The set of all real values of x satisfying the inequality (x−1)(2x+1)3x2(x+1)>0 is
A(−∞,−1) U (−21,0) U (1,+∞)
B(−1,−21) U (1,+∞)
C(−1,0) U (1,+∞)
D(−1,−21) U (0,+∞)
Pick an option to attempt
The Setup: We must find the set of all real values of x that satisfy the rational inequality (x−1)(2x+1)3x2(x+1)>0.
Step 1: Identify the critical points.
Set the numerator and denominator factors to zero to find the critical points where the expression changes sign or becomes undefined:
Numerator: x=0 (multiplicity 2, even), x=−1 (multiplicity 1, odd).
Denominator: x=1 (multiplicity 1, odd), x=−1/2 (multiplicity 3, odd).
The critical points in ascending order are: −1,−1/2,0,1.
Step 2: Evaluate the sign intervals using a test-point method.
Because the factor x2 is always non-negative and touches 0 at x=0, the sign does not change across x=0. We test the regions partitioned by the odd-multiplicity critical points (−1,−1/2,1):
* **Region x>1:** Test x=2. Expression is (+)(+)/(+)(+)>0. (Valid)
* **Region −1/2<x<1 (excluding 0):** Since sign alternates at x=1, this region is negative. (Invalid)
* **Region −1<x<−1/2:** Since sign alternates at x=−1/2, this region is positive. (Valid)
* **Region x<−1:** Since sign alternates at x=−1, this region is negative. (Invalid)
Step 3: Combine the valid intervals.
The strict inequality requires the expression to be strictly greater than 0, so the critical points themselves are not included.
The valid intervals are (−1,−1/2) and (1,+∞).
Set=(−1,−21)∪(1,+∞)Final Answer:(−1,−21)∪(1,+∞)
Consider the following statements:
(i) When (0<x<1), then (1+x1<1−x+x2)
(ii) When (0<x<1), then (1+x1>1−x+x2)
(iii) When (−1<x<0), then (1+x1<1−x+x2)
(iv) When (−1<x<0), then (1+x1>1−x+x2)
Then the correct statements are:
A(i) and (ii)
B(ii) and (iv)
C(i) and (iv)
D(ii) and (iii)
Pick an option to attempt
The Setup: Cross-multiplying an inequality whose denominator changes sign is how this question eats people alive. We dodge that entirely with the sum of cubes identity a3+b3=(a+b)(a2−ab+b2), taking a=1 and b=x, which turns the whole comparison into a single fraction whose sign we can read off.
Step 0: Note the structure. Statements (i) and (ii) are exact opposites on the same interval, as are (iii) and (iv). So exactly one of each pair must be true, and the answer is guaranteed to be one statement from each pair - that alone kills option 1 ((i) and (ii) cannot both hold) before any algebra.
Step 1: Exploit the identity. With a=1, b=x:
1+x3=(1+x)(1−x+x2)
On both intervals x>−1, so 1+x>0 - it is never zero, and we may safely divide by it:
1−x+x2=1+x1+x3Step 2: Find the delta. Subtract the two quantities being compared and put them over the common denominator:
(1−x+x2)−1+x1=1+x1+x3−1+x1=1+xx3
Everything now rides on the sign of 1+xx3, and since 1+x>0 throughout, that sign is simply the sign of x3 - which is the sign of x.
**Step 3: Evaluate Case 1, 0<x<1.** The numerator x3 is positive and the denominator 1+x is positive, so the delta is positive:
1+xx3>0⟹1−x+x2>1+x1
Flipping that around to the form the paper uses gives 1+x1<1−x+x2, which is exactly Statement (i) - and therefore (ii) is false.
*Spot check at x=21:* 1+x1=32 and 1−x+x2=43, and indeed 32<43, with the gap 121 matching 1+xx3=3/21/8=121.
**Step 4: Evaluate Case 2, −1<x<0.** Here the numerator x3 is negative (a negative number cubed stays negative), while the denominator 1+x is still positive - for instance 1+(−0.5)=0.5. A negative over a positive is negative:
1+xx3<0⟹1−x+x2<1+x1
Flipping it round gives 1+x1>1−x+x2, which is exactly Statement (iv) - and therefore (iii) is false.
*Spot check at x=−21:* 1+x1=2 and 1−x+x2=47, and indeed 2>47, with the gap −41 matching 1+xx3=1/2−1/8=−41.
