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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 201. |
If `z = -5 + 3i`, find the value of `(z^(4)+9z^(3)+26z^(2)-14z + 8)`. |
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Answer» We have `z = -5 + 3i rArr (z + 5) = 3i rArr (z+5)^(2) = 9i^(2) rArr z^(2) + 10z + 34 = 0." "...(i)` Now, `z^(4) + 9z^(3) + 26z^(2) - 14z + 8` `= z^(2)(z^(2) + 10z + 34) -z(z^(2) + 10z + 34) + 2(z^(2) + 10z + 34) - 60" "["using (i)"]` Hence, the value of `z^(4) + 9z^(3) + 26z^(2) - 14z + 8` is -60. |
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| 202. |
Solve the equation `25 x^2-30 x+11=0`by using the general expression for roots quadratic equation `a x^2+b x+c=0,`we get: `a=25 ,b=-30a d nc-11`. |
| Answer» Correct Answer - `{(3)/(5)+(sqrt(2))/(5)i,(3)/(5)-(sqrt(2))/(5)i}` | |
| 203. |
0, The equation formed by decreasing each root of the equation ` a x^2 + bx + c = 0` by 1 is ` 2 x^2 + 8x + 2 = 0` then |
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Answer» Let `alpha,beta ` are the roots of equation `ax^2+bc+c = 0` Then, `alpha-1` and `beta-1` are the roots of equation `2x^2+8x+2 = 0` Then, sum of roots, ` (alpha-1)+(beta-1) = -8/2` `=>alpha+beta = -4+2` `=>alpha+beta = -2->(1)` Product of roots, ` (alpha-1)(beta-1) = 2/2` `=>alphabeta-alpha-beta+1 = 1` `=>alphabeta -(alpha+beta) = 0` `=>alphabeta = alpha+beta` `=>alphabeta = -2->(2)` As, `alpha` and `beta` are the roots of `ax^2+bx+c = 0` `:. -b/a = alpha+beta and c/a = alphabeta` `=> -b/a = -2 and c/a = -2` `=>b = 2a and ca= -2a` `=>b=-c` So, option `2` is the correct option. |
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| 204. |
The number of integral solutions of `x^2+9 |
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Answer» `x^2+9< (x+3)^2` `x^2+9`6x>0` `x>0` `x^2+9+6x<8x+25` `x^2-2x-16<0` `x=(2+-sqrt(4+64))/2` `x=(2+-sqrt(17))/2` `=1+-sqrt17` `1-sqrt17 < x <1+sqrt17 ` `0 < x <1+sqrt17` `0 < x < 5......` `x= 1,2,3,4,5,` `` `` `` `` `` `` `` `` `` `` `` `` `` `` `` `` `` `` `` |
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| 205. |
Solve the following quadratic: `17 x^2-8x+1=0` |
| Answer» Correct Answer - `{(4)/(17)+(1)/(17)i,(4)/(17)-(1)/(17)i}` | |
| 206. |
If `z = 3 + 2i`, prove that `z^(2) - 6z + 13 = 0` and hence deduce that `3z^(3) - 13z^(2) + 9z + 65 = 0`. |
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Answer» `z = 3 + 2i rArr z -3 = 2i rArr (z - 3)^(2) = 4i^(2)` `rArr" "z^(2) - 6z + 13 = 0" "...(i)`. Thus, `z^(2) -6z + 13 = 0`. Now, `3z^(3) - 13z^(2) + 9z + 65 = 3z(z^(2) - 6z + 13)+5(z^(2) - 6z + 13)=(3z xx 0) + (5 xx 0) = 0" "["using (i)"]` Hence, `3z^(2) - 13z^(2) + 9z + 65 = 0`. |
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| 207. |
Solve the equation:`x^2+3x+5=0` |
| Answer» Correct Answer - `{(-3)/(2)+(sqrt(11))/(2)i,(-3)/(2)-(sqrt(11))/(2)i}` | |
| 208. |
Solve: `8x^(2) + 2x + 1 = 0` |
| Answer» Correct Answer - `{(-1)/(8)+(sqrt(7))/(8)i,(-1)/(8)-(sqrt(7))/(8)i}` | |
| 209. |
Solve: `2x^(2)-sqrt(3) x + 1 = 0` |
| Answer» Correct Answer - `{(sqrt(3))/(4)+(sqrt(5))/(4)i,(sqrt(3))/(4)-(sqrt(5))/(4)i}` | |
| 210. |
`sum_(n = 1)^(13)(i^(n)+i^(n+1))=(-1+i), n in N`. |
