{"id":17463,"date":"2021-01-28T10:43:12","date_gmt":"2021-01-28T05:13:12","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=17463"},"modified":"2021-01-28T10:43:12","modified_gmt":"2021-01-28T05:13:12","slug":"mcq-questions-for-class-11-maths-chapter-5","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/mcq-questions-for-class-11-maths-chapter-5\/","title":{"rendered":"MCQ Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations with Answers"},"content":{"rendered":"

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Complex Numbers and Quadratic Equations Class 11 MCQs Questions with Answers<\/h2>\n

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Explore numerous MCQ Questions of Complex Numbers and Quadratic Equations Class 11 with answers provided with detailed solutions by looking below.<\/p>\n

Question 1.
\nLet z1<\/sub> and z2<\/sub> be two roots of the equation z\u00b2 + az + b = 0, z being complex. Further assume that the origin, z1<\/sub> and z2<\/sub> form an equilateral triangle. Then
\n(a) a\u00b2 = b
\n(b) a\u00b2 = 2b
\n(c) a\u00b2 = 3b
\n(d) a\u00b2 = 4b<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) a\u00b2 = 3b
\nGiven, z1<\/sub> and z1<\/sub> be two roots of the equation z\u00b2+ az + b = 0
\nNow, z1<\/sub> + z2<\/sub> = -a and z1<\/sub> \u00d7 z2<\/sub> = b
\nSince z1<\/sub> and z2<\/sub> and z3<\/sub> from an equilateral triangle.
\n\u21d2 z1<\/sub>\u00b2 + z2<\/sub>\u00b2 + z3<\/sub>\u00b2 = z1<\/sub> \u00d7 z2<\/sub> + z2<\/sub> \u00d7 z3<\/sub> + z1<\/sub> \u00d7 z3<\/sub>
\n\u21d2 z1<\/sub>\u00b2 + z2<\/sub>\u00b2 = z1<\/sub> \u00d7 z2<\/sub> {since z3<\/sub> = 0}
\n\u21d2 (z1<\/sub> + z2<\/sub>)\u00b2 – 2z1<\/sub> \u00d7 z2<\/sub> = z1<\/sub> \u00d7 z2<\/sub>
\n\u21d2 (z1<\/sub> + z2<\/sub>)\u00b2 = 2z1<\/sub> \u00d7 z2<\/sub> + z1<\/sub> \u00d7 z2<\/sub>
\n\u21d2 (z1<\/sub> + z2<\/sub>)\u00b2 = 3z1<\/sub> \u00d7 z2<\/sub>
\n\u21d2 (-a)\u00b2 = 3b
\n\u21d2 a\u00b2 = 3b<\/p>\n<\/details>\n

\n

Question 2.
\nThe value of ii<\/sup> is
\n(a) 0
\n(b) e-\u03c0<\/sup>
\n(c) 2e-\u03c0\/2<\/sup>
\n(d) e-\u03c0\/2<\/sup><\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) e-\u03c0\/2<\/sup>
\nLet A = ii<\/sup>
\n\u21d2 log A = i log i
\n\u21d2 log A = i log(0 + i)
\n\u21d2 log A = i [log 1 + i tan-1<\/sup> \u221e]
\n\u21d2 log A = i [0 + i \u03c0\/2]
\n\u21d2 log A = -\u03c0\/2
\n\u21d2 A = e-\u03c0\/2<\/sup><\/p>\n<\/details>\n

\n

Question 3.
\nThe value of \u221a(-25) + 3\u221a(-4) + 2\u221a(-9) is
\n(a) 13 i
\n(b) -13 i
\n(c) 17 i
\n(d) -17 i<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 17 i
\nGiven, \u221a(-25) + 3\u221a(-4) + 2\u221a(-9)
\n= \u221a{(-1) \u00d7 (25)} + 3\u221a{(-1) \u00d7 4} + 2\u221a{(-1) \u00d7 9}
\n= \u221a(-1) \u00d7 \u221a(25) + 3{\u221a(-1) \u00d7 \u221a4} + 2{\u221a(-1) \u00d7 \u221a9}
\n= 5i + 3 \u00d7 2i + 2 \u00d7 3i {since \u221a(-1) = i}
\n= 5i + 6i + 6i
\n= 17 i
\nSo, \u221a(-25) + 3\u221a(-4) + 2\u221a(-9) = 17 i<\/p>\n<\/details>\n

