Chapter 3: Real Numbers, Complex Numbers and Quadratic Expressions (Set-4)
If xx lies in the interval [−2,5)[−2,5), which statement is always true
A x > 5
B x ≤ 5
C x ≥ −2
D x < −2
The interval [−2,5)[−2,5) includes −2 (closed bracket) and excludes 5 (open bracket). So every number in this set must be at least −2, while being strictly less than 5.
For real numbers a
A a excluded, b included
B a included, b excluded
C both included
D both excluded
Parenthesis “(” means endpoint is not included, while bracket “]” means included. So (a,b](a,b] contains numbers greater than a and up to b including b.
Which statement about irrational numbers is correct
A Always repeating decimals
B Can be p/q
C Non-repeating decimals
D Only negative numbers
Irrational numbers cannot be expressed as a fraction of integers. Their decimal expansion neither terminates nor repeats, like 22 or ππ. Repeating or terminating decimals are rational.
If ∣x∣=∣y∣∣x∣=∣y∣, then which must be true for real numbers
A x=y always
B x=y or x=−y
C x=−y always
D xy=1 always
Absolute values being equal means distances from zero are equal. This happens when the numbers are the same or exact opposites. So ∣x∣=∣y∣⇒x=y∣x∣=∣y∣⇒x=y or x=−yx=−y.
Simplify 8228
A 2
B √4
C √2
D 4
82=82=4=228=28=4=2. Dividing surds under the same root is valid for positive numbers and often simplifies into an integer.
Rationalize 35+25+23
A 3(5−2)333(5−2)
B 3(5+2)773(5+2)
C 3(5−2)773(5−2)
D 3(5−2)5−25−23(5−2)
Multiply by conjugate (5−2)(5−2). Denominator becomes (5)2−(2)2=5−2=3(5)2−(2)2=5−2=3. So the rationalized form is 3(5−2)3=5−233(5−2)=5−2.
If am=anam=an for a>0,a≠1a>0,a=1, then
A m+n=0
B mn=1
C m=n
D m−n=1
For positive base not equal to 1, exponential function is one-to-one. So equal powers imply equal exponents. Example: 2m=2n⇒m=n2m=2n⇒m=n. This is used in solving exponential equations.
Solve 5x−1=255x−1=25
A 3
B 1
C 2
D 0
Rewrite 2525 as 5252. Then 5x−1=525x−1=52 gives x−1=2x−1=2, so x=3x=3. Matching bases makes solving straightforward.
If log2(x)=log2(8)log2(x)=log2(8), then xx equals
A 2
B 8
C 4
D 16
If logs with the same base are equal, their arguments are equal (with valid domain). Since log2(x)=log2(8)log2(x)=log2(8), we must have x=8x=8 because x>0x>0 also holds.
The expression loga(xk)loga(xk) equals (for x>0)
A log_a x + k
B log_a x − k
C k log_a x
D log_a (x+k)
Power rule: loga(xk)=klogaxloga(xk)=klogax for x>0x>0. It converts exponents into multipliers and is widely used in simplifying logs and solving equations.
If z=4−7iz=4−7i, then zˉzˉ is
A 4+7i
B −4+7i
C 7−4i
D −4−7i
The conjugate of a+bia+bi is a−bia−bi. So conjugate of 4−7i4−7i is 4+7i4+7i. Conjugates reflect points across the real axis.
If z=3+2iz=3+2i, then zzˉzzˉ equals
A 5
B 9
C 1
D 13
zzˉ=(3+2i)(3−2i)=32+22=9+4=13zzˉ=(3+2i)(3−2i)=32+22=9+4=13. This equals ∣z∣2∣z∣2. Multiplying by conjugate removes imaginary parts and produces a real number.
For z=a+biz=a+bi, the value of zzˉzˉz has modulus
A 0
B |z|
C 1
D |z|²
∣zzˉ∣=∣z∣∣zˉ∣=∣z∣∣z∣=1zˉz=∣zˉ∣∣z∣=∣z∣∣z∣=1 for z≠0z=0. Conjugate keeps modulus same, so the ratio lies on unit circle.