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Answer» `underset(n = 1)overset(13)sum (i^(n)+i^(n+1))=underset(n = 1)overset(13)sum i^(n)(1+i)=(1+i).underset(n = 1)overset(13)sum i^(n)` `=(1+i).(i+i^(2)+i^(3)+....+i^(13)}` `=(1+i)(a(1-r^(n)))/((1-r))`, where a=i, r=i, and n = 13 `(1+i).(i(1-i^(13)))/((1-i))= i(1+i).((1-i))/((1-i))=(-1 + i)." "[because i^(13) = (i^(13) = (i^(4))^(13) xx i = 1 xx i = i]` |
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| 211. |
Solve: `x^(2) + x + 1 = 0` |
| Answer» Correct Answer - `{(-1)/(2)+(sqrt(3))/(2)i,(-1)/(2)-(sqrt(3))/(2)i}` | |
| 212. |
Evaluate `sum_(n=1)^(13)(i^n+i^(n+1)),`where `n in Ndot` |
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Answer» Given that, `sum_(n=1)^(13) (i^(n) + i^(n+1)), "where n" in N. ` ` =(i+i^(2)+ i ^(3) + i^(4)+ i^(5)+ i^(6)+ i^(7)+ i^(8)+ i^(9)+ i^(10)+ i^(11)+ i^(12)+ i^(13))` + ` (i+i^(2)+ i ^(3) + i^(4)+ i^(5)+ i^(6)+ i^(7)+ i^(8)+ i^(9)+ i^(10)+ i^(11)+ i^(12)+ i^(13)+i^(14))` ` =(i+2i^(2)+ 2i ^(3) + 2i^(4)+ 2i^(5)+ 2i^(6)+ 2i^(7)+ 2i^(8)+ 2i^(9)+ 2i^(10)+ 2i^(11)+ 2i^(12)+ 2i^(13)+i^(14))` =`i-2-2i+2+2i+2(i^(4))+i^(2)+2(i)^(4)i^(3)+2(i^2)^(4)i+2(i^(2))^(5)+2(i^(2))^(5).i+2(i^(2))^(6)+2(i^(2))^(6).i+(i^(2))^(7)` `=i-2-2i+2+2i-2-2i+2+2i-2-2i+2+2i-1-1-i` "Alternate Method" `sum_(n=1)^(13) (i^(n) + i^(n+1)), " n" in N =` `=(1 -i)[i+i^(2)+i^(3)+i^(4)+i^(5)+i^(6)+i^(7)+i^(8)+i^(9)+i^(10)+i^(11)+i^(13)] ` `=0(1-i)[i^(13)]" "[:.i^(n)+i^(n+1)+i^(n+2)+i^(n+3)= 0. where n in N i.e., `sum_(n=1)^(13) i^(n) =0` =`(+i)i" "[:.(i^(4)^(3).i = i] ` =`(i^(2)+i)=i-1` |
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| 213. |
Solve the equation :`27 x^2-10 x+1=0` |
| Answer» Correct Answer - `{(-5)/(27)+(sqrt(2))/(27)i,(-5)/(27)-(sqrt(2))/(27)i}` | |
| 214. |
Solve: `2x^(2) - 4x + 3 = 0` |
| Answer» Correct Answer - `{1+(1)/(sqrt(2))i,1-(1)/(sqrt(2))i}` | |
| 215. |
Solve: `2x^(2) + 1 = 0` |
| Answer» Correct Answer - `{(i)/(sqrt(2)),(-i)/(sqrt(2))}` | |
| 216. |
Let `omega` be a complex number such that `2omega+1=z` where `z=sqrt(-3)`. If `|{:(1,1,1),(1,-omega^(2)-1,omega^(2)),(1,omega^(2),omega^(7))|=3k`, then k is equal to |
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Answer» Here, `z = sqrt (-3) = sqrt3i` `:. 2omega+1 = sqrt3i` `=>omega = (sqrt3i-1)/2` Now, we will find the value of the given determinant. `|[1,1,1],[1,-omega^2-1,omega^2],[1,omega^2,omega^7]|` Now, we know, `1+omega+omega^2 = 0` amd `omega^3 = 1`So, our determinant becomes, `|[1,1,1],[1,omega,omega^2],[1,omega^2,omega]|` `=[omega^2-omega^4-omega+omega^2+omega^2-omega]` `=[omega^2-omega-omega+omega^2+omega^2-omega]` `=3(omega^2 - omega)` `=3(-1/2-(sqrt3i)/2 + 1/2 - (sqrt3i)/2))` `= -3sqrt3i` `:. -3sqrt3i = 3k` `=> k = -sqrt3i` `=> k = -z.` |
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| 217. |
If `x^2+x+1=0` then the value of `(x+1/x)^2+(x^2+1/(x^2))^2+...+(x^27+1/(x^27))^2` is |
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Answer» `1 + omega + omega^2 = 0` `1 + x + x^2 = 0 ` `omega^3 = 1 , 1 + omega + omega^2 = 0` `(x + 1/x)^2 + ( x^2 + 1/x^2)^2 + ....... (x^27 + 1/x^27)^2` when `x= omega` `( omega + omega^2/omega^3)^2 + ( omega^2 + omega/omega^3) + (1+1)^2 + ( omega + omega^2/omega^6)^2 + (omega^2 + omega/omega^6)^2 + (1+1)^2` = `(-1)^2 + (-1)^2 + (2)^2 + (-1)^2 + (-1)^2 + (2)^2 ` `= 9[(-1)^2 + (-1)^2 + 2^2]` `= 9 xx 6 = 54` Answer |
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| 218. |
Solve the equation : `3x^2-4x+(20)/3=0` |
| Answer» Correct Answer - `{(2)/(3)+(4)/(3)i,(2)/(3)-(4)/(3)i}` | |
| 219. |
