Chapter 3: Real Numbers, Complex Numbers and Quadratic Expressions (Set-5)
A number x satisfies the distance condition ∣x−1∣+∣x−5∣=10∣x−1∣+∣x−5∣=10; what are the possible values of x
A x = −2 only
B x = −2, 8
C x = 8 only
D No real solution
∣x−1∣+∣x−5∣∣x−1∣+∣x−5∣ is the sum of distances from 1 and 5. It equals 10 only outside the segment [1,5]. Solving gives x = −2 and x = 8.
For real x, the inequality ∣x−1∣+∣x−5∣<6∣x−1∣+∣x−5∣<6 gives which solution set
A (1,5)
B (−∞,6)
C (0,5)
D (0,6)
Between 1 and 5, the sum is constant 4, so it satisfies the inequality. Outside, solve piecewise: for x<1 gives x>0, and for x>5 gives x<6. So (0,6).
Simplify 5+25−25−25+2 to a form without surds in the denominator
A (7−2√10)/3
B (5+√10)/3
C (7+2√10)/3
D (5−√10)/3
Multiply numerator and denominator by 5+25+2. Denominator becomes 5−2=35−2=3. Numerator becomes (5+2)2=7+210(5+2)2=7+210. So result is (7+210)/3(7+210)/3.
When expanding (6+2)2(6+2)2, what simplified result is obtained
A 8+2√12
B 8+4√3
C 4+8√3
D 8−4√3
(a+b)2=a2+b2+2ab(a+b)2=a2+b2+2ab. Here a2=6a2=6, b2=2b2=2, and 2ab=212=432ab=212=43. So total is 8+438+43.
Evaluate 2x⋅4x−1=322x⋅4x−1=32 by reducing everything to base 2
A x = 7/3
B x = 2
C x = 3
D x = 5/3
4x−1=(22)x−1=22x−24x−1=(22)x−1=22x−2. Multiply with 2x2x gives 23x−2=2523x−2=25. So 3x−2=5⇒x=7/33x−2=5⇒x=7/3.
A logarithmic equation log3(x−1)+log3(x−4)=2log3(x−1)+log3(x−4)=2 has which valid solution
A (5−3√5)/2
B 4 only
C 5 only
D (5+3√5)/2
Combine logs: log3((x−1)(x−4))=2⇒(x−1)(x−4)=9log3((x−1)(x−4))=2⇒(x−1)(x−4)=9. Solve x2−5x−5=0x2−5x−5=0. Domain needs x>4, so take (5+35)/2(5+35)/2.
Solve log2(x2−5x+6)>0log2(x2−5x+6)>0 using the fact that base 2 is increasing
A 2<x<3
B x<r1 or x>r2
C x>r2 only
D x<r1 only
log2(A)>0⇒A>1log2(A)>0⇒A>1. So x2−5x+6>1⇒x2−5x+5>0x2−5x+6>1⇒x2−5x+5>0. Roots are r1=(5−5)/2r1=(5−5)/2, r2=(5+5)/2r2=(5+5)/2. Also x2−5x+6>0x2−5x+6>0 holds automatically for those intervals.
Find the principal argument of 1+i1−i1−i1+i after simplifying the complex fraction
A 0
B π
C π/2
D −π/2
1+i1−i⋅1+i1+i=(1+i)22=2i2=i1−i1+i⋅1+i1+i=2(1+i)2=22i=i. The point i lies on the positive imaginary axis, so its principal argument is π/2π/2.
If ∣1+i∣=2∣1+i∣=2, what is the modulus of (1+i)5(1+i)5
A 8
B 2√2
C 16
D 4√2
Modulus of a power follows ∣zn∣=∣z∣n∣zn∣=∣z∣n. Here ∣1+i∣=2∣1+i∣=2, so ∣(1+i)5∣=(2)5=25/2=42∣(1+i)5∣=(2)5=25/2=42.
For z=2−3iz=2−3i, what is the real part of z2z2
A −5
B 5
C −12
D 12
(2−3i)2=4−12i+9i2=4−12i−9=−5−12i(2−3i)2=4−12i+9i2=4−12i−9=−5−12i. The real part is the non-i part, which is −5.
