JEE Mathematics Solutions of Equations and Inequations MCQs Set 05

Answer: (b) \( x=0,\ 8 \)

Explanation:
1. \( x\le0 \): \( -x-2x+12=12 \) gives \( x=0 \).
2. \( 0\lt x\le6 \): \( x+12-2x=12 \) gives \( x=0 \), not in the interval.
3. \( x\gt 6 \): \( 3x-12=12 \) gives \( x=8 \).
4. Solutions: \( x=0 \) and \( x=8 \).

Teacher's Note:
1. Critical points of the moduli are 0 and 6; check each root against its interval.

Answer: (a) \( x=0,\ 1 \)

Explanation:
1. Squaring both sides: \( (x^2-x+1)^2=(x^2-x-1)^2 \), i.e. \( 4(x^2-x)=0 \).
2. \( x(x-1)=0 \), so \( x=0 \) or \( x=1 \).

Teacher's Note:
1. |a| = |b| is equivalent to a^2 = b^2, and the difference of squares simplifies the work.

Answer: (a) \( 4 \)

Explanation:
1. Squaring: \( (2x^2+x-1)^2=(x^2+4x+1)^2 \), i.e. \( (2x^2+x-1-x^2-4x-1)(2x^2+x-1+x^2+4x+1)=0 \).
2. \( (x^2-3x-2)(3x^2+5x)=0 \).
3. \( x^2-3x-2=0 \) gives two irrational roots \( \tfrac{3\pm\sqrt{17}}2 \); \( x(3x+5)=0 \) gives \( x=0,-\tfrac53 \).
4. These four roots are distinct, so there are 4 real roots.

Teacher's Note:
1. Use a^2 - b^2 = (a - b)(a + b) after squaring.

Answer: (b) \( 0 \)

Explanation:
1. The left side depends on \( |x| \) and is strictly increasing in \( |x| \), so \( |x|=1 \) is the only possibility (\( 2+3+4=9 \)).
2. The solutions are \( x=\pm1 \) with sum 0.

Teacher's Note:
1. Spot the integer solution |x| = 1 and use monotonicity to show it is the only one.

Answer: (c) \( 0 \)

Explanation:
1. \( x^3-6x^2+11x-6=(x-1)(x-2)(x-3) \), a product of three consecutive integers for integer \( x \).
2. Such a product is divisible by 6, so its modulus is either 0 or at least 6 and divisible by 6, hence never prime.
3. So there are 0 such integers.

Teacher's Note:
1. The product of three consecutive integers is always a multiple of 6.

Answer: (d) \( (-\infty,-4]\cup[-1,1]\cup[4,\infty) \)

Explanation:
1. The inequality is equivalent to \( 9x^2\le(x^2-4)^2 \) with \( x\ne\pm2 \).
2. This gives \( (x^2-4-3x)(x^2-4+3x)\ge0 \), i.e. \( (x-4)(x+1)(x+4)(x-1)\ge0 \).
3. The sign diagram gives \( x\le-4 \), \( -1\le x\le1 \) or \( x\ge4 \) (none of these includes \( \pm2 \)).

Teacher's Note:
1. Square both sides to remove the modulus and factorise the difference of squares.

Answer: (d) \( (2,\infty) \)

Explanation:
1. Since \( (x+2)+(x-3)=2x-1 \) and \( (x+2)-(x-3)=5 \), compare with \( 2x+1 \) piece by piece.
2. \( x\ge3 \): \( 2x-1\lt2x+1 \) is always true. \( 2\lt x\lt3 \): \( (x+2)+(3-x)=5\lt2x+1 \) gives \( x\gt2 \), true.
3. \( x\lt-2 \): the left side is \( 1-2x \), larger than the right side \( -2x-1 \). For \( -2\le x\le2 \) the left side is 5 while \( |2x+1|\le5 \), so the strict inequality fails (at \( x=2 \) both sides equal 5).
4. Solution set: \( (2,\infty) \).

Teacher's Note:
1. Check x = 2: both sides equal 5, so 2 is excluded.

Answer: (c) \( \left(0,\dfrac14\right]\cup[2,\infty) \)

Explanation:
1. Put \( t=\log_2x \): the inequality is \( t^2\ge|t-2| \).
2. \( t\ge2 \): \( t^2-t+2\ge0 \) is always true (negative discriminant), so every \( t\ge2 \) works.
3. \( t\lt2 \): \( t^2+t-2\ge0 \), i.e. \( (t+2)(t-1)\ge0 \), so \( t\le-2 \) or \( 1\le t\lt2 \).
4. Combining: \( t\le-2 \) or \( t\ge1 \), i.e. \( 0\lt x\le\tfrac14 \) or \( x\ge2 \).

Teacher's Note:
1. t >= 1 means log2 x >= 1, which gives x >= 2 (not x >= 1).

