Chapter 3: Real Numbers, Complex Numbers and Quadratic Expressions (Set-3)
The decimal 0.375 can be written as a fraction, so it belongs to which set
A Natural numbers
B Integers
C Rational numbers
D Irrational numbers
Any number that can be written as p/qp/q (with q≠0q=0) is rational. 0.375=375/1000=3/80.375=375/1000=3/8, so it clearly fits the rational number set.
A number with non-terminating and non-repeating decimal expansion is called
A Rational number
B Irrational number
C Integer
D Natural number
Irrational numbers cannot be expressed as a fraction of integers. Their decimal form never ends and never repeats in a fixed pattern, unlike rational decimals which terminate or repeat.
For a negative real number xx, the value of ∣x∣∣x∣ becomes
A −x value
B x itself
C x squared
D 1/x value
Absolute value is distance from zero, so it is always non-negative. If xx is negative, distance is obtained by changing sign: ∣x∣=−x∣x∣=−x, making it positive.
Simplify the surd product (35)(220)(35)(220)
A 30
B 12√5
C 60
D 6√20
20=4⋅5=2520=4⋅5=25. Then (35)(2⋅25)=(3⋅4)(5⋅5)=12⋅5=60(35)(2⋅25)=(3⋅4)(5⋅5)=12⋅5=60. Surds multiply under root.
Rationalize and simplify 12−32−31
A 2 − √3
B 2 + √3
C (2+√3)/7
D 1 + √3
Multiply by the conjugate (2+3)(2+3): 12−3⋅2+32+3=2+34−3=2+32−31⋅2+32+3=4−32+3=2+3. Denominator becomes rational.
Simplify 27−1227−12
A 5√3
B √15
C 1/√3
D √3
27=3327=33 and 12=2312=23. Subtract like surds: 33−23=333−23=3. Only surds with same root part can be combined.
For nonzero aa, the expression a−2a−2 equals
A a²
B −a²
C 1/a²
D 1/2a
A negative exponent means reciprocal: a−2=1/a2a−2=1/a2. This rule keeps exponent laws consistent, such as am⋅a−m=a0=1am⋅a−m=a0=1 when a≠0a=0.
Evaluate 163/4163/4
A 8
B 4
C 16
D 2
161/4=2161/4=2 because 24=1624=16. Then 163/4=(161/4)3=23=8163/4=(161/4)3=23=8. Fractional powers mean roots and powers together.
Solve the logarithmic equation log3(x)=2log3(x)=2
A 6
B 9
C 3
D 12
log3(x)=2log3(x)=2 means 32=x32=x. Since 32=932=9, the solution is x=9x=9. Logarithms convert an exponent question into a simple value.
Change of base formula for logablogab can be written as
A ln a / ln b
B b ln a
C ln b / ln a
D a ln b
logab=lnblnalogab=lnalnb for a>0,a≠1,b>0a>0,a=1,b>0. This allows evaluation of any base log using natural log (or common log) safely.
Conjugate of the complex number −2+5i−2+5i is
A 2 + 5i
B −2 + 5i
C 2 − 5i
D −2 − 5i
Conjugate changes only the sign of the imaginary part. If z=a+biz=a+bi, then zˉ=a−bizˉ=a−bi. So −2+5i−2+5i becomes −2−5i−2−5i, useful in division and modulus.
Add the complex numbers (4−3i)+(−1+7i)(4−3i)+(−1+7i)
A 5 + 10i
B 3 + 4i
C 3 − 10i
D 5 + 4i
Add real parts and imaginary parts separately: (4−1)=3(4−1)=3 and (−3+7)i=4i(−3+7)i=4i. So the sum is 3+4i3+4i. This keeps complex addition clear and error-free.
Multiply (1−2i)(3+i)(1−2i)(3+i)
A 5 + 5i
B −5 + 5i
C 5 − 5i
D −5 − 5i
Expand: 1⋅3+1⋅i−2i⋅3−2i⋅i=3+i−6i−2i21⋅3+1⋅i−2i⋅3−2i⋅i=3+i−6i−2i2. Since i2=−1i2=−1, result is 5−5i5−5i.
