JEE Mathematics Linear Inequalities MCQs Set 03

Answer: (b) \((-\infty,-1)\cup(-1,2]\)

Explanation:
1. Use \(x^3+1=(x+1)(x^2-x+1)\). The numerator over the common denominator is \(2(x+1)-(x^2-x+1)-(2x-1)=-x^2+x+2=-(x-2)(x+1)\).
2. So for \(x\ne-1\) the expression is \(\dfrac{-(x-2)}{x^2-x+1}\), and \(x^2-x+1>0\). It is non-negative iff \(x\le2\).
3. Excluding x = -1 gives (-inf, -1) U (-1, 2].

Teacher's Note:
1. Find the critical points first and test the sign of the expression on each interval.
2. With modulus, split into cases at the points where the expression inside changes sign.

Answer: (c) \([-2,0]\cup[2,4]\)

Explanation:
1. \(|x-1|\le3\) gives \(-2\le x\le4\); \(|x-1|\ge1\) gives \(x\le0\) or \(x\ge2\).
2. The intersection is [-2, 0] U [2, 4].

Teacher's Note:
1. Find the critical points first and test the sign of the expression on each interval.
2. With modulus, split into cases at the points where the expression inside changes sign.

Answer: (d) a unique solution

Explanation:
1. From the second equation \(x=2+|y-1|\ge2\), so \(|x-1|=x-1=1+|y-1|\).
2. The first equation becomes \(|y-1|=3-3y\). For \(y\le1\): \(1-y=3-3y\) gives \(y=1\); for \(y>1\): \(y-1=3-3y\) gives \(y=1\), rejected.
3. So y = 1, x = 2 is the only solution.

Teacher's Note:
1. Find the critical points first and test the sign of the expression on each interval.
2. With modulus, split into cases at the points where the expression inside changes sign.

Answer: (a) 5

Explanation:
1. The equation requires \(2-|1-|x||=\pm1\), so \(|1-|x||=1\) or \(3\).
2. \(|1-|x||=1\) gives \(|x|=0\) or \(2\); \(|1-|x||=3\) gives \(|x|=4\) (the case \(|x|=-2\) is impossible).
3. The roots are 0, 2, -2, 4, -4: five roots.

Teacher's Note:
1. Find the critical points first and test the sign of the expression on each interval.
2. With modulus, split into cases at the points where the expression inside changes sign.

Answer: (c) \(\left(\dfrac92,\infty\right)\)

Explanation:
1. For \(x\ge1\): \((x+2)-(x-1)=3<x-\tfrac32\) gives \(x>\tfrac92\).
2. For \(-2\le x<1\): \(2x+1<x-\tfrac32\) gives \(x<-\tfrac52\), outside this range. For \(x<-2\): \(-3<x-\tfrac32\) gives \(x>-\tfrac32\), impossible.
3. So the solution set is (9/2, infinity).

Teacher's Note:
1. Find the critical points first and test the sign of the expression on each interval.
2. With modulus, split into cases at the points where the expression inside changes sign.

Answer: (b) \(\dfrac32\)

Explanation:
1. \(12|x|\ge4x^2+9\) means \(4x^2-12|x|+9\le0\), i.e. \((2|x|-3)^2\le0\).
2. So \(2|x|=3\) and \(|x|=\tfrac32\).

Teacher's Note:
1. Find the critical points first and test the sign of the expression on each interval.
2. With modulus, split into cases at the points where the expression inside changes sign.

Answer: (d) \((-\infty,4]\)

Explanation:
1. \(|x-1|+a=\pm4\) gives \(|x-1|=4-a\) or \(|x-1|=-4-a\).
2. A real solution exists when one of these is non-negative: \(4-a\ge0\) or \(-4-a\ge0\). The union is \(a\le4\).

Teacher's Note:
1. Find the critical points first and test the sign of the expression on each interval.
2. With modulus, split into cases at the points where the expression inside changes sign.

Answer: (a) \([6,7)\)

Explanation:
1. \(|f|=-f\) means \(f\le0\), where \(f=\dfrac{(x-2)(x-6)}{(x-3)(x-7)}\).
2. Sign analysis with critical points 2, 3, 6, 7 gives \(f\le0\) on \([2,3)\cup[6,7)\). So \([6,7)\) is a solution interval.

