JEE Mathematics Linear Inequalities MCQs Set 01

Answer: (b) \( (-\infty,-5)\cup[1,2]\cup(6,\infty)\cup\{0\} \)

Explanation:
1. Factorise: \( \dfrac{x^2(x-1)(x-2)}{(x-6)(x+5)}\ge0 \).
2. Critical points: \( -5,1,2,6 \) and the double root \( 0 \); the denominator zeros \( -5,6 \) are excluded.
3. Sign scheme gives \( x\lt-5 \), \( 1\le x\le2 \), \( x\gt6 \), and \( x=0 \) (numerator zero).

Teacher's Note:
1. A squared factor never changes the sign, but its zero is still a solution of a '≥ 0' inequality.
2. Zeros of the denominator are always excluded.

Answer: (d) 4

Explanation:
1. For \( x\gt0 \) the denominator is positive, so \( x^2-1\lt6x+15 \), i.e. \( x^2-6x-16\lt0 \).
2. This gives \( (x-8)(x+2)\lt0 \), so \( 0\lt x\lt8 \).
3. Primes: \( 2,3,5,7 \), i.e. 4.

Teacher's Note:
1. When the sign of the denominator is known you may multiply across without flipping.
2. Only positive x matter here because primes are positive.

Answer: (b) \( x\in\left[1,\frac{5+\sqrt{17}}{2}\right] \)

Explanation:
1. \( \dfrac{x^2-1}{x+3}\ge0 \) gives \( x\in(-3,-1]\cup[1,\infty) \).
2. The quadratic gives \( \frac{5-\sqrt{17}}2\le x\le\frac{5+\sqrt{17}}2 \), i.e. about \( [0.44,\,4.56] \).
3. Common part: \( \left[1,\frac{5+\sqrt{17}}{2}\right] \).

Teacher's Note:
1. Solve each inequality separately and intersect on the number line.
2. Approximate the surds to see the order of end points.

Answer: (d) 5

Explanation:
1. \( x^2+9\lt(x+3)^2 \) gives \( 6x\gt0 \), so \( x\gt0 \).
2. \( (x+3)^2\lt8x+25 \) gives \( x^2-2x-16\lt0 \), so \( 1-\sqrt{17}\lt x\lt1+\sqrt{17} \), about \( x\lt5.12 \).
3. Integers: \( 1,2,3,4,5 \), i.e. 5.

Teacher's Note:
1. Split a double inequality into two and intersect.
2. Estimate surds to find the integers in an interval.

Answer: (d) 3

Explanation:
1. Since \( x^3+1=(x+1)(x^2-x+1) \), combine over this denominator: numerator \( 2(x+1)-(x^2-x+1)-(2x-1)=-x^2+x+2=-(x-2)(x+1) \).
2. So the expression is \( \dfrac{-(x-2)}{x^2-x+1}\ge0 \) for \( x\ne-1 \), i.e. \( x\le2 \).
3. Non-negative integers: \( 0,1,2 \), i.e. 3.

Teacher's Note:
1. Always factorise sums of cubes to find the common denominator.
2. Check that the excluded point x = -1 is not a non-negative integer.

Answer: (d) 0

Explanation:
1. The square root requires \( x^2\ge16 \), i.e. \( |x|\ge4 \). Then \( x^2-9\gt0 \), so the denominator is positive.
2. The fraction is non-positive only when the numerator is zero: \( x^2=16 \), so \( x=\pm4 \).
3. Sum \( =4+(-4)=0 \).

Teacher's Note:
1. Check the domain before sign analysis.
2. When one factor is strictly positive on the domain, only zeros of the numerator can satisfy '≤ 0'.

Answer: (a) a unit set

Explanation:
1. Here \( \frac{10}7(\log_25-1)\approx1.89\gt1 \), so the logarithm to base 0.3 is negative.
2. The root needs \( 2\le x\le8 \); a non-negative fraction with a negative denominator requires the numerator to be 0, so \( x=2 \) or \( x=8 \).
3. The second inequality: \( x\gt3+\log_231\approx7.95 \), so only \( x=8 \) works. The set is a unit set.

Teacher's Note:
1. Decide the sign of the denominator first.
2. Then test each candidate against the remaining condition.

Answer: (a) \( [9,\infty) \)

Explanation:
1. \( \log_4\{\cdot\}\ge0 \) gives \( \log_3\{\cdot\}\ge1 \), then \( \log_2(x^2-2x+a)\ge3 \), so \( x^2-2x+a\ge8 \).
2. The minimum of \( x^2-2x+a \) is \( a-1 \), so we need \( a-1\ge8 \).
3. Hence \( a\ge9 \).

Teacher's Note:
1. Peel off the logarithms from the outside, one at a time.
2. 'For all x' means the minimum value of the quadratic must satisfy the bound.