Simplify the division 2+i1+i1+i2+i
A (3 + i)/2
B (3 − i)/2
C (1 − i)/2
D (2 − i)/3
Multiply top and bottom by conjugate (1−i)(1−i). Denominator becomes 12+12=212+12=2. Numerator (2+i)(1−i)=2−2i+i−i2=3−i(2+i)(1−i)=2−2i+i−i2=3−i. So value is (3−i)/2(3−i)/2.
If a complex number zz satisfies z+zˉ=10z+zˉ=10, then its real part equals
A 5
B 10
C −5
D 0
Let z=a+biz=a+bi. Then zˉ=a−bizˉ=a−bi. Adding gives z+zˉ=2az+zˉ=2a. So 2a=10⇒a=52a=10⇒a=5. This quickly extracts real part without extra steps.
A complex number is purely real when it satisfies
A z = −z̄
B |z| = 0
C z = z̄
D arg(z) = π/2
If z=a+biz=a+bi equals its conjugate a−bia−bi, then bi=−bibi=−bi so b=0b=0. That means imaginary part is zero, making zz purely real.
Find the modulus of 1−i1−i
A 1
B √2
C 2
D √3
For a+bia+bi, modulus is a2+b2a2+b2. Here a=1a=1, b=−1b=−1. So ∣1−i∣=12+(−1)2=2∣1−i∣=12+(−1)2=2. Modulus is always non-negative.
For any complex number zz, the value of ∣zˉ∣∣zˉ∣ is always
A −|z|
B 1/|z|
C |z|²
D |z|
Conjugation reflects a point across the real axis but does not change its distance from the origin. Since modulus is distance, ∣zˉ∣=∣z∣∣zˉ∣=∣z∣ always, making it a dependable identity.
Compute the value of i2026i2026
A 1
B i
C −1
D −i
Powers of ii repeat every 4. 2026=4⋅506+22026=4⋅506+2, so i2026=i2i2026=i2. Since i2=−1i2=−1, the value is −1−1.
If z=cosθ+isinθz=cosθ+isinθ, then ∣z∣∣z∣ equals
A 1
B θ
C cosθ
D sinθ
∣z∣=cos2θ+sin2θ=1=1∣z∣=cos2θ+sin2θ=1=1. This is a key idea behind unit circle and polar form, where cosine and sine represent direction with modulus 1.
Using De Moivre, (cos60∘+isin60∘)2(cos60∘+isin60∘)2 equals
A cos30+i sin30
B cos120+i sin120
C cos60+i sin60
D cos240+i sin240
De Moivre says (cosθ+isinθ)n=cos(nθ)+isin(nθ)(cosθ+isinθ)n=cos(nθ)+isin(nθ). Here n=2n=2, θ=60∘θ=60∘, so result is cos120∘+isin120∘cos120∘+isin120∘.
Solve the equation w2=−16w2=−16 in complex numbers
A ±4
B ±8i
C ±4i
D ±2i
w2=−16⇒w=±−16=±4−1=±4iw2=−16⇒w=±−16=±4−1=±4i. Squares of 4i4i and −4i−4i both give −16−16, so two solutions exist.
The argument of the complex number −5−5 (on real axis) is
A 0
B π
C π/2
D 3π/2
−5−5 lies on the negative real axis. The angle from the positive real axis to the negative real axis is ππ radians. That angle is the principal argument here.
If z=(2i)(3−4i)z=(2i)(3−4i), then ∣z∣∣z∣ equals
A 10
B 8
C 6
D 5
First multiply: 2i(3−4i)=6i−8i2=8+6i2i(3−4i)=6i−8i2=8+6i. Then ∣z∣=82+62=64+36=100=10∣z∣=82+62=64+36=100=10. Modulus uses Pythagoras.
The exponential form of 2(cosπ6+isinπ6)2(cos6π+isin6π) is
A e^{iπ/6}
B 2e^{−iπ/6}
C 2e^{iπ/6}
D e^{−iπ/6}
Euler’s formula gives eiθ=cosθ+isinθeiθ=cosθ+isinθ. So r(cosθ+isinθ)=reiθr(cosθ+isinθ)=reiθ. Here r=2r=2, θ=π/6θ=π/6, so it becomes 2eiπ/62eiπ/6.
The locus of points satisfying Re(z) = Im(z) is the line
A y = −x
B x = 0
C y = 0
D y = x
Write z=x+iyz=x+iy. Then Re(z)=x and Im(z)=y. Condition x=yx=y gives the straight line y=xy=x through origin, making equal real and imaginary coordinates.