Write zz in a+bi form if ∣z∣=2∣z∣=2 and arg(z)=3π/4arg(z)=3π/4
A √2 + i√2
B −2 + 0i
C −√2 + i√2
D 0 − 2i
z=r(cosθ+isinθ)z=r(cosθ+isinθ). With r=2r=2, θ=3π/4θ=3π/4, cosθ=−2/2cosθ=−2/2, sinθ=2/2sinθ=2/2. So z=2(−2/2)+i2(2/2)=−2+i2z=2(−2/2)+i2(2/2)=−2+i2.
One square root of −4i−4i can be written as which complex number
A √2 + i√2
B √2 − i√2
C −2i
D 2
−4i=4(cos(−π/2)+isin(−π/2))−4i=4(cos(−π/2)+isin(−π/2)). Square roots have modulus 2 and angles −π/4−π/4 and 3π/43π/4. One root is 2(cos(−π/4)+isin(−π/4))=2−i22(cos(−π/4)+isin(−π/4))=2−i2.
Using De Moivre, the argument of (cos15∘+isin15∘)4(cos15∘+isin15∘)4 becomes
A 45°
B 30°
C 75°
D 60°
De Moivre gives (cosθ+isinθ)n=cos(nθ)+isin(nθ)(cosθ+isinθ)n=cos(nθ)+isin(nθ). Here nθ=4×15∘=60∘nθ=4×15∘=60∘. So the new argument is 60°.
The product of all 5th roots of unity equals
A −1
B 1
C 5
D 0
The 5th roots solve x5−1=0x5−1=0. Product of roots of a monic polynomial equals (−1)5⋅(−1)=(−1)6=1(−1)5⋅(−1)=(−1)6=1. So the product is 1.
If ωω is a non-real cube root of unity, then ω+ω2ω+ω2 equals
A −1
B 0
C 1
D 2
Cube roots of unity are 1,ω,ω21,ω,ω2 with 1+ω+ω2=01+ω+ω2=0. Rearranging gives ω+ω2=−1ω+ω2=−1. This identity is used repeatedly in simplification.
Simplify 11−ω1−ω1 for a non-real cube root ωω
A (1+ω)/3
B (1−ω)/3
C (1+ω²)/3
D (1−ω²)/3
Multiply numerator and denominator by (1−ω2)(1−ω2). Denominator becomes (1−ω)(1−ω2)=1−(ω+ω2)+ω3=1−(−1)+1=3(1−ω)(1−ω2)=1−(ω+ω2)+ω3=1−(−1)+1=3. So value is (1−ω2)/3(1−ω2)/3.
The locus ∣z−2∣=2∣z∣∣z−2∣=2∣z∣ forms a circle; its radius is
A 2/3
B 3/4
C 4/3
D 2
Square both sides: (x−2)2+y2=4(x2+y2)(x−2)2+y2=4(x2+y2). Rearranging gives x2+y2+43x−43=0x2+y2+34x−34=0. Completing square: (x+23)2+y2=169(x+32)2+y2=916, so radius 4/34/3.
The inequality ∣z−i∣<∣z+i∣∣z−i∣<∣z+i∣ represents which region in the Argand plane
A Im(z)<0
B Im(z)>0
C Re(z)>0
D Re(z)<0
Points closer to ii than to −i−i lie above the perpendicular bisector of the segment joining them. That bisector is the real axis y=0y=0. Closer to ii means y>0y>0.
The condition Re((1−i)z)=0((1−i)z)=0 describes which straight line
A y = x
B x = 0
C y = 0
D y = −x
Let z=x+iyz=x+iy. Then (1−i)z=(x+y)+i(y−x)(1−i)z=(x+y)+i(y−x). Its real part is x+yx+y. Setting x+y=0x+y=0 gives y=−xy=−x, a line through the origin.
A point z=1−2iz=1−2i is mapped by w=izw=iz; the image ww equals
A −2 + i
B 2 − i
C 2 + i
D −2 − i
Multiply: iz=i(1−2i)=i−2i2=i+2=2+iiz=i(1−2i)=i−2i2=i+2=2+i. Multiplying by ii rotates every point by 90∘90∘ anticlockwise without changing modulus.