Compute ∣3i−4∣∣3i−4∣
A 7
B 5
C 1
D √7
3i−4=−4+3i3i−4=−4+3i. Modulus is (−4)2+32=16+9=25=5(−4)2+32=16+9=25=5. It represents distance from origin in Argand plane.
If z1=1+iz1=1+i and z2=1−iz2=1−i, then z1z2z1z2 equals
A 2
B 0
C −2
D i
(1+i)(1−i)=1−i+i−i2=1+1=2(1+i)(1−i)=1−i+i−i2=1+1=2. This is a conjugate product. Imaginary parts cancel and i2=−1i2=−1 makes the result real.
Argument of ii (principal) is
A 0
B π
C π/2
D −π/2
ii lies on positive imaginary axis, at point (0,1). The angle from positive real axis to this point is 90∘90∘ or π/2π/2. This is its principal argument.
If z=r(cosθ+isinθ)z=r(cosθ+isinθ), then zˉzˉ equals
A r(cosθ+i sinθ)
B r(sinθ+i cosθ)
C r(cos(−θ)+i sinθ)
D r(cosθ−i sinθ)
Conjugate changes sign of imaginary part. In polar form r(cosθ+isinθ)r(cosθ+isinθ), conjugate becomes r(cosθ−isinθ)r(cosθ−isinθ). It also corresponds to angle −θ−θ geometrically.
Using De Moivre, (cosθ+isinθ)3(cosθ+isinθ)3 equals
A cos3θ+i sin3θ
B cosθ+i sin3θ
C cos3θ+i sinθ
D cos2θ+i sin2θ
De Moivre’s theorem: (cosθ+isinθ)n=cos(nθ)+isin(nθ)(cosθ+isinθ)n=cos(nθ)+isin(nθ) for integer nn. For n=3n=3, it becomes cos3θ+isin3θcos3θ+isin3θ.
Square root of −9−9 in complex numbers is
A ±3
B ±9i
C ±3i
D ±i
−9=9−1=3i−9=9−1=3i. Since both 3i3i and −3i−3i square to −9−9, we write the square roots as ±3i±3i.
If z=cosθ+isinθz=cosθ+isinθ, then z4z4 equals
A cos2θ+i sin2θ
B cos4θ+i sin4θ
C cosθ+i sinθ
D cos3θ+i sin3θ
Apply De Moivre with n=4n=4: z4=(cosθ+isinθ)4=cos4θ+isin4θz4=(cosθ+isinθ)4=cos4θ+isin4θ. It simplifies powers of complex numbers on unit circle.
The locus ∣z−(1+i)∣=∣z−(1−i)∣∣z−(1+i)∣=∣z−(1−i)∣ is
A Imaginary axis
B Line x=1
C Real axis
D Line y=1
Points equidistant from 1+i1+i and 1−i1−i lie on perpendicular bisector of the segment joining them. The segment is vertical through x=1, so bisector is horizontal line y=0 (real axis).
The locus ∣z∣=∣z−2i∣∣z∣=∣z−2i∣ is the line
A y=1
B x=1
C y=0
D x=0
Equidistant from 0 and 2i means perpendicular bisector of segment joining (0,0) and (0,2). Midpoint is (0,1) and bisector is horizontal line y=1.
The region ∣z−2∣>3∣z−2∣>3 represents
A Interior of circle
B Circle boundary
C Exterior of circle
D Vertical strip
∣z−a∣>r∣z−a∣>r means points whose distance from center aa is greater than rr. So it is outside the circle centered at 2 on real axis with radius 3, excluding boundary.
The region Re(z) ≤ 2 represents
A Right half-plane
B Left half-plane
C Upper half-plane
D Disc region
Re(z)=x. Condition x≤2x≤2 includes all points to the left of vertical line x=2, including the line itself. It’s a half-plane bounded by x=2.
The set ∣z∣=2∣z∣=2 and Arg(z)=π/2 gives the point
A 2i
B −2i
C 2
D −2
∣z∣=2∣z∣=2 means distance 2 from origin. Arg =π/2=π/2 means positive imaginary direction. So the point lies at (0,2), i.e., 2i2i.