The locus ∣z−2∣=∣z+2∣∣z−2∣=∣z+2∣ represents
A y = 0
B x = 0
C x = 2
D y = 2
Points equidistant from 2 and −2 lie on the perpendicular bisector of segment joining them. That segment lies on the real axis, so the bisector is the imaginary axis x=0x=0.
The region 1<Re(z)<31<Re(z)<3 forms a
A Horizontal strip region
B Circular disc
C Vertical strip region
D Right half-plane
Re(z)=x. Condition 1<x<31<x<3 means x lies between two vertical lines x=1 and x=3. All y values are allowed, forming an open vertical strip.
The set ∣z−3i∣≤2∣z−3i∣≤2 describes a
A Circle centre 3i
B Annulus region
C Left half-plane
D Disc centre 3i
∣z−a∣≤r∣z−a∣≤r means all points within distance rr of aa, including boundary. Here center is 3i3i and radius 2, so it is a closed disc.
The condition Arg(z) = 0 represents the
A Negative real axis
B Positive real axis
C Positive imaginary axis
D Unit circle
Arg(z)=0 means the point lies along the positive real direction from origin. That includes all positive real numbers (and sometimes origin depending on convention), forming a ray on x-axis.
The condition Arg(z) = π represents the
A Negative real axis
B Positive real axis
C Upper half-plane
D Lower half-plane
Angle ππ from positive real axis points directly left. So Arg(z)=π corresponds to negative real numbers on the real axis, excluding the positive direction.
The locus ∣z∣=∣z−4∣∣z∣=∣z−4∣ is the line
A y = 2
B x = 4
C x = 2
D y = 0
Points equidistant from 0 and 4 lie on the perpendicular bisector of the segment joining them on real axis. Midpoint is 2, so bisector is vertical line x=2x=2.
The mapping w=z+(1−2i)w=z+(1−2i) moves every point
A Left 1, up 2
B Right 1, down 2
C Right 2, up 1
D Left 2, down 1
Adding 1−2i1−2i increases real part by 1 and decreases imaginary part by 2. So every point shifts right by 1 unit and down by 2 units in Argand plane.
A quadratic graph opens upward when its coefficient aa is
A a > 0
B a < 0
C a = 0
D any real a
In y=ax2+bx+cy=ax2+bx+c, sign of aa controls opening direction. If a>0a>0, parabola opens upward and has a minimum at the vertex. If a<0a<0, it opens downward.
For x2+6x+k=0x2+6x+k=0 to have equal roots, kk equals
A 12
B 6
C 9
D 3
Equal roots require discriminant b2−4ac=0b2−4ac=0. Here a=1,b=6,c=ka=1,b=6,c=k. So 36−4k=0⇒k=936−4k=0⇒k=9. This makes the quadratic a perfect square.
The equation x2−4x+13=0x2−4x+13=0 has roots that are
A Two real distinct
B Two real equal
C No solution
D Complex conjugate roots
Discriminant is Δ=b2−4ac=(−4)2−4(1)(13)=16−52=−36<0Δ=b2−4ac=(−4)2−4(1)(13)=16−52=−36<0. Negative discriminant means no real roots; roots are complex and occur as conjugate pair.
If sum of roots is −2 and product is −15, the monic quadratic is
A x²−2x−15
B x²+2x−15
C x²+2x+15
D x²−2x+15
For monic quadratic with roots α,βα,β: x2−(α+β)x+αβ=0x2−(α+β)x+αβ=0. Here α+β=−2α+β=−2 and αβ=−15αβ=−15, giving x2+2x−15=0x2+2x−15=0.
If roots are 3 and 1/3, an integer-coefficient equation is
A x²−10x+3
B 3x²+10x+3
C 3x²−10x+3
D 3x²−10x−3
(x−3)(x−13)=x2−103x+1(x−3)(x−31)=x2−310x+1. Multiply by 3 to remove fraction: 3×2−10x+3=03×2−10x+3=0. This keeps roots the same but coefficients integer.
The discriminant of x2−8x+12=0x2−8x+12=0 is
A 4
B 16
C −16
D 0
Δ=b2−4ac=(−8)2−4(1)(12)=64−48=16Δ=b2−4ac=(−8)2−4(1)(12)=64−48=16. Positive discriminant means two distinct real roots. Discriminant value helps decide root nature quickly.