The equation x2+(k−3)x+k>0x2+(k−3)x+k>0 holds for all real x only when
A k<1
B 1<k<9
C k>9
D k=1 or 9
For ax2+bx+c>0ax2+bx+c>0 for all real x, need a>0a>0 and discriminant <0<0. Here a=1a=1 and Δ=(k−3)2−4k=k2−10k+9=(k−1)(k−9)Δ=(k−3)2−4k=k2−10k+9=(k−1)(k−9). For Δ<0Δ<0, 1<k<91<k<9.
For x2−px+1=0x2−px+1=0 to have non-real roots of modulus 1, p must satisfy
A p≤−2
B p≥2
C p=0 only
D −2<p<2
Roots have product 1. For both to lie on unit circle and be non-real, they must be e±iθe±iθ, so sum =2cosθ=2cosθ. Hence pp must be between −2 and 2, excluding endpoints.
If α,βα,β are roots of x2−5x+2=0x2−5x+2=0, then α2+β2α2+β2 equals
A 21
B 25
C 17
D 9
α+β=5α+β=5 and αβ=2αβ=2. Use α2+β2=(α+β)2−2αβ=25−4=21α2+β2=(α+β)2−2αβ=25−4=21. This avoids solving the quadratic fully.
For roots of x2−3x+1=0x2−3x+1=0, the value of 1α2+1β2α21+β21 is
A 5
B 9
C 7
D 3
Here α+β=3α+β=3 and αβ=1αβ=1. First α2+β2=32−2(1)=7α2+β2=32−2(1)=7. Then 1α2+1β2=α2+β2(αβ)2=71=7α21+β21=(αβ)2α2+β2=17=7.
Solve the inequality (x−2)(x−3)x−4≤0x−4(x−2)(x−3)≤0 correctly using a sign chart
A [2,3]
B (−∞,2]∪[3,4)
C (2,3)∪(4,∞)
D (−∞,4)
Critical points are 2, 3 (zeros) and 4 (undefined). Check signs on intervals: negative on (−∞,2](−∞,2] and [3,4)[3,4), positive elsewhere. Include 2 and 3; exclude 4.
The inequality ∣x−2∣≥∣x+1∣∣x−2∣≥∣x+1∣ holds for which real x
A x ≥ 1/2
B x < −1
C x > 2
D x ≤ 1/2
Square both sides: (x−2)2≥(x+1)2⇒x2−4x+4≥x2+2x+1⇒−6x≥−3⇒x≤1/2(x−2)2≥(x+1)2⇒x2−4x+4≥x2+2x+1⇒−6x≥−3⇒x≤1/2. Squaring is safe since both sides are nonnegative.
Solve ∣x−3∣≤2x+1∣x−3∣≤2x+1 considering the necessary domain for the right side
A x ≥ −4
B x ≤ 2/3
C x ≥ 2/3
D all real x
Need 2x+1≥0⇒x≥−1/22x+1≥0⇒x≥−1/2. Then ∣x−3∣≤2x+1∣x−3∣≤2x+1 gives −(2x+1)≤x−3≤2x+1−(2x+1)≤x−3≤2x+1. Left inequality gives x≥2/3x≥2/3; right gives x≥−4x≥−4. Combine to get x≥2/3x≥2/3.
Solve ∣2x−5∣<3∣x−1∣∣2x−5∣<3∣x−1∣ by converting to a quadratic inequality
A −2<x<8/5
B x<−2 or x>8/5
C x>−2 only
D x<8/5 only
Square both sides: (2x−5)2<9(x−1)2⇒5×2+2x−16>0(2x−5)2<9(x−1)2⇒5×2+2x−16>0. Roots are −2−2 and 8/58/5. Since parabola opens up, it’s positive outside the roots: x<−2x<−2 or x>8/5x>8/5.
A circle ∣z−1∣=1∣z−1∣=1 intersects the vertical line Re(z)=1/2 in how many points
A 2 points
B 1 point
C 0 points
D Infinite points
Circle center is (1,0) with radius 1. Distance from center to line x=1/2 is 1/2, which is less than radius. So the line cuts the circle at two points.