Answer: (a) 0.10

Explanation:
1. Put \( u=\log_{10}x \): \( u^2-u-2\ge0 \) gives \( (u-2)(u+1)\ge0 \).
2. So \( u\le-1 \) or \( u\ge2 \), i.e. \( 0\lt x\le\frac1{10} \) or \( x\ge100 \).
3. Hence \( a=0.10 \).

Teacher's Note:
1. Substitution u = log x reduces this to a quadratic inequality.
2. Convert back to x with powers of 10.

Answer: (b) 5.67

Explanation:
1. Base \( \frac12\lt1 \), so \( 0\lt(x+5)^2\lt(3x-1)^2 \) with \( x\ne-5 \) and \( x\ne\frac13 \).
2. This gives \( (x+1)(x-3)\gt0 \), i.e. \( x\lt-1 \) or \( x\gt3 \); removing \( x=-5 \) we get \( (-\infty,-5)\cup(-5,-1)\cup(3,\infty) \).
3. So \( p=-5,\ q=-5,\ r=-1,\ s=3 \) and the value is \( \frac{25+25+1}{9}=\frac{51}{9}\approx5.67 \).

Teacher's Note:
1. Reversing the inequality for a base less than 1 is the key step.
2. Always exclude the points where the arguments of the logarithms vanish.

Answer: (b) 5

Explanation:
1. Since \( a\gt2 \), \( a\lt2a\lt a^2 \). The inequality holds for \( x\lt a \) and for \( 2a\lt x\lt a^2 \).
2. Positive integers: \( (a-1)+(a^2-2a-1)=a^2-a-2 \).
3. Set \( a^2-a-2=18 \): \( a^2-a-20=0 \), so \( a=5 \).

Teacher's Note:
1. Order the roots first, then use the wavy-curve method.
2. Count integers strictly between the roots carefully.

Answer: (d) \( \{-2\}\cup[1,8) \)

Explanation:
1. Rewrite \( (1-x)^3=-(x-1)^3 \): the inequality becomes \( \dfrac{(x-1)^3(x+2)^4}{(x+9)^2(x-8)}\le0 \).
2. Even powers do not change sign: the sign is that of \( \frac{x-1}{x-8} \), which is \( \le0 \) for \( 1\le x\lt8 \).
3. The zero of \( (x+2)^4 \) gives the extra solution \( x=-2 \); \( x=-9 \) is excluded.

Teacher's Note:
1. Convert factors of the form (1 - x) to (x - 1) carefully.
2. Isolated zeros from even-power factors must be added.

Answer: (a) \( [-\sqrt2,-1)\cup(-1,\sqrt2]\cup[3,4) \)

Explanation:
1. Factorise: \( x^2-3x-4=(x-4)(x+1) \), so the denominator is \( (x+1)^2(x-4) \) with \( x\ne-1,4 \).
2. Multiply by \( -1 \): \( \dfrac{(x^2-2)(x-3)^3}{(x+1)^2(x-4)}\le0 \), giving \( (x+1)^2 \) positive: sign of \( (x^2-2)(x-3)/(x-4) \) (odd power of \( x-3 \)).
3. Solution: \( -\sqrt2\le x\le\sqrt2 \) (with \( x\ne-1 \)) and \( 3\le x\lt4 \).

Teacher's Note:
1. A repeated factor in the denominator still excludes its root.
2. Check the end points where the numerator is zero.

Answer: (a) \( (-1,1)\cup(4,6) \)

Explanation:
1. Move everything to the left: \( \dfrac{14x(x-4)-(9x-30)(x+1)}{(x+1)(x-4)}\lt0 \).
2. The numerator is \( 14x^2-56x-9x^2+21x+30=5x^2-35x+30=5(x-1)(x-6) \).
3. So \( \dfrac{(x-1)(x-6)}{(x+1)(x-4)}\lt0 \), which holds on \( (-1,1) \) and \( (4,6) \). (Check: \( x=0 \) gives \( 0\lt7.5 \).)

Teacher's Note:
1. Test one value from each region to confirm the sign scheme.
2. Expand and simplify the numerator carefully.

Answer: (c) \( (-3,-2)\cup(-1,\infty) \)

Explanation:
1. Numerator of the difference: \( (x-1)(x-2)(x-3)-(x+1)(x+2)(x+3)=(x^3-6x^2+11x-6)-(x^3+6x^2+11x+6)=-12(x^2+1) \).
2. So the inequality is \( \dfrac{-12(x^2+1)}{(x+1)(x+2)(x+3)}\le0 \), i.e. \( (x+1)(x+2)(x+3)\gt0 \).
3. This holds on \( (-3,-2) \) and \( (-1,\infty) \).

Teacher's Note:
1. Cubic expansions here are nearly mirror images, so the cubic terms cancel.
2. The denominator zeros are excluded.

