Chapter 3: Real Numbers, Complex Numbers and Quadratic Expressions (Set-2)
Which statement defines rational numbers?
A B. Non-terminating non-repeating
B C. Only positive numbers
C D. Only whole numbers
D A. Terminating or repeating
A rational number can be written as p/q (q≠0). Its decimal expansion either terminates (like 0.125) or repeats (like 0.333…). Non-terminating non-repeating decimals are irrational.
Which is a real number?
A B. 5
B A. √(-1)
C C. 3+2i
D D. 2i
Real numbers include all rationals and irrationals on the number line. 5 is real. √(-1), 2i, and 3+2i involve i, so they are complex, not purely real.
Which interval represents x between 1 and 4 inclusive?
A A. (1,4)
B B. [1,4]
C C. [1,4)
D D. (1,4]
Inclusive means endpoints are included, shown with square brackets. So x can be 1, 4, or any number between them. Hence the closed interval is [1,4].
If |x−2| = 0, then x equals
A A. 0
B C. −2
C D. 1
D B. 2
Absolute value equals zero only when the inside is zero. So |x−2|=0 implies x−2=0, giving x=2. No other value makes the distance from 2 equal to zero.
Simplify (√3)(√12)
A B. 3√4
B C. √36
C A. 6
D D. 2√9
√3·√12 = √(3×12) = √36 = 6. Multiplying surds under one radical is valid for nonnegative numbers, then simplify using perfect squares.
Simplify √18 + √8
A B. 3√10
B C. √26
C D. 6√2
D A. 5√2
√18=√(9×2)=3√2 and √8=√(4×2)=2√2. Add like surds: 3√2+2√2=5√2. Only like radicals can be added.
Value of 2^0 is
A B. 1
B A. 0
C C. 2
D D. Undefined
Any nonzero number raised to power 0 equals 1. This rule keeps exponent laws consistent, like a^m/a^m = a^(m−m)=a^0=1 for a≠0.
log_10(100) equals
A A. 1
B B. 2
C C. 10
D D. 0
log_10(100)=2 because 10^2=100. A logarithm asks “what power of the base gives the number,” so base 10 giving 100 needs exponent 2.
If a>0, a≠1, then log_a(a) equals
A A. 0
B C. a
C D. −1
D B. 1
log_a(a)=1 because a^1=a. This is a basic identity of logarithms. It holds for any valid base a (positive and not equal to 1).
Which is true for all real x?
A C. |x| ≥ x
B A. |x| = x always
C B. |x| = −x always
D D. |x| ≤ x
If x≥0, then |x|=x so equality holds. If x<0, then |x|=−x which is greater than x. Therefore |x| is always at least x.
Real part of 7−3i is
A A. −3
B B. 3
C C. 7
D D. −7
In a+bi, the real part is a and imaginary part coefficient is b. For 7−3i, a=7 and b=−3. Real part is the x-coordinate on Argand plane.
Imaginary part of 4+9i is
A A. 4
B C. i
C D. −9
D B. 9
Imaginary part means the coefficient of i in a+bi form. For 4+9i, the coefficient of i is 9. The number i itself is not the imaginary part.
Add (2+3i) + (1−5i)
A B. 1−2i
B A. 3−2i
C C. 3+8i
D D. 2−8i
Add real parts and imaginary parts separately: (2+1) + (3i−5i) = 3 − 2i. This method follows algebraic rules treating i like a variable.
Multiply i(3−2i) equals
A B. 3i+2
B A. 3i−2
C C. 3−2i
D D. −3i−2
i(3−2i)=3i−2i^2. Since i^2=−1, −2i^2 = −2(−1)=+2. So the result is 2+3i.
(1+i)^2 equals
A B. 2
B C. 1+i
C D. −2i
D A. 2i
(1+i)^2 = 1 + 2i + i^2 = 1 + 2i − 1 = 2i. The real parts cancel because i^2 = −1.
Simplify (2+i)(2−i)
A A. 3
B B. 4
C C. 5
D D. 6
(2+i)(2−i) = 2^2 − (i)^2 = 4 − (−1) = 5. Conjugate multiplication removes imaginary parts and gives a real number.
If z = 3+4i, then |z| equals
A A. 5
B B. 7
C C. 1
D D. √7
|3+4i| = √(3²+4²) = √(9+16) = √25 = 5. Modulus is the distance from origin in the Argand plane.
If z = a+bi, then z + z̄ equals
A A. 2bi
B C. a
C D. b
D B. 2a
z̄ = a−bi. Adding gives (a+bi)+(a−bi)=2a. Imaginary parts cancel. This identity is used to extract the real part from a complex number.