\n

Question 4.
\nIf the cube roots of unity are 1, \u03c9 and \u03c9\u00b2, then the value of (1 + \u03c9 \/ \u03c9\u00b2)\u00b3 is
\n(a) 1
\n(b) -1
\n(c) \u03c9
\n(d) \u03c9\u00b2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) -1
\nGiven, the cube roots of unity are 1, \u03c9 and \u03c9\u00b2
\nSo, 1 + \u03c9 + \u03c9\u00b2 = 0
\nand \u03c9\u00b3 = 1
\nNow, {(1 + \u03c9)\/ \u03c9\u00b2}\u00b3 = {-\u03c9\u00b2\/ \u03c9\u00b2}\u00b3 = {-1}\u00b3 = -1<\/p>\n<\/details>\n

\n

Question 5.
\nIf {(1 + i)\/(1 – i)}\u207f = 1 then the least value of n is
\n(a) 1
\n(b) 2
\n(c) 3
\n(d) 4<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 4
\nGiven, {(1 + i)\/(1 – i)}\u207f = 1
\n\u21d2 [{(1 + i) \u00d7 (1 + i)}\/{(1 – i) \u00d7 (1 + i)}]\u207f = 1
\n\u21d2 [{(1 + i)\u00b2}\/{(1 – i\u00b2)}]\u207f = 1
\n\u21d2 [(1 + i\u00b2 + 2i)\/{1 – (-1)}]\u207f = 1
\n\u21d2 [(1 – 1 + 2i)\/{1 + 1}]\u207f = 1
\n\u21d2 [2i\/2]\u207f = 1
\n\u21d2 i\u207f = 1
\nNow, i\u207f is 1 when n = 4
\nSo, the least value of n is 4<\/p>\n<\/details>\n

\n

Question 6.
\nThe value of [i19<\/sup> + (1\/i)25<\/sup>]\u00b2 is
\n(a) -1
\n(b) -2
\n(c) -3
\n(d) -4<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) -4
\nGiven, [i19<\/sup> + (1\/i)25<\/sup>]\u00b2
\n= [i19<\/sup> + 1\/i25<\/sup>]\u00b2
\n= [i16<\/sup> \u00d7 i\u00b3 + 1\/(i24<\/sup> \u00d7 i)]\u00b2
\n= [1 \u00d7 i\u00b3 + 1\/(1 \u00d7 i)]\u00b2 {since i4<\/sup> = 1}
\n= [i\u00b3 + 1\/i]\u00b2
\n= [i\u00b2 \u00d7 i + 1\/i]\u00b2
\n= [(-1) \u00d7 i + 1\/i]\u00b2 {since i\u00b2 = -1}
\n= [-i + 1\/i]\u00b2
\n= [-i + i4<\/sup> \/i]\u00b2
\n= [-i + i\u00b3]\u00b2
\n= [-i + i\u00b2 \u00d7 i]\u00b2
\n= [-i + (-1) \u00d7 i]\u00b2
\n= [-i – i]\u00b2
\n= [-2i]\u00b2
\n= 4i\u00b2
\n= 4 \u00d7 (-1)
\n= -4
\nSo, [i19<\/sup> + (1\/i)25<\/sup>]\u00b2 = -4<\/p>\n<\/details>\n

\n

Question 7.
\nIf z and w be two complex numbers such that |z| \u2264 1, |w| \u2264 1 and |z + iw| = |z – iw| = 2, then z equals {w is congugate of w}
\n(a) 1 or i
\n(b) i or – i
\n(c) 1 or – 1
\n(d) i or – 1<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 1 or – 1
\nGiven |z + iw| = |z – iw| = 2 {w is congugate of w}
\n\u21d2 |z – (-iw)| = |z – (iw)| = 2
\n\u21d2 |z – (-iw)| = |z – (-iw)|
\nSo, z lies on the perpendicular bisector of the line joining -iw and -iw.
\nSince, -iw is the mirror in the x-axis, the locus of z is the x-axis.
\nLet z = x + iy and y = 0
\n\u21d2 |z| < 1 and x\u00b2 + 0\u00b2 < 0
\n\u21d2 -1 \u2264 x \u2264 1
\nSo, z may take value 1 or -1<\/p>\n<\/details>\n

\n

Question 8.
\nThe value of {-\u221a(-1)}4n+3<\/sup>, n \u2208 N is
\n(a) i
\n(b) -i
\n(c) 1
\n(d) -1<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) i
\nGiven, {-\u221a(-1)}4n+3<\/sup>
\n= {-i}4n+3<\/sup> {since \u221a(-1) = i}
\n= {-i}4n<\/sup> \u00d7 {-i}\u00b3
\n= {(-i)4<\/sup>}\u207f \u00d7 (-i\u00b3) {since i4<\/sup> = 1}
\n= 1\u207f \u00d7(-i \u00d7 i\u00b2)
\n= -i \u00d7 (-1) {since i\u00b2 = -1}
\n= i<\/p>\n<\/details>\n