Answer: (c) \( (-\infty,-20)\cup(23,\infty) \)

Explanation:
1. Since \( x^2+1\gt0 \), multiply: \( 2x^2-3x-459\gt x^2+1 \).
2. So \( x^2-3x-460\gt0 \), i.e. \( (x-23)(x+20)\gt0 \).
3. Hence \( x\lt-20 \) or \( x\gt23 \).

Teacher's Note:
1. A denominator that is always positive can be cleared directly.
2. Check the factorisation: 23 times 20 is 460 and 23 minus 20 is 3.

Answer: (a) \( \left(\frac12,3\right) \)

Explanation:
1. Here \( x^2-4x+5=(x-2)^2+1\gt0 \), so \( x^2-5x+12\gt3x^2-12x+15 \).
2. This gives \( 2x^2-7x+3\lt0 \), i.e. \( (2x-1)(x-3)\lt0 \).
3. Hence \( \frac12\lt x\lt3 \).

Teacher's Note:
1. Complete the square to prove the denominator is positive.
2. The quadratic is negative between its roots.

Answer: (c) \( \left(\frac{25}4,\infty\right) \)

Explanation:
1. \( \dfrac{x-1}{x+2}\gt1 \) gives \( \dfrac{-3}{x+2}\gt0 \), so \( x\lt-2 \).
2. \( \dfrac{x-1}{x+2}\lt7 \) gives \( \dfrac{-6x-15}{x+2}\lt0 \), i.e. \( \dfrac{2x+5}{x+2}\gt0 \), so \( x\lt-\frac52 \) or \( x\gt-2 \).
3. Together \( x\lt-\frac52 \), so \( x^2\gt\frac{25}4 \).

Teacher's Note:
1. Take the intersection of the two conditions before squaring.
2. Squaring a number below -5/2 gives a value above 25/4.

Answer: (c) 2

Explanation:
1. For positive \( x \): \( 3^x-5^x\lt0 \) and \( x^2+5x+2\gt0 \).
2. So we need \( x-2\le0 \), i.e. \( x\le2 \).
3. Positive integers: \( 1,2 \): 2 values.

Teacher's Note:
1. For x > 0 the larger base gives the larger power.
2. x = 2 is included because the numerator vanishes there.

Answer: (c) \( (-1,0)\cup(0,1) \)

Explanation:
1. Add 2: \( \dfrac{x^2+2+2x^2-2}{x^2-1}\lt0 \), i.e. \( \dfrac{3x^2}{x^2-1}\lt0 \).
2. The numerator is non-negative, so we need \( x^2-1\lt0 \) and \( x\ne0 \) (the numerator must not be zero for a strict inequality).
3. Hence \( x\in(-1,0)\cup(0,1) \).

Teacher's Note:
1. For a strict inequality a zero of the numerator is not a solution.
2. Use 'add 2 and combine' instead of cross-multiplying.

Answer: (b) \( (-\infty,-1)\cup(5,\infty) \)

Explanation:
1. The numerator \( x^4+x^2+1 \) is always positive.
2. So the sign is that of \( x^2-4x-5=(x-5)(x+1) \).
3. It is positive for \( x\lt-1 \) or \( x\gt5 \) (end points excluded since they make the denominator zero).

Teacher's Note:
1. Spot expressions that are always positive and ignore them.
2. Zeros of the denominator can never be included.

Answer: (a) \( [-2,-1]\cup[1,2] \)

Explanation:
1. Factorise: \( (x^2-1)(x^2-4)\le0 \).
2. This means \( 1\le x^2\le4 \).
3. So \( 1\le|x|\le2 \): \( [-2,-1]\cup[1,2] \).

Teacher's Note:
1. Treat x² as a single variable and then return to x.
2. Remember that |x| between two positive numbers gives two intervals.

Answer: (c) \( [-3,3] \)

Explanation:
1. Factorise: \( (x^2-9)(x^2+7)\le0 \).
2. Since \( x^2+7\gt0 \), we need \( x^2\le9 \).
3. Hence \( -3\le x\le3 \).

Teacher's Note:
1. Look for the factor pair 9 and -7 when factorising -63.
2. The factor x² + 7 never changes sign.

Answer: (a) \( \left(-\infty,-\frac52\right) \)

Explanation:
1. \( \dfrac{x-1}{x+2}\gt1 \) gives \( x\lt-2 \).
2. \( \dfrac{x-1}{x+2}\lt7 \) gives \( \dfrac{2x+5}{x+2}\gt0 \), i.e. \( x\lt-\frac52 \) or \( x\gt-2 \).
3. The common part is \( x\lt-\frac52 \).

Teacher's Note:
1. Solve the two parts separately and intersect.
2. Compare -2.5 with -2 on the number line.