Step 5: Assemble. The true statements are (i) and (iv), one from each contradictory pair as Step 0 predicted.
Final Answer: (i) and (iv)
The smallest possible number of students in a class if the girls in the class are less than 50% but more than 48% is
A27
B100
C200
D25
Pick an option to attempt
The Setup: We need to find the smallest possible total number of students in a class, given the percentage of girls is strictly between 48% and 50%.
Step 1: Formulate the inequality constraint.
Let G be the integer number of girls and N be the total integer number of students.
48%<NG<50%2512<NG<21Step 2: Test values for N starting from the lower bound constraints.
For the fraction to sit strictly between 0.48 and 0.50, N must be minimally larger than 25.
If N=25: 2512=0.48 (Fails the strict greater-than inequality).
If N=26: The maximum integer G strictly less than half of 26 is 12. 2612≈0.461 (Fails, as it is less than 0.48).
If N=27: The maximum integer G strictly less than half of 27 is 13. 2713≈0.48148 (Valid).
Step 3: Conclude the minimum total.
Since 0.48<0.48148<0.50, the minimum total class size N that supports a valid integer amount of girls is 27.
Final Answer: 27
If x∈(a,b) satisfies the inequality x2+3x+2x−3≥1, then the largest possible value of b−a is
A3
B1
C2
DNo real values of x satisfies the inequality
Pick an option to attempt
The Setup: A rational inequality. We cannot cross-multiply, because the denominator's sign is unknown and multiplying by a negative would flip the inequality. Everything goes to one side instead.
Step 1: Reframe. Move the 1 across and combine over a common denominator:
x2+3x+2x−3−1≥0⟹x2+3x+2x−3−(x2+3x+2)≥0⟹x2+3x+2−x2−2x−5≥0Step 2: Kill the numerator's sign ambiguity. Multiply by −1 and reverse the inequality:
x2+3x+2x2+2x+5≤0
Check the numerator's discriminant:
Δ=22−4(1)(5)=4−20=−16<0
With Δ<0 and a positive leading coefficient, x2+2x+5 is **strictly positive for every real x - it never touches zero. So the numerator can never make the fraction zero or negative on its own.
Step 3: Force the denominator negative.** A positive over something, being ≤0, requires the denominator to be strictly negative (zero is not allowed - the expression would be undefined):
(x+1)(x+2)<0
An upward parabola sits below zero strictly between its roots:
x∈(−2,−1)
Note the interval is open because the endpoints −2 and −1 make the denominator zero, where the original expression is undefined. Separately worth seeing: the numerator x2+2x+5 never vanishes, so the fraction is never exactly zero - meaning equality in the original ≥1 is never attained, and the condition behaves throughout as a strict >1.
Step 4: Compute the width. With a=−2 and b=−1:
b−a=−1−(−2)=1
Option 4 - claiming no real x works - is the trap for anyone who mishandles the sign flip in Step 2 and concludes the fraction can never be negative.
Final Answer: 1
The set of values of x which satisfy the inequality 0.7(2x2−3x+4)<0.343 is
A(21,1)
B(21,∞)
C(−∞,21)
D(−∞,21)∪(1,∞)
Pick an option to attempt
The Setup: An exponential inequality with a trap built into the base. Because 0.7<1, the function 0.7t is decreasing, so comparing exponents reverses the inequality - an UNO reverse card most people forget to play.
Step 1: Match the bases. Since 73=343, scaling the decimal places gives:
0.343=(0.7)3⟹0.7(2x2−3x+4)<0.73Step 2: Drop the bases and flip. For a base in (0,1), a smaller value corresponds to a larger exponent:
2x2−3x+4>3Step 3: Solve the quadratic inequality.2x2−3x+1>0
Factor by splitting the middle term:
2x2−2x−x+1>0⟹2x(x−1)−1(x−1)>0⟹(2x−1)(x−1)>0
The roots are x=21 and x=1. The leading coefficient is positive, so the parabola opens upward and is positive outside its roots and negative between them.
Step 4: Write the interval.x<21orx>1
Option 1, the interval (21,1), is precisely what you get by not flipping the sign in Step 2: that error yields 2x2−3x+1<0, whose solution is the region *between* the roots rather than outside them. The whole question turns on that single step.
Final Answer:(−∞,21)∪(1,∞)