\n

Question 9.
\nFind real \u03b8 such that (3 + 2i \u00d7 sin \u03b8)\/(1 – 2i \u00d7 sin \u03b8) is real
\n(a) \u03c0
\n(b) n\u03c0
\n(c) n\u03c0\/2
\n(d) 2n\u03c0<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) n\u03c0
\nGiven,
\n(3 + 2i \u00d7 sin \u03b8)\/(1 – 2i \u00d7 sin \u03b8) = {(3 + 2i \u00d7 sin \u03b8)\u00d7(1 – 2i \u00d7 sin \u03b8)}\/(1 – 4i\u00b2 \u00d7 sin\u00b2 \u03b8)
\n(3 + 2i \u00d7 sin \u03b8)\/(1 – 2i \u00d7 sin \u03b8) = {(3 – 4sin\u00b2 \u03b8) + 8i \u00d7 sin \u03b8}\/(1 + 4sin\u00b2 \u03b8) …………. 1
\nNow, equation 1 is real if sin \u03b8 = 0
\n\u21d2 sin \u03b8 = sin n\u03c0
\n\u21d2 \u03b8 = n\u03c0<\/p>\n<\/details>\n

\n

Question 10.
\nIf i = \u221a(-1) then 4 + 5(-1\/2 + i\u221a3\/2)334<\/sup> + 3(-1\/2 + i\u221a3\/2)365<\/sup> is equals to
\n(a) 1 – i\u221a3
\n(b) -1 + i\u221a3
\n(c) i\u221a3
\n(d) -i\u221a3<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) i\u221a3
\nGiven, 4 + 5(-1\/2 + i\u221a3\/2)334<\/sup> + 3(-1\/2 + i\u221a3\/2)365<\/sup>
\n= 4 + 5w334<\/sup> + 3w365<\/sup> {since w = -1\/2 + i\u221a3\/2}
\n= 4 + 5w + 3w\u00b2 {since w\u00b3 = 1}
\n= 4 + 5(-1\/2 + i\u221a3\/2) + 3(-1\/2 – i\u221a3\/2) {since w\u00b2 = (-1\/2 – i\u221a3\/2)}
\n= i\u221a3<\/p>\n<\/details>\n

\n

Question 11.
\nThe real part of the complex number \u221a9 + \u221a(-16) is
\n(a) 3
\n(b) -3
\n(c) 4
\n(d) -4<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 3
\nGiven, \u221a9 + \u221a(-16) = \u221a9 + \u221a(16) \u00d7 \u221a(-1)
\n= 3 + 4i {since i = \u221a(-1)}
\nSo, the real part of the complex number is 3<\/p>\n<\/details>\n

\n

Question 12.
\nThe modulus of 5 + 4i is
\n(a) 41
\n(b) -41
\n(c) \u221a41
\n(d) -\u221a41<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) \u221a41
\nLet Z = 5 + 4i
\nNow modulus of Z is calculated as
\n|Z| = \u221a(5\u00b2 + 4\u00b2)
\n\u21d2 |Z| = \u221a(25 + 16)
\n\u21d2 |Z| = \u221a41
\nSo, the modulus of 5 + 4i is \u221a41<\/p>\n<\/details>\n

\n

Question 13.
\nThe modulus of 1 + i\u221a3 is
\n(a) 1
\n(b) 2
\n(c) 3
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 2
\nLet Z = 1 + i\u221a3
\nNow modulus of Z is calculated as
\n|Z| = \u221a{1\u00b2 + (\u221a3)\u00b2}
\n\u21d2 |Z| = \u221a(1 + 3)
\n\u21d2 |Z| = \u221a4
\n\u21d2 |Z| = 2
\nSo, the modulus of 1 + i\u221a3 is 2<\/p>\n<\/details>\n

\n

Question 14.
\nThe value of {-\u221a(-1)}4n+3<\/sup>, n \u2208 N is
\n(a) i
\n(b) -i
\n(c) 1
\n(d) -1<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) i
\nGiven, {-\u221a(-1)}4n+3<\/sup>
\n= {-i}4n+3<\/sup> {since \u221a(-1) = i}
\n= {-i}4n<\/sup> \u00d7 {-i}\u00b3
\n= {(-i)4<\/sup>}\u207f \u00d7(-i\u00b3) {since i4<\/sup> = 1}
\n= 1\u207f \u00d7 (-i \u00d7 i\u00b2)
\n= -i \u00d7 (-1) {since i\u00b2 = -1}
\n= i<\/p>\n<\/details>\n