If a quadratic has roots 4 and 6, its monic equation is
A x²+10x+24
B x²−10x−24
C x²+10x−24
D x²−10x+24
For roots 4 and 6, equation is (x−4)(x−6)=x2−10x+24(x−4)(x−6)=x2−10x+24. Sum of roots is 10, product is 24, matching −b−b and cc in monic form.
If α,βα,β are roots, then α2+β2α2+β2 equals
A (α+β)²+2αβ
B (αβ)²
C (α+β)²−2αβ
D αβ−(α+β)
Identity: α2+β2=(α+β)2−2αβα2+β2=(α+β)2−2αβ. This is useful with quadratic relations where α+βα+β and αβαβ are known from coefficients.
If roots are 2±3i2±3i, the real-coefficient quadratic is
A x²−4x+13
B x²+4x+13
C x²−4x−13
D x²+4x−13
Sum of roots: (2+3i)+(2−3i)=4(2+3i)+(2−3i)=4. Product: (2+3i)(2−3i)=4+9=13(2+3i)(2−3i)=4+9=13. So equation is x2−4x+13=0x2−4x+13=0 for monic quadratic.
The quadratic 2×2−3x−2=02×2−3x−2=0 factors as
A (2x−1)(x+2)
B (2x+1)(x−2)
C (x−1)(2x+2)
D (2x+2)(x−1)
Multiply gives 2×2−4x+x−2=2×2−3x−22×2−4x+x−2=2×2−3x−2. Correct factorization gives roots x=2x=2 and x=−1/2x=−1/2. Factorization must match all coefficients exactly.
If x2−5x+m=0x2−5x+m=0 has roots 2 and 3, then m is
A 5
B −6
C 6
D 1
Product of roots equals c/ac/a. Here a=1a=1, so m=2⋅3=6m=2⋅3=6. Also sum is 5 matching coefficient −5, confirming correctness.
The vertex of y=x2−6x+5y=x2−6x+5 has x-coordinate
A −3
B 6
C −6
D 3
For y=ax2+bx+cy=ax2+bx+c, vertex x-coordinate is −b/(2a)−b/(2a). Here a=1, b=−6, so x=−(−6)/(2)=3. This gives axis of symmetry x=3.
Minimum value of x2−6x+5×2−6x+5 equals
A −4
B 4
C 0
D 5
Complete square: x2−6x+5=(x−3)2−9+5=(x−3)2−4×2−6x+5=(x−3)2−9+5=(x−3)2−4. Since (x−3)2≥0(x−3)2≥0, minimum value is −4−4 at x=3.
For x2+px+q=0x2+px+q=0, if roots are equal, then
A p²=2q
B p²=4q
C q²=4p
D p=0 always
Equal roots mean discriminant p2−4q=0p2−4q=0 (since a=1). So p2=4qp2=4q. This condition tells when a quadratic becomes a perfect square.
If one root of x2−7x+10=0x2−7x+10=0 is 5, the other root is
A 10
B −2
C 2
D 7
Sum of roots is 7. If one root is 5, the other must be 7−5=2. Also product 5×2=10 matches constant term, confirming the second root.
Solve the inequality 4−3x≥14−3x≥1
A x ≤ 1
B x ≥ 1
C x < 1
D x > 1
4−3x≥1⇒−3x≥−34−3x≥1⇒−3x≥−3. Divide by −3 and reverse sign: x≤1x≤1. Always reverse inequality when dividing by a negative number.
Solve 2
A 2
B 1≤x<5
C 1
D 0
Subtract 1 throughout: 2−1
The solution of x≥0x≥0 and x<3x<3 is
A (0,3]
B (−∞,3)
C [0,∞)
D [0,3)
Intersection of conditions keeps values satisfying both. x≥0 includes 0, and x<3 excludes 3. So solution set is all numbers from 0 up to but not including 3: [0,3).