Answer: (b) \( \left[\dfrac12,2\right)\cup(5,\infty) \)

Explanation:
1. Domain: \( 2x-1\ge0 \) gives \( x\ge\tfrac12 \), and \( x\ne2 \).
2. \( \tfrac12\le x\lt2 \): the denominator is negative and the numerator non-negative, so the fraction is \( \le0\lt1 \): all of this interval works (including \( x=\tfrac12 \)).
3. \( x\gt2 \): need \( \sqrt{2x-1}\lt x-2 \), i.e. \( 2x-1\lt x^2-4x+4 \), i.e. \( x^2-6x+5\gt0 \), so \( x\gt5 \) (since \( x\gt2 \)).
4. Solution set: \( [\tfrac12,2)\cup(5,\infty) \).

Teacher's Note:
1. Treat the sign of the denominator separately, and include the endpoint 1/2 where the root is zero.

Answer: (a) \( (-\infty,-3] \)

Explanation:
1. Domain: \( (x-8)(x+3)\ge0 \) gives \( x\le-3 \) or \( x\ge8 \).
2. \( x+2\lt0 \) (i.e. \( x\lt-2 \)): holds for every domain value there, so \( x\le-3 \).
3. \( x\ge-2 \): squaring \( x^2-5x-24\gt x^2+4x+4 \) gives \( -9x\gt28 \), i.e. \( x\lt-\tfrac{28}9 \), which contradicts \( x\ge-2 \).
4. Solution set: \( (-\infty,-3] \) (the point \( x=-3 \) gives \( \sqrt{0}=0\gt-1 \)).

Teacher's Note:
1. Check the endpoint x = -3: the root is zero and x + 2 = -1 is negative, so it works.

Answer: (a) \( 2 \)

Explanation:
1. Rewrite as \( x(2^x-1)=1 \). The function \( h(x)=x(2^x-1) \) is non-negative for all real \( x \) (both factors have the same sign), with \( h(0)=0 \).
2. For \( x\gt0 \): \( h \) is increasing from 0 to infinity, so it equals 1 exactly once; indeed \( h(1)=1 \), so \( x=1 \).
3. For \( x\lt0 \): \( h(x)=|x|(1-2^x) \) tends to infinity as \( x\to-\infty \) and equals 0 at 0; it is decreasing on \( x\lt0 \), so it equals 1 exactly once, at about \( x\approx-1.53 \).
4. Hence 2 roots.

Teacher's Note:
1. x(2^x - 1) is never negative, which tells you immediately where roots can occur.

Answer: (d) \( \{-10,-6,0,4\} \)

Explanation:
1. \( |x+3|-5=\pm2 \) gives \( |x+3|=7 \) or \( |x+3|=3 \).
2. \( |x+3|=7 \): \( x=4,-10 \). \( |x+3|=3 \): \( x=0,-6 \).
3. Solution set \( \{-10,-6,0,4\} \).

Teacher's Note:
1. Peel the outer modulus first: ||A| - 5| = 2 means |A| - 5 = 2 or -2.

Answer: (d) \( x=0,\ 4 \)

Explanation:
1. \( x^3-6x^2+11x-6=(x-1)(x-2)(x-3) \).
2. Case \( +6 \): \( x^3-6x^2+11x-12=0 \) gives \( (x-4)(x^2-2x+3)=0 \), so \( x=4 \) (the quadratic has no real root).
3. Case \( -6 \): \( x^3-6x^2+11x=0 \) gives \( x(x^2-6x+11)=0 \), so \( x=0 \).
4. Solutions: \( x=0 \) and \( x=4 \).

Teacher's Note:
1. Write |f| = 6 as f = 6 or f = -6 and factorise each cubic.

Answer: (c) \( 0 \)

Explanation:
1. The equation depends only on \( |x| \) (since \( x^2=|x|^2 \)), so the solutions come in pairs \( \pm x \).
2. Put \( t=|x| \): \( t^2-5t-4=0 \) has one positive root, so there are two solutions, \( \pm t \).
3. Their sum is 0.

Teacher's Note:
1. An even equation always has roots symmetric about zero.

Answer: (b) \( 2 \)

Explanation:
1. Left side: \( x\lt 1: 6-3x \); \( 1\le x\lt 2: 4-x \); \( 2\le x\lt 3: x \); \( x\ge3: 3x-6 \).
2. \( 6-3x=9 \) gives \( x=-1 \) (valid); \( 3x-6=9 \) gives \( x=5 \) (valid). The middle pieces take values at most 4, so give nothing.
3. Exactly 2 solutions.

Teacher's Note:
1. A sum of distances is V-shaped (convex): a horizontal line above the minimum meets it twice.

Answer: (a) \( \{-1,5\} \)

Explanation:
1. A modulus is never negative, so \( \le0 \) means \( =0 \).
2. \( |x-2|=3 \) gives \( x=5 \) or \( x=-1 \).