Teacher's Note:
1. Find the critical points first and test the sign of the expression on each interval.
2. With modulus, split into cases at the points where the expression inside changes sign.

Answer: (b) \([-3,3]\cup(-\infty,-4)\cup(4,\infty)\)

Explanation:
1. With \(t=|x|\): \(\dfrac{3-t}{4-t}\ge0\) holds for \(t\le3\) or \(t>4\).
2. So |x| <= 3 or |x| > 4.

Teacher's Note:
1. Find the critical points first and test the sign of the expression on each interval.
2. With modulus, split into cases at the points where the expression inside changes sign.

Answer: (c) A <= B

Explanation:
1. \(x^2+y^2\ge2xy\) gives \(\dfrac{x+y}{x^2+y^2}\le\dfrac{x+y}{2xy}=\dfrac12\left(\dfrac1x+\dfrac1y\right)\).
2. Adding the three such inequalities gives A <= B (equality when x = y = z), so A < B is not always true.

Teacher's Note:
1. Find the critical points first and test the sign of the expression on each interval.
2. With modulus, split into cases at the points where the expression inside changes sign.

Answer: (d) \(u\le-2\)

Explanation:
1. \(x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)\ge-2(xy+yz+zx)\).
2. Dividing by the negative number \(xy+yz+zx\) reverses the inequality: \(u\le-2\).

Teacher's Note:
1. Find the critical points first and test the sign of the expression on each interval.
2. With modulus, split into cases at the points where the expression inside changes sign.

Answer: (a) \(n(2c)^{1/n}\)

Explanation:
1. Apply AM-GM to the n numbers \(a_1,\ldots,a_{n-1},2a_n\): their sum is at least \(n(a_1a_2\cdots a_{n-1}\cdot2a_n)^{1/n}\).
2. The product is \(2c\), so the minimum is \(n(2c)^{1/n}\).

Teacher's Note:
1. Find the critical points first and test the sign of the expression on each interval.
2. With modulus, split into cases at the points where the expression inside changes sign.

Answer: (c) \(\left(-\infty,-\dfrac52\right]\cup(0,\infty)\)

Explanation:
1. The argument must be positive: \(\dfrac{x+2}x>0\) gives \(x<-2\) or \(x>0\).
2. Base 0.2 < 1 reverses the inequality: \(\dfrac{x+2}x\ge0.2\), so \(\dfrac{0.8x+2}x\ge0\), giving \(x\le-\tfrac52\) or \(x>0\).
3. Both conditions together give the stated set.

Teacher's Note:
1. Find the critical points first and test the sign of the expression on each interval.
2. With modulus, split into cases at the points where the expression inside changes sign.

Answer: (c) 7

Explanation:
1. The inequality means \(5(x^2+1)\le ax^2+4x+a\), i.e. \((a-5)x^2+4x+(a-5)\ge0\) for all real x.
2. This needs \(a>5\) and discriminant \(16-4(a-5)^2\le0\), i.e. \((a-5)^2\ge4\), so \(a\ge7\).
3. The least integer is 7.

Teacher's Note:
1. Find the critical points first and test the sign of the expression on each interval.
2. With modulus, split into cases at the points where the expression inside changes sign.

Answer: (a) 41

Explanation:
1. \(\dfrac{3x^2+9x+17}{3x^2+9x+7}=1+\dfrac{10}{3x^2+9x+7}\).
2. \(3x^2+9x+7=3\left(x+\tfrac32\right)^2+\tfrac14\) has minimum \(\tfrac14\), so the maximum is \(1+10\cdot4=41\).

Teacher's Note:
1. Find the critical points first and test the sign of the expression on each interval.
2. With modulus, split into cases at the points where the expression inside changes sign.

Answer: (b) 1

Explanation:
1. Cancel \(\sqrt{3-x}\) (defined for \(x\le3\)): \(x\ge3\).
2. Together with x <= 3 this leaves only x = 3.

Teacher's Note:
1. Find the critical points first and test the sign of the expression on each interval.
2. With modulus, split into cases at the points where the expression inside changes sign.