If z = a+bi, then z − z̄ equals
A A. 2a
B C. 2bi
C B. −2a
D D. 2b
z − z̄ = (a+bi) − (a−bi) = 2bi. Real parts cancel. This helps isolate imaginary part since Im(z) = (z−z̄)/(2i).
Value of i^9 is
A B. i
B A. 1
C C. −1
D D. −i
Powers of i repeat every 4. i^9 = i^(8+1) = (i^4)^2 · i = 1·i = i. Always reduce exponent modulo 4.
Point representing 2−i is
A B. (−2,1)
B C. (1,2)
C A. (2,−1)
D D. (−1,2)
In Argand plane, a+bi corresponds to (a, b). For 2−i, a=2 and b=−1, so the point is (2, −1), right and downward.
Locus |z−(2+i)| = 3 is
A A. Line
B C. Parabola
C D. Ray
D B. Circle
|z−z0| = r represents all points at distance r from fixed point z0. So it forms a circle centered at 2+i with radius 3 in the complex plane.
Region Im(z) > 0 means
A B. Upper half-plane
B A. Lower half-plane
C C. Left half-plane
D D. Right half-plane
Im(z)>0 means y-coordinate is positive, so points lie above the real axis. This is called the upper half-plane, excluding the real axis itself.
Region Re(z) < 0 means
A A. Right half-plane
B B. Left half-plane
C C. Upper half-plane
D D. Origin only
Re(z)<0 means x-coordinate is negative. These points lie left of the imaginary axis. The imaginary axis itself has Re(z)=0 and is not included.
Locus |z| = 1 is
A B. Unit square
B C. Real axis
C D. Imaginary axis
D A. Unit circle
|z| is distance from origin. |z|=1 gives all points one unit away from origin, forming a circle of radius 1 centered at origin, called the unit circle.
Region 1 < |z| < 2 is
A A. Annulus region
B B. Disc region
C C. Half-plane
D D. Line segment
1<|z|<2 means points whose distance from origin lies between 1 and 2. This creates a ring-shaped region between two concentric circles, excluding both boundaries.
Locus |z−1| = |z+1| is
A A. Real axis
B C. Circle center 1
C B. Imaginary axis
D D. Line y=x
Points equidistant from 1 and −1 lie on the perpendicular bisector of segment joining them. That segment is on real axis, so bisector is imaginary axis (x=0).
Roots of x²+1=0 are
A B. ±1
B C. 0 and 1
C D. ±√1
D A. ±i
x²+1=0 gives x²=−1, so x=±√(−1)=±i. This is the simplest example showing quadratics can have complex roots.
If roots are 2 and −3, equation is
A A. x²+x−6
B B. x²−x−6
C C. x²−x+6
D D. x²+x+6
For roots α,β, equation is x² −(α+β)x + αβ = 0. Here α+β = −1 and αβ = −6, so x² + x − 6 = 0.
Which factorization is correct? x²−9
A B. (x−9)(x+1)
B A. (x−3)(x+3)
C C. (x−3)²
D D. (x+3)²
x²−9 is a difference of squares: x²−3² = (x−3)(x+3). This identity is commonly used to quickly solve and simplify expressions.
Minimum value of (x−4)² is
A B. 4
B C. −4
C D. 16
D A. 0
A square is always ≥0, and it becomes 0 when the inside is 0. So (x−4)² is minimum at x=4 and the minimum value is 0.
If a quadratic has equal roots, then
A B. Δ=0
B A. Δ>0
C C. Δ<0
D D. a=0
Equal roots occur when the discriminant is zero. Then √Δ=0, both solutions become the same value −b/(2a). Graphically, the parabola touches the x-axis once.
For x²+bx+9=0 with equal roots, b equals
A B. 9 or −9
B C. 3 or −3
C A. 6 or −6
D D. 18 or −18
Equal roots means Δ=b²−4ac=0. Here a=1, c=9, so b²−36=0 ⇒ b²=36 ⇒ b=±6. Both values make the roots repeated.
If α,β are roots, then αβ equals
A A. −b/a
B C. b/a
C D. −c/a
D B. c/a
For ax²+bx+c=0, product of roots αβ = c/a. This comes from expanding a(x−α)(x−β) = ax² −a(α+β)x + aαβ and comparing coefficients.