Solve the inequality 2x−1x−3>0x−32x−1>0
A (−∞,1/2)∪(3,∞)
B (1/2,3)
C (−∞,3)
D (3,1/2)
Critical points: numerator zero at x=1/2x=1/2, denominator zero at x=3x=3 (excluded). Sign chart shows fraction positive when both numerator and denominator have same sign: x<1/2x<1/2 or x>3x>3.
Solve ∣x+2∣≥5∣x+2∣≥5
A −7
B x≤−3 or x≥7
C x≤−7 or x≥3
D −3
∣x+2∣≥5∣x+2∣≥5 means distance from −2 is at least 5. So x+2≥5⇒x≥3x+2≥5⇒x≥3 or x+2≤−5⇒x≤−7x+2≤−5⇒x≤−7. Union gives final set.
If ∣x−4∣<∣x−1∣∣x−4∣<∣x−1∣, then x lies
A x<2.5
B x>2.5
C x=2.5
D all real x
Points closer to 4 than to 1 lie on the side of the perpendicular bisector of 1 and 4. Midpoint is 2.5. Closer to 4 means x>2.5 on the number line.
The ceiling value ⌈3.01⌉⌈3.01⌉ equals
A 3
B 3.01
C 4
D 2
Ceiling is the smallest integer greater than or equal to the number. Since 3.01 is slightly more than 3, the next integer is 4. Ceiling always rounds up to an integer.
If zz is nonzero, the value of ∣1z∣z1 equals
A 1/|z|
B |z|
C |z|²
D 1/|z|²
Modulus of reciprocal follows ∣1z∣=1∣z∣z1=∣z∣1 for z≠0z=0. Because modulus behaves like distance, inversion flips magnitude to its reciprocal.
If ∣z∣=2∣z∣=2, then ∣z3∣∣z3∣ equals
A 8
B 6
C 4
D 2
Modulus property: ∣zn∣=∣z∣n∣zn∣=∣z∣n for integer n. So ∣z3∣=23=8∣z3∣=23=8. This helps simplify magnitudes without expanding complex powers.
If z2=1z2=1, then possible values of z are
A ±i
B 0 and 1
C ±1
D 1 only
Solutions to z2=1z2=1 are z=±1z=±1, because both 1 and −1 square to 1. These are real solutions and are also complex numbers.
The sum of cube roots of unity equals
A 1
B 0
C 3
D −1
Cube roots of unity are 1,ω,ω21,ω,ω2 where ω≠1ω=1. They satisfy 1+ω+ω2=01+ω+ω2=0. Geometrically, they form an equilateral triangle on unit circle.
If xx satisfies 2x=2−32x=2−3, then x equals
A −3
B 3
C 0
D 1
Same base implies equal exponents: 2x=2−3⇒x=−32x=2−3⇒x=−3. This is a basic exponential property and avoids taking logs when bases already match.
If log5(x)>1log5(x)>1, then x must satisfy
A x<5
B 0
C x>5
D x<0
Base 5 is greater than 1, so log function is increasing. log5(x)>1log5(x)>1 implies x>51=5x>51=5, with domain x>0x>0. So the solution is x>5.
If x2+1=0x2+1=0, then x lies on
A Real axis
B Unit circle only
C Line y=x
D Imaginary axis
Solutions are x=±ix=±i, which have real part 0 and nonzero imaginary part. Points ±i lie on imaginary axis at (0,1) and (0,−1) in Argand plane.
For complex numbers, the inequality ∣z1+z2∣≤∣z1∣+∣z2∣∣z1+z2∣≤∣z1∣+∣z2∣ is called
A Modulus identity
B Triangle inequality
C Conjugate rule
D De Moivre rule
Triangle inequality states the modulus of a sum is at most the sum of moduli. It matches the geometric idea that one side of a triangle is not longer than the other two sides combined.
If z=eiθz=eiθ, then zˉzˉ equals
A e^{−iθ}
B e^{iθ}
C e^{θ}
D θe^{i}
Using Euler form, eiθ=cosθ+isinθeiθ=cosθ+isinθ. Conjugate changes sign of imaginary part, giving cosθ−isinθ=e−iθcosθ−isinθ=e−iθ. So eiθ‾=e−iθeiθ=e−iθ.