Solve: `x^(2) + 5 = 0` |
| Answer» Correct Answer - `{isqrt(5),-isqrt(5)}` | |
| 220. |
If `omega` is a complex cube root of unity, then the value of the expression `1(2-omega)(2-omega^2)+2(3-omega)(3-omega^2) +...+(n-1) (n-omega)(n-omega^2) (n>=2) ` is equal to (A) `(n^2(n+1)^2)/4 - n` (B) `(n^2(n+1)^2)/4 +n` (C) `(n^2(n+1))/4 -n` (D) `(n(n+1)^2)/4 -n` |
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Answer» We can write the given expression as , `S = sum_(m=1)^n(m-1)(m-omega)(m-omega^2)` So , general term in this expression can be given as, `T_m = (m-1)(m-omega)(m-omega^2)` `= (m-1)(m^2-(omega+omega^2)m+omega^3)` As, `1+omega+omega^2 = 0=>omega+omega^2 = -1` `:. T_m = (m-1)(m^2+m+1)` (As `omega^3 = 1`) We know, `(a-1)(a^2+a+1) = a^3-1``:. T_m=m^3-1` `:. S = sum_(m=1)^n (m^3-1) = sum_(m=1)^n m^3- sum_(m=1)^n 1` `=> S= ((n(n+1))/2)^2+n = 1/4(n^2)(n+1)^2-n` So, option `A` is the correct option. |
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| 221. |
Solve: `x^(2) + 2 = 0` |
| Answer» Correct Answer - `{isqrt(2),-isqrt(2)}` | |
| 222. |
Find the modulus and argument of the following complex number and henceexpress each of them in the polar form:` 1-i` |
| Answer» Correct Answer - `sqrt(2),(-pi)/(4),sqrt(2){cos((-pi)/(4))+i sin((-pi)/(4))}` | |
| 223. |
If z is a complex number such that `|z|=4` and `arg(z) =(5pi)/6` then z is equal to |
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Answer» `z= |z|(e^(img(z)))` `= 4(e^(i5 pi/6))` `= 4(cos 5 pi/6 + i sin 5 pi/6)` `= 4[cos(pi-pi/6) + i sin( pi - pi/6)]` `= 4[- sqrt3/2 + i 1/2]` `= 2[- sqrt3 + i]` `z = - 2 sqrt 3 + 2i` option 1 is correct Answer |
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| 224. |
The area bounded by the curves ` arg z = pi/3 and arg z = 2 pi /3 and arg(z-2-2isqrt3) = pi` in the argand plane is (in sq. units) |
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Answer» `arg(Z-(2+2sqrt3i))=pi` `tantheta=(2sqrt3)/2=sqrt3` Area=`1/2*(4)^2*2sqrt3` `=4sqrt3Unit^2`. |
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| 225. |
Find the modulus and argument of each of the following complex numbers and hence express each of them in polar form: `(sin 120^(@) - i cos 120^(@))` |
| Answer» Correct Answer - `(cos 30^(@)+ i sin 30^(@))` | |
| 226. |
Find the modulus and argument of each of the following complex numbers and hence express each of them in polar form: -i+i |
| Answer» Correct Answer - `sqrt(2),(3pi)/(4), sqrt(2)("cos"(3pi)/(4)+ "i sin"(3 pi)/(4))` | |
| 227. |
Find the modulus and argument of the following complex number and henceexpress each of them in the polar form: ` sqrt(3)+i` |
| Answer» Correct Answer - `2,(pi)/(6),2("cos"(pi)/(6)+"i sin"(pi)/(6))` | |
| 228. |
Find the modulus and argument of the following complex number and henceexpress each of them in the polar form: `(1-i)/(1+i)` |
| Answer» Correct Answer - `1,(-pi)/(2),{cos((-pi)/(2))+i sin((-pi)/(2))}` | |
| 229. |
Find the modulus and argument of each of the following complex numbers and hence express each of them in polar form: `1- sqrt(3)i` |
| Answer» Correct Answer - `2,(-pi)/(3),2{cos((-pi)/(3))+i sin((-pi)/(3))}` | |
| 230. |
Find the modulus and argument of each of the following complex numbers and hence express each of them in polar form: 4 |
| Answer» Correct Answer - `4, 0, 4(cos 0 + i sin 0)` | |
| 231. |