For ax2+bx+c=0ax2+bx+c=0 with roots α,βα,β, the value of 1α+1βα1+β1 equals
A −b/c
B b/c
C −c/b
D c/b
1α+1β=α+βαβα1+β1=αβα+β. Using relations: α+β=−b/aα+β=−b/a and αβ=c/aαβ=c/a. So ratio becomes (−b/a)/(c/a)=−b/c(−b/a)/(c/a)=−b/c, provided c≠0c=0.
To make x2+8x+kx2+8x+k a perfect square (x+4)2(x+4)2, kk must be
A 8
B 4
C 16
D 32
(x+4)2=x2+8x+16(x+4)2=x2+8x+16. Comparing with x2+8x+kx2+8x+k, we must have k=16k=16. Completing the square uses half the x-coefficient, squared.
Solve the quadratic inequality x2+2x−3≤0x2+2x−3≤0
A (−3,1)
B x≤−3 or ≥1
C x<−3 or >1
D [−3,1]
Factor x2+2x−3=(x+3)(x−1)x2+2x−3=(x+3)(x−1). Since parabola opens upward, expression is ≤0 between roots including endpoints. So solution is −3≤x≤1−3≤x≤1, written [−3,1][−3,1].
If a quadratic has roots rr and −r−r, then in ax2+bx+c=0ax2+bx+c=0, bb must be
A a
B 0
C c
D 2a
Roots rr and −r−r have sum 00. For ax2+bx+c=0ax2+bx+c=0, sum of roots is −b/a−b/a. So −b/a=0⇒b=0−b/a=0⇒b=0. This symmetry forces middle term absent.
In x2+px+1=0x2+px+1=0, the product of roots is always
A p
B −p
C 1
D 0
For ax2+bx+c=0ax2+bx+c=0, product of roots is c/ac/a. Here a=1a=1 and c=1c=1, so product is 11. This means if roots exist, they are reciprocals of each other.
For x2+4x+k=0x2+4x+k=0 to have real roots, the greatest possible kk is
A 4
B 0
C 8
D 16
Real roots require discriminant Δ≥0Δ≥0. Here Δ=42−4(1)(k)=16−4k≥0⇒k≤4Δ=42−4(1)(k)=16−4k≥0⇒k≤4. The greatest value satisfying this is k=4k=4.
Solve the absolute inequality ∣x−3∣>1∣x−3∣>1
A 2<x<4
B x<2 or x>4
C x≤2 or ≥4
D x<4 only
∣x−3∣>1∣x−3∣>1 means distance from 3 is greater than 1. So xx lies outside the interval [2,4][2,4]. Hence x<2x<2 or x>4x>4; endpoints are excluded because “>”.
Solve the fractional inequality x−1x+2<0x+2x−1<0
A (−∞,−2)∪(1,∞)
B (−∞,1)
C (−2,1)
D (1,∞)
The expression is negative when numerator and denominator have opposite signs. Critical points are x=1x=1 and x=−2x=−2. Testing intervals shows negativity only between them, so solution is (−2,1)(−2,1).
Solve the logarithmic inequality log2(x)<1log2(x)<1
A (0,1)
B (2,∞)
C (−∞,2)
D (0,2)
Domain requires x>0x>0. Since log2(x)<1log2(x)<1 means x<21=2x<21=2 (base 2 is increasing), the solution is 0<x<20<x<2, written (0,2)(0,2).
Evaluate the floor value ⌊−2.3⌋⌊−2.3⌋
A −2
B −3
C 2
D 3
Floor function gives the greatest integer less than or equal to the number. For −2.3−2.3, the integers below it are −3,−4,…−3,−4,…. The greatest among them is −3−3.
The region ∣z−1∣≤∣z+1∣∣z−1∣≤∣z+1∣ corresponds to
A Re(z) ≥ 0
B Re(z) ≤ 0
C Im(z) ≥ 0
D |z| ≤ 1
∣z−1∣≤∣z+1∣∣z−1∣≤∣z+1∣ means points at least as close to 1 as to −1 on the real axis. The boundary is the perpendicular bisector x=0x=0. Closer-to-1 side is x≥0x≥0.