The locus ∣z∣=∣z−2∣∣z∣=∣z−2∣ is a straight line; its equation is
A Im(z)=1
B Re(z)=0
C Im(z)=0
D Re(z)=1
Points equidistant from 0 and 2 lie on the perpendicular bisector of the segment from 0 to 2 on the real axis. Midpoint is 1, so the bisector is x=1, i.e., Re(z)=1.
If z≠0z=0 satisfies z+1z=2z+z1=2, then z equals
A −1
B i
C 1
D 2
Multiply by z: z2+1=2z⇒z2−2z+1=0⇒(z−1)2=0z2+1=2z⇒z2−2z+1=0⇒(z−1)2=0. So the only solution is z=1z=1. Nonzero condition is satisfied.
Evaluate (1+i)81616(1+i)8 using power patterns of (1+i)
A −1
B 1
C i
D −i
(1+i)2=2i(1+i)2=2i, so (1+i)4=(2i)2=−4(1+i)4=(2i)2=−4, and (1+i)8=(−4)2=16(1+i)8=(−4)2=16. Dividing by 16 gives 1.
Compute (1+i)6(1+i)6 in standard a+bi form
A 8i
B −8
C 8
D −8i
Use (1+i)2=2i(1+i)2=2i. Then (1+i)6=((1+i)2)3=(2i)3=8i3=8(−i)=−8i(1+i)6=((1+i)2)3=(2i)3=8i3=8(−i)=−8i. This avoids long expansion.
The quadratic x2−2kx+(k2+1)=0x2−2kx+(k2+1)=0 always has which nature of roots
A Complex conjugates
B Equal real roots
C Distinct real roots
D One real root
Discriminant Δ=(−2k)2−4(1)(k2+1)=4k2−4k2−4=−4Δ=(−2k)2−4(1)(k2+1)=4k2−4k2−4=−4, which is always negative. So roots are always non-real and occur as a conjugate pair.
If one root of x2+ax+1=0x2+ax+1=0 is 2+32+3, the other root must be
A −2−√3
B 2+√3
C 2−√3
D −2+√3
Product of roots is c/a=1c/a=1. So the other root is the reciprocal of 2+32+3. Since 12+3=2−32+31=2−3, the second root is 2−32−3.
A new quadratic has roots α+1α+1 and β+1β+1 where α,βα,β solve x2+px+q=0x2+px+q=0; the new equation is
A x²+(p+2)x+(1+p+q)
B x²+(p−2)x+(1−p+q)
C x²+(p−2)x+(1+p+q)
D x²+(p+2)x+(1−p+q)
If αα is a root, then α+1α+1 is a root of the shifted polynomial f(x−1)=0f(x−1)=0. Substitute x−1x−1: (x−1)2+p(x−1)+q=0⇒x2+(p−2)x+(1−p+q)=0(x−1)2+p(x−1)+q=0⇒x2+(p−2)x+(1−p+q)=0.
For real parameter k, the inequality ∣x−k∣≤∣x∣∣x−k∣≤∣x∣ holds for all real x only when
A k = 1
B k = −1
C k = 2
D k = 0
Squaring gives (x−k)2≤x2⇒−2kx+k2≤0(x−k)2≤x2⇒−2kx+k2≤0. If k≠0k=0, choosing x with opposite sign makes −2kx−2kx large positive, breaking the inequality. Only k=0k=0 works for every x.
Points satisfying ∣x−2∣≤∣x−5∣∣x−2∣≤∣x−5∣ lie in which set on the real line
A x ≥ 7/2
B x < 2
C x ≤ 7/2
D x > 5
Being closer to 2 than to 5 means x is on the 2-side of the midpoint. Midpoint is (2+5)/2=7/2(2+5)/2=7/2. Including “≤” means the midpoint is included, so x≤7/2x≤7/2.
Evaluate log2(8)log2(8) exactly without decimals
A 2
B 3/2
C 1/2
D 3
8=(23)1/2=23/28=(23)1/2=23/2. Therefore log2(8)=log2(23/2)=3/2log2(8)=log2(23/2)=3/2. Converting to a power of 2 makes it immediate.