Teacher's Note:
1. |A| <= 0 is the same as A = 0, which gives isolated points, not an interval.

Answer: (a) \( \left(-\dfrac23,4\right) \)

Explanation:
1. Squaring: \( (x+3)^2\gt(2x-1)^2 \), i.e. \( (x+3-2x+1)(x+3+2x-1)\gt0 \).
2. \( (4-x)(3x+2)\gt0 \) gives \( -\tfrac23\lt x\lt4 \).

Teacher's Note:
1. For |a| > |b| use (a - b)(a + b) > 0.

Answer: (b) \( [1,2]\cup[3,4] \)

Explanation:
1. Let \( a=x^2-4x+3=(x-1)(x-3) \) and \( b=x^2-6x+8=(x-2)(x-4) \). Then \( a-b=2x-5 \).
2. So the equation reads \( |a|+|b|=|a-b| \), which holds exactly when \( ab\le0 \).
3. \( (x-1)(x-2)(x-3)(x-4)\le0 \) gives \( 1\le x\le2 \) or \( 3\le x\le4 \).

Teacher's Note:
1. |a| + |b| = |a - b| exactly when a and b have opposite signs (or one is zero).

Answer: (c) \( \left(0,\dfrac1{27}\right]\cup[3,\infty) \)

Explanation:
1. Put \( t=\log_3x \): \( 2|t|+t\ge3 \).
2. \( t\ge0 \): \( 3t\ge3 \) gives \( t\ge1 \).
3. \( t\lt0 \): \( -2t+t=-t\ge3 \) gives \( t\le-3 \).
4. So \( t\le-3 \) or \( t\ge1 \), i.e. \( 0\lt x\le3^{-3}=\tfrac1{27} \) or \( x\ge3 \).

Teacher's Note:
1. Handle the sign of t separately, then convert t back using x = 3^t.

Answer: (d) \( \left[-1,\dfrac{\sqrt5-1}2\right) \)

Explanation:
1. Domain: \( |x|\le1 \). The inequality is \( x\lt\sqrt{1-|x|} \).
2. For \( -1\le x\le0 \): the left side is at most 0 and the right side is at least 0, and they are not both 0 (at \( x=0 \) the right side is 1; at \( x=-1 \) the left side is \( -1 \)), so the inequality holds on all of \( [-1,0] \).
3. For \( 0\lt x\le1 \): square both sides: \( x^2\lt1-x \) gives \( x^2+x-1\lt0 \), so \( 0\lt x\lt\tfrac{\sqrt5-1}2 \).
4. Solution set: \( \left[-1,\tfrac{\sqrt5-1}2\right) \).

Teacher's Note:
1. Squaring is allowed only when both sides are non-negative, i.e. for x > 0 here.

Answer: (d) \( [-2,0)\cup\left[\sqrt{2-\sqrt3},\ \sqrt{2+\sqrt3}\right] \)

Explanation:
1. Domain: \( -2\le x\le2 \), \( x\ne0 \).
2. \( -2\le x\lt0 \): the right side is negative, so the inequality holds on all of \( [-2,0) \).
3. \( 0\lt x\le2 \): multiply by \( x\gt0 \) and square: \( x^2(4-x^2)\ge1 \), i.e. \( x^4-4x^2+1\le0 \), so \( 2-\sqrt3\le x^2\le2+\sqrt3 \).
4. That gives \( \sqrt{2-\sqrt3}\le x\le\sqrt{2+\sqrt3} \) (about \( 0.52 \) to \( 1.93 \)), within \( (0,2] \).

Teacher's Note:
1. For x < 0 the reciprocal is negative while the square root is non-negative.

Answer: (c) \( 1\lt k\lt\dfrac54 \)

Explanation:
1. Let \( h(x)=|x^2-1|+x \): for \( x\le-1 \) it is \( x^2+x-1 \) (decreasing from infinity to \( -1 \)); for \( -1\le x\le1 \) it is \( 1+x-x^2 \) (rising from \( -1 \) to the peak \( \tfrac54 \) at \( x=\tfrac12 \), then falling to 1 at \( x=1 \)); for \( x\ge1 \) it is \( x^2+x-1 \) (rising from 1).
2. A line \( y=k \) with \( 1\lt k\lt\tfrac54 \) meets the left branch once, the middle hump twice and the right branch once: 4 solutions.
3. For comparison: \( k=1 \) and \( k=\tfrac54 \) give 3 solutions, \( -1\lt k\lt1 \) and \( k\gt\tfrac54 \) give 2 solutions, and \( k=-1 \) gives 1 solution.

Teacher's Note:
1. Draw the three pieces and note the hump height 5/4 at x = 1/2 and the junction values -1 and 1.