Find the modulus and argument of each of the following complex numbers and hence express each of them in polar form: -i |
| Answer» Correct Answer - `1,(-pi)/(2),{cos((-pi)/(2)+i sin((-pi)/(2))}` | |
| 232. |
Find the modulus and argument of each of the following complex numbers and hence express each of them in polar form: `-1+sqrt(3)i` |
| Answer» Correct Answer - `2,(2pi)/(3),2("cos"(2pi)/(3)+"i sin"(2pi)/(3))` | |
| 233. |
Find the modulus and argument of each of the following complex numbers and hence express each of them in polar form: `(i^(25))^(3)` |
| Answer» Correct Answer - `1, (-pi)/(2),{cos((-pi)/(2))+i sin((-pi)/(2))}` | |
| 234. |
Find the modulus and argument of each of the following complex numbers and hence express each of them in polar form: `-4+4 sqrt(3)i` |
| Answer» Correct Answer - `8,(2pi)/(3)8{"cos"(2pi)/(3)+"i sin"(2pi)/(3)}` | |
| 235. |
Find the modulus and argument of each of the following complex numbers and hence express each of them in polar form: `(1+i)/(1-i)` |
| Answer» Correct Answer - `1,(pi)/(2),("cos"(pi)/(2)+"i sin"(pi)/(2))` | |
| 236. |
Find the modulus and argument of each of the following complex numbers and hence express each of them in polar form: -2 |
| Answer» Correct Answer - `2, pi, 2(cos pi + i sin pi)` | |
| 237. |
Find the modulus and argument of each of the following complex numbers and hence express each of them in polar form: `sqrt((1+i)/(1-i))` |
| Answer» Correct Answer - `1,(pi)/(4),("cos"(pi)/(4)+"i sin"(pi)/(4))` | |
| 238. |
The complex number z having least positive argument which satisfy the condition `|z - 25i| |
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Answer» With the given condition `|z-25i| le 15`, we can draw a circle with center `(0,25)` and radius `15`. Please refer to video for the figure. From the figure, we can see that, `P` point will have least positive argument. Here, `CP = 15, OC = 25 and /_CPO = 90^@` Then, `OP^2 =OC^2- CP^2 = 25^2-15^2 = 400` `=>OP = 20` Let `/_POC = theta` Then, `sintheta = 15/25 = 3/5, cos theta = 20/25 = 4/5` So, coordinates of point `P` will be `(OP sintheta,OPcostheta)`. Coordinates of `P` are `(20(3/5),20(4/5))` that is `(12,16)`. If we convert it into a complex number, then it will be `12+16i`, which is the required number. |
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| 239. |
Find the modulus and argument of each of the following complex numbers and hence express each of them in polar form: `(2+6 sqrt(3)i)/(5+sqrt(3)i)` |
| Answer» Correct Answer - `2,(pi)/(3),2("cos"(pi)/(3)+"i sin"(pi)/(3))` | |
| 240. |
Find the modulus and argument of each of the following complex numbers and hence express each of them in polar form: `-3 sqrt(2)+3 sqrt(2)i` |
| Answer» Correct Answer - `6,(3pi)/(4),6("cos"(3pi)/(4)+"i sin"(3pi)/(4))` | |
| 241. |
Find the modulus and argument of each of the following complex numbers and hence express each of them in polar form: `(5-i)/(2-3i)` |
| Answer» Correct Answer - `sqrt(2),(pi)/(4),sqrt(2)("cos"(pi)/(4)+ "i sin"(pi)/(4))` | |
| 242. |
Evaluate `((i^(180)+i^(178)+i^(176)+i^(174)+i^(172))/(i^(170)+i^(168)+i^(166)+i^(164)+i^(162)))`. |
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Answer» Correct Answer - -1 Given expression `=(i^(172)(i^(8)+i^(6)+i^(4)+i^(2)+1))/(i^(162)(i^(8)+i^(6)+i^(4)+i^(2)+1))=i^((172-162))=i^(10)=i^(8)xxi^(2) = -1`. |
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