\n

Question 15.
\nIf \u03c9 is cube root of unity (\u03c9 \u2260 1) , then the least value of n where n is a positive integer such that (1 + \u03c9\u00b2)\u207f = (1 + \u03c94<\/sup>)\u207f is
\n(a) 2
\n(b) 3
\n(c) 5
\n(d) 6<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 3
\nGiven \u03c9 is an imaginary cube root of unity.
\nSo 1 + \u03c9 + \u03c9\u00b2 = 0 and \u03c9\u00b3 = 1
\nNow, (1 + \u03c9\u00b2)\u207f = (1 + \u03c94<\/sup>)\u207f
\n\u21d2 (-1)\u207f \u00d7(\u03c9)\u207f = (1 + \u03c9 \u00d7 \u03c9\u00b3)\u207f
\n\u21d2 (-1)\u207f \u00d7 (\u03c9)\u207f = (1 + \u03c9)\u207f
\n\u21d2 (-1)\u207f \u00d7 (\u03c9)\u207f = (-\u03c9\u00b2)\u207f
\n\u21d2 (-1)\u207f \u00d7 (\u03c9)\u207f = (-1)\u207f \u00d7 \u03c9\u00b2\u207f
\n\u21d2 \u03c9\u207f = \u03c9\u00b2\u207f
\nSince \u03c9\u00b3 = 1, So the least value of n is 3<\/p>\n<\/details>\n

\n

Question 16.
\nThe value of i9<\/sup> + i10<\/sup> + i11<\/sup> + i12<\/sup> is
\n(a) i
\n(b) 2i
\n(c) 0
\n(d) 1<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 0
\nGiven, i9<\/sup> + i10<\/sup> + i11<\/sup> + i12<\/sup>
\n= i9 (1 + i + i2 + i3 )
\n= i9 (1 + i – 1 – i ) {since i2 = (-1) and i4 = 1}
\n= i9 \u00d7 0
\n= 0<\/p>\n<\/details>\n

\n

Question 17.
\nIf a = cos \u03b1 + i sin \u03b1 and b = cos \u03b2 + i sin \u03b2 , then the value of 1\/2(ab + 1\/ ab) is
\n(a) sin (\u03b1 + \u03b2)
\n(b) cos (\u03b1 + \u03b2)
\n(c) sin (\u03b1 – \u03b2)
\n(d) cos (\u03b1 – \u03b2)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) cos (\u03b1 + \u03b2)
\nGiven a = cos \u03b1 + i sin \u03b1 and b = cos \u03b2 + i sin \u03b2
\nNow, 1\/a = 1\/(cos \u03b1 + i sin \u03b1)
\n\u21d2 1\/a = {1 \u00d7 (cos \u03b1 – i sin \u03b1)\/{(cos \u03b1 + i sin \u03b1) \u00d7 (cos \u03b1 + i sin \u03b1)}
\n\u21d2 1\/a = (cos \u03b1 – i sin \u03b1)\/(cos\u00b2 \u03b1 + i sin\u00b2 \u03b1)
\n\u21d2 1\/a = (cos \u03b1 – i sin \u03b1)
\nAgain, 1\/b = 1\/(cos \u03b2 + i sin \u03b2)
\n\u21d2 1\/b = {1 \u00d7 (cos \u03b2 – i sin \u03b2)\/{(cos \u03b2 + i sin \u03b2) \u00d7 (cos \u03b2 + i sin \u03b2)}
\n\u21d2 1\/b = (cos \u03b2 – i sin \u03b2)\/(cos\u00b2 \u03b2 + i sin\u00b2 \u03b2)
\n\u21d2 1\/b = (cos \u03b2 – i sin \u03b2)
\nNow, ab = (cos \u03b1 + i sin \u03b1) \u00d7 (cos \u03b2 + i sin \u03b2)
\n\u21d2 ab = cos \u03b1 \u00d7 cos \u03b2 + i cos \u03b1 \u00d7 sin \u03b2 + i sin \u03b1 \u00d7 cos \u03b2 – sin \u03b1 \u00d7 sin \u03b2
\nAgain, 1\/ab = (cos \u03b1 – i sin \u03b1) \u00d7 (cos \u03b2 – i sin \u03b2)
\n\u21d2 1\/ab = cos \u03b1 \u00d7 cos \u03b2 – i cos \u03b1 \u00d7 sin \u03b2 – i sin \u03b1 \u00d7 cos \u03b2 – sin \u03b1 \u00d7 sin \u03b2
\nNow, ab + 1\/ab = cos \u03b1 \u00d7 cos \u03b2 + i cos \u03b1 \u00d7 sin \u03b2 + i sin \u03b1 \u00d7 cos \u03b2 – sin \u03b1 \u00d7 sin \u03b2 + cos \u03b1 \u00d7 cos \u03b2 – i cos \u03b1 \u00d7 sin \u03b2 – i sin \u03b1 \u00d7 cos \u03b2 – sin \u03b1 \u00d7 sin \u03b2
\n\u21d2 ab + 1\/ab = 2(cos \u03b1 \u00d7 cos \u03b2 – sin \u03b1 \u00d7 sin \u03b2)
\n\u21d2 1\/2(ab + 1\/ ab) = 2(cos \u03b1 \u00d7 cos \u03b2 – sin \u03b1 \u00d7 sin \u03b2)\/2
\n\u21d2 1\/2(ab + 1\/ ab) = cos \u03b1 \u00d7 cos \u03b2 – sin \u03b1 \u00d7 sin \u03b2
\n\u21d2 1\/2(ab + 1\/ ab) = cos(\u03b1 + \u03b2)<\/p>\n<\/details>\n

