≤5,∣x∣≥2\displaystyle |x-1| \leq 5,|x| \geq 2

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    NCERT’s answer
    \(\displaystyle [-4,-2] \cup[2,6]\)
    Solve each inequality, then take the common values. \[|x-1| \le 5 \Rightarrow -5 \le x-1 \le 5 \Rightarrow -4 \le x \le 6 \] \[|x| \ge 2 \Rightarrow x \le -2 \quad \text{or} \quad x \ge 2 \] \[[-4,\,6] \cap \big((-\infty,\,-2] \cup [2,\,\infty)\big) = [-4,\,-2] \cup [2,\,6] \] Answer: \(\displaystyle x \in [-4,\,-2] \cup [2,\,6]\).

  5. Exercise 5

    −5≤2−3x4≤9\displaystyle -5 \leq \frac{2-3 x}{4} \leq 9

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    NCERT’s answer
    \(\displaystyle \left[\frac{-34}{3}, \frac{22}{3}\right]\)
    Multiply by $\displaystyle 4$, then subtract $\displaystyle 2$, then divide by \(\displaystyle -3\) (reversing the signs). \[-20 \le 2-3x \le 36 \] \[-22 \le -3x \le 34 \] \[-\frac{34}{3} \le x \le \frac{22}{3} \] Answer: \(\displaystyle x \in \left[-\tfrac{34}{3},\,\tfrac{22}{3}\right]\).

  6. Exercise 6

    4x+3≥2x+17,3x−5<−2\displaystyle 4 x+3 \geq 2 x+17,3 x-5<-2.

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    NCERT’s answer
    No Solution
    \[4x+3 \ge 2x+17 \Rightarrow 2x \ge 14 \Rightarrow x \ge 7 \] \[3x-5 < -2 \Rightarrow 3x < 3 \Rightarrow x < 1 \] No number is both \(\displaystyle \ge 7\) and \(\displaystyle <1\). \[[7,\,\infty) \cap (-\infty,\,1) = \varnothing \] Answer: no solution (the solution set is empty).

  7. Exercise 7

    A company manufactures cassettes. Its cost and revenue functions are C(x)=26,000+30x\displaystyle \mathrm{C}(x)=26,000+30 x and R(x)=43x\displaystyle \mathrm{R}(x)=43 x, respectively, where x\displaystyle x is the number of cassettes produced and sold in a week. How many cassettes must be sold by the company to realise some profit?

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    NCERT’s answer
    More than 2000.
    Profit needs revenue greater than cost. \[R(x) > C(x) \] \[43x > 26000 + 30x \] \[13x > 26000 \] \[x > 2000 \] Answer: more than $\displaystyle 2000$ cassettes a week, i.e. at least 2001.

  8. Exercise 8

    The water acidity in a pool is considerd normal when the average pH reading of three daily measurements is between 8.2\displaystyle 8.2 and 8.5. If the first two pH readings are 8.48\displaystyle 8.48 and 8.35\displaystyle 8.35, find the range of pH value for the third reading that will result in the acidity level being normal.

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    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    Let the third reading be \(\displaystyle x\); the average lies between $\displaystyle 8.2$ and 8.5. \[8.2 < \frac{8.48+8.35+x}{3} < 8.5 \] \[24.6 < 16.83 + x < 25.5 \] \[7.77 < x < 8.67 \] Answer: the third reading must lie between $\displaystyle 7.77$ and 8.67.

  9. Exercise 9

    A solution of 9\displaystyle 9% acid is to be diluted by adding 3\displaystyle 3% acid solution to it. The resulting mixture is to be more than 5\displaystyle 5% but less than 7\displaystyle 7% acid. If there is 460\displaystyle 460 litres of the 9\displaystyle 9% solution, how many litres of 3\displaystyle 3% solution will have to be added?

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    NCERT’s answer
    More than $\displaystyle 230$ litres but less than $\displaystyle 920$ litres.
    Let \(\displaystyle x\) litres of the \(\displaystyle 3\%\) solution be added.\[\text{acid} = 0.09(460) + 0.03x = 41.4 + 0.03x, \qquad \text{mixture} = 460 + x \]\[0.05(460 + x) < 41.4 + 0.03x < 0.07(460 + x) \]\[23 + 0.05x < 41.4 + 0.03x \;\Rightarrow\; 0.02x < 18.4 \;\Rightarrow\; x < 920 \]\[41.4 + 0.03x < 32.2 + 0.07x \;\Rightarrow\; 9.2 < 0.04x \;\Rightarrow\; x > 230 \]Answer: more than $\displaystyle 230$ litres and less than $\displaystyle 920$ litres, i.e. \(\displaystyle 230 < x < 920\).

  10. Exercise 10

    A solution is to be kept between 40∘C\displaystyle 40^{\circ} \mathrm{C} and 45∘C\displaystyle 45^{\circ} \mathrm{C}. What is the range of temperature in degree fahrenheit, if the conversion formula is F=95C+32\displaystyle \mathrm{F}=\frac{9}{5} \mathrm{C}+32 ?

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    NCERT’s answer
    Between \(\displaystyle 104^{\circ} \mathrm{F}\) and \(\displaystyle 113^{\circ} \mathrm{F}\)
    \(\displaystyle F\) increases with \(\displaystyle C\), so the inequality keeps its direction.\[40 < C < 45 \]\[\tfrac{9}{5}(40) + 32 < F < \tfrac{9}{5}(45) + 32 \]\[72 + 32 < F < 81 + 32 \]\[104 < F < 113 \]Answer: between \(\displaystyle 104^{\circ}\mathrm{F}\) and \(\displaystyle 113^{\circ}\mathrm{F}\).