Quadratic inequality x²−4x+3 > 0 holds for
A B. 1<x<3
B A. x<1 or x>3
C C. x≤1 or ≥3
D D. All real x
Factor x²−4x+3=(x−1)(x−3). Product is positive when both factors are positive or both negative, giving x<1 or x>3. Endpoints excluded due to “>”.
Solve 3x−5 ≤ 7
A A. x ≤ 4
B B. x ≥ 4
C C. x < 4
D D. x > 4
3x−5 ≤ 7 ⇒ 3x ≤ 12 ⇒ x ≤ 4. Division by positive 3 keeps the inequality sign unchanged. Always isolate x step-by-step.
Solve 5−2x > 1
A A. x > 2
B C. x ≥ 2
C D. x ≤ 2
D B. x < 2
5−2x>1 ⇒ −2x>−4. Divide by −2 (negative), reverse sign: x<2. Sign reversal is essential whenever dividing by a negative number.
Solve |x−1| ≤ 3
A B. −2 < x < 4
B A. −2 ≤ x ≤ 4
C C. x ≤ −2 or ≥4
D D. x < −2 or >4
|x−1|≤3 means −3 ≤ x−1 ≤ 3. Add 1 throughout: −2 ≤ x ≤ 4. Because “≤” includes endpoints, the interval is closed.
Solve |2x| < 6
A A. −3 < x < 3
B B. −6 < x < 6
C C. x < −3 or >3
D D. x ≤ −3 or ≥3
|2x|<6 implies −6<2x<6. Divide by 2 (positive): −3<x<3. Absolute inequality becomes a double inequality for “<”.
Solve |x| = 4 gives
A A. x = 4 only
B B. x = −4 only
C D. x = 0
D C. x = ±4
|x|=4 means the distance of x from 0 is 4. That happens at two points: x=4 and x=−4. Absolute value equations usually give two symmetric solutions.
Interval (−∞,3) represents
A A. x ≤ 3
B C. x ≥ 3
C B. x < 3
D D. x > 3
Parenthesis means endpoint not included. So (−∞,3) means all real numbers strictly less than 3. Infinity is always written with a parenthesis.
Solution set of x≥−1 and x<2 is
A A. [−1,2)
B B. (−1,2)
C C. [−1,2]
D D. (−∞,2)
x≥−1 includes −1, and x<2 excludes 2. Intersection gives all x from −1 up to but not including 2, written as [−1,2).
Solve 2^x = 8 gives x
A A. 2
B C. 4
C D. 8
D B. 3
8 equals 2^3. So 2^x=8 implies x=3. Exponential equations often become easy when rewriting numbers with the same base.
Solve log_2(x) = 5 gives x
A A. 10
B C. 32
C B. 16
D D. 64
log_2(x)=5 means 2^5=x. Since 2^5=32, x=32. Logs convert exponent questions into direct values using base powers.
Which is true for a>0, a≠1?
A B. log_a(x+y)=log_a x + log_a y
B C. log_a(xy)=log_a x − log_a y
C A. log_a(xy)=log_a x + log_a y
D D. log_a(x)=a^x
Product rule of logarithms states log_a(xy)=log_a x + log_a y for x>0,y>0. There is no simple rule for log of a sum.
Euler form connects polar form as
A B. e^{iθ}=cosθ−i sinθ
B C. e^{iθ}=sinθ+i cosθ
C D. e^{iθ}=θ+i
D A. e^{iθ}=cosθ+i sinθ
Euler’s formula states e^{iθ}=cosθ+i sinθ. It links exponential and trigonometric forms and helps write complex numbers compactly in polar/exponential form.
If z is purely imaginary, then Re(z) is
A B. 0
B A. 1
C C. Nonzero
D D. Undefined
A purely imaginary number has the form bi where b≠0. Its real part is 0. On Argand plane, such numbers lie on the imaginary axis.
Cube roots of unity satisfy
A B. z²=1
B C. z³=0
C D. z=1 only
D A. z³=1
Cube roots of unity are the complex numbers that satisfy z³=1. They are 1, ω, and ω², equally spaced by 120° on the unit circle.
Distance between z1 and z2 in plane is
A A. |z1+z2|
B B. |z1−z2|
C C. |z1·z2|
D D. |z1/z2|
In the complex plane, distance between points z1 and z2 is the modulus of their difference: |z1−z2|. This matches usual distance formula in coordinate geometry.
For any complex z, |z|² equals
A C. z·z̄
B A. z+z̄
C B. z−z̄
D D. z/z̄
If z=a+bi, then z̄=a−bi. Multiplying gives z·z̄ = (a+bi)(a−bi)=a²+b² = |z|². This identity is widely used in simplification.