Solve 3x=2x+13x=2x+1 and express x in a clean logarithmic form
A log_{2}3/2
B log_{3}2
C log_{2}3
D log_{3/2}2
3x=2x+1=2⋅2x⇒(3/2)x=23x=2x+1=2⋅2x⇒(3/2)x=2. Taking logs gives x=ln2ln(3/2)=log3/22x=ln(3/2)ln2=log3/22. This is the exact solution.
If ∣z∣=2∣z∣=2 and zz also satisfies ∣z−1∣=∣z+1∣∣z−1∣=∣z+1∣, then possible values of Im(z) are
A 2 or −2
B 1 or −1
C 0 only
D 2 only
∣z−1∣=∣z+1∣∣z−1∣=∣z+1∣ means z lies on the perpendicular bisector of −1 and 1, i.e., Re(z)=0. With ∣z∣=2∣z∣=2, the points are z=±2iz=±2i. So Im(z)=±2.
One non-real cube root of 88 (i.e., a solution of z3=8z3=8) is
A −1 − i√3
B −1 + i√3
C 2
D 1 + i
Write 8=23=8(cos0+isin0)8=23=8(cos0+isin0). Cube roots are 2(cos2kπ3+isin2kπ3)2(cos32kπ+isin32kπ). For k=1k=1, 2(cos120∘+isin120∘)=−1+i32(cos120∘+isin120∘)=−1+i3.
The region described by ∣z−2∣>3∣z−2∣>3 is best described as
A Inside the circle
B Circle boundary
C Vertical strip
D Outside the circle
∣z−a∣>r∣z−a∣>r means points whose distance from center a is greater than r. So it is the exterior of the circle centered at 2 (on real axis) with radius 3, excluding boundary.
If z is purely imaginary and ∣z∣=5∣z∣=5, then z can be
A 5 or −5
B 3+4i only
C 5i or −5i
D 4+3i only
Purely imaginary numbers have real part 0, so z=bi. Modulus is ∣z∣=∣b∣∣z∣=∣b∣. If ∣z∣=5∣z∣=5, then ∣b∣=5⇒b=±5∣b∣=5⇒b=±5, so z=±5i.
If z=3+4iz=3+4i, then the value of z∣z∣∣z∣z lies on
A Real axis
B Unit circle
C Imaginary axis
D Circle radius 5
∣z∣=5∣z∣=5. Dividing by modulus gives ∣z∣z∣∣=∣z∣∣z∣=1∣z∣z=∣z∣∣z∣=1. Any complex number with modulus 1 lies on the unit circle.
The number of distinct solutions of z4=16z4=16 in complex numbers is
A 4
B 2
C 3
D 1
16=16(cos0+isin0)16=16(cos0+isin0). Fourth roots have modulus 2 and arguments 0+2kπ440+2kπ for k=0,1,2,3. That gives four distinct complex solutions.
If x2−8x+12=0x2−8x+12=0 has roots α,βα,β, then α−βα−β (magnitude) equals
A 2
B √16
C 6
D 4
Discriminant Δ=b2−4ac=64−48=16Δ=b2−4ac=64−48=16. For a monic quadratic, difference of roots magnitude is ΔΔ. So ∣α−β∣=16=4∣α−β∣=16=4.
A quadratic has roots whose sum is 0 and product is −9; the monic equation is
A x² + 9
B x² + 9x
C x² − 9
D x² − 9x.
Using x2−(α+β)x+αβ=0x2−(α+β)x+αβ=0 with sum 0 and product −9 gives x2−9=0x2−9=0. This matches roots 33 and −3−3, sum 0 and product −9.
If (x−1)(x−4)>0(x−1)(x−4)>0, the correct solution set is
A 1<x<4
B x<1 or x>4
C x≤1 or ≥4
D x>1 only
A product is positive when both factors are positive or both are negative. So either x>4x>4 (both positive) or x<1x<1 (both negative). Endpoints are excluded due to “>”.
When a complex number zz satisfies ∣z−1∣=∣z∣∣z−1∣=∣z∣, the value of Re(z) must be
A 1
B 0
C 2
D 1/2
∣z−1∣=∣z∣∣z−1∣=∣z∣ means point z is equidistant from 1 and 0 on the real axis. The locus is the perpendicular bisector of segment [0,1], which is the vertical line x=1/2, so Re(z)=1/2.