\n

Question 18.
\nThe polar form of -1 + i is
\n(a) \u221a2(cos \u03c0\/2 + i \u00d7 sin \u03c0\/2)
\n(b) \u221a2(cos \u03c0\/4 + i \u00d7 sin \u03c0\/4)
\n(c) \u221a2(cos 3\u03c0\/2 + i \u00d7 sin 3\u03c0\/2)
\n(d) \u221a2(cos 3\u03c0\/4 + i \u00d7 sin 3\u03c0\/4)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) \u221a2(cos 3\u03c0\/4 + i \u00d7 sin 3\u03c0\/4)
\nThe polar form of a com plex number = r(cos \u03b8 + i \u00d7 sin \u03b8)
\nGiven, complex number = -1 + i
\nLet x + iy = -1 + i
\nNow, x = -1, y = 1
\nNow, r = \u221a{(-1)\u00b2 + 1\u00b2} = \u221a(1 + 1) = \u221a2
\nand tan \u03b8 = y\/x
\n\u21d2 tan \u03b8 = 1\/(-1)
\n\u21d2 tan \u03b8 = -1
\n\u21d2 \u03b8 = 3\u03c0\/4
\nNow, polar form is \u221a2(cos 3\u03c0\/4 + i \u00d7 sin 3\u03c0\/4)<\/p>\n<\/details>\n

\n

Question 19.
\nFor all complex numbers z1<\/sub>, z2<\/sub> satisfying |z1<\/sub>| = 12 and |z2<\/sub> – 3 – 4i| = 5, the minimum value of |z1<\/sub> – z2<\/sub>| is
\n(a) 0
\n(b) 2
\n(c) 7
\n(d) 17<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 2
\nGiven For all complex numbers z1<\/sub>, z2<\/sub> satisfying |z1<\/sub>| = 12 and |z2<\/sub> – 3 – 4i| = 5
\nNow, mod(z1<\/sub>) = 12 represents a circle centred at 0 and radius 12
\nmod(z2<\/sub> – 3 – 4i) = 5 represents a circle centred at (3, 4) and radius 5
\nThis circle passes through the origin. Distance of diametrically opposite end is 10
\nSo, the minimum value (z1<\/sub> – z2<\/sub>) = 2<\/p>\n<\/details>\n

\n

Question 20.
\nThe value of (1 – i)\u00b2 is
\n(a) i
\n(b) -i
\n(c) 2i
\n(d) -2i<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) -2i
\nGiven, (1 – i)\u00b2 = 1 + i\u00b2 – 2i
\n= 1 + (-1) – 2i
\n= 1 – 1 – 2i
\n= -2i<\/p>\n<\/details>\n

\n

We believe the knowledge shared regarding NCERT MCQ Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations with Answers Pdf free download has been useful to the possible extent. If you have any other queries regarding CBSE Class 11 Maths Complex Numbers and Quadratic Equations MCQs Multiple Choice Questions with Answers, feel free to reach us via the comment section and we will guide you with the possible solution.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the …<\/p>\n

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