Test: Complex Numbers - JEE MCQ

# Test: Complex Numbers - JEE MCQ

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## 29 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Complex Numbers

Test: Complex Numbers for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The Test: Complex Numbers questions and answers have been prepared according to the JEE exam syllabus.The Test: Complex Numbers MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Complex Numbers below.
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Test: Complex Numbers - Question 1
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Test: Complex Numbers - Question 2
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Test: Complex Numbers - Question 3
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Test: Complex Numbers - Question 7

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Test: Complex Numbers - Question 8

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Test: Complex Numbers - Question 9

Find the value of

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Test: Complex Numbers - Question 10

If the roots of the equation x2 - 5x + 16 = 0 are a, b and the roots of the equation x2 + px + q = 0 are (a2 + b2) and , then-

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Test: Complex Numbers - Question 11

The number of the integer solutions of x2 + 9 < (x + 3)2 < 8x + 25 is

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Test: Complex Numbers - Question 12

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Test: Complex Numbers - Question 13

The equatiobn πx = –2x2 + 6x – 9 has

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The RHS of the expression has a<0 which means the graph will lie below the x-axis and πx lies above the x-axis.Therefore,no solution.

Test: Complex Numbers - Question 14

Let a > 0, b > 0 and c > 0. Then both the roots of the equation ax2 + bx + c = 0

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Test: Complex Numbers - Question 15

The set of all solutions of the inequality  < 1/4 contains the set

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(1/2)(x2 - 2x) < (1/4)
(1/2)(x2 - 2x) < (1/2)2
x2 − 2x > 2......(as after multiplicative inverse sign of inequality changes)
x2 − 2x − 2 > 0
x2  -  2x + 1 - 3  >
(x-1)2  - 3  > 0
(x-1)2  > 3
So for the above to hold good both the expression must be positive or both must be negative. After finding the solution the range of the solution will be
either x > 3
(3,¥)

Test: Complex Numbers - Question 16

Consider y =, where x is real, then the range of expression y2 + y – 2 is

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Test: Complex Numbers - Question 17

The values of k, for which the equation x2 + 2(k – 1) x + k + 5 = 0 possess atleast one positive root, are

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Test: Complex Numbers - Question 18

If  <= 4, then least and the highest values of 4x2 are

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Test: Complex Numbers - Question 19

If coefficients of the equation ax2 + bx + c = 0, a < 0 are real and roots of the equation are non–real complex and a + c + b < 0, then

Detailed Solution for Test: Complex Numbers - Question 19

Puting x=-1
a-b+c
but
a+c<b or a+c-b<0
hence f(x)<0 for all real values of x
therefore
putting x=-2
we get f(X)=4a+c-2b<0
or 4a+c<2b

Test: Complex Numbers - Question 20

If the roots of the equation x2 + 2ax + b = 0 are real and distinct and they differ by at most 2m, then b lies in the interval

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Test: Complex Numbers - Question 21

If the roots of the quadratic equation x2 + px + q = 0 are tan 30º and tan 15º respectively, then the value of 2 + q – p is

Test: Complex Numbers - Question 22

If (1 – p) is root of quadratic equation x2 + px + (1 – p) = 0, then its roots are

Detailed Solution for Test: Complex Numbers - Question 22

As (1 – p) is root of the equation: x2 + px + (1 – p) = 0

(1 – p)2 + p(1 – p) + (1 – p) = 0

(1 – p)[1 – p + p + 1] = 0

(1 – p) = 0

p = 1

Therefore, given equation now becomes

x2 + x = 0

x(x + 1) = 0

x = 0, -1

Test: Complex Numbers - Question 23

The values of x and y besides y can satisfy the equation (x, y ∈ real numbers) x2– xy + y2 – 4x – 4y + 16 = 0

Test: Complex Numbers - Question 24

For what value of a and b the equation x4 – 4x3 + ax2 + bx + 1 = 0 has four real positive roots ?

Detailed Solution for Test: Complex Numbers - Question 24

Since all roots are positive,
coeff. of x shld be negatuve i.e., b is negative
let roots be m(>0), n(>0), p(>0), q(>0)
m+n+p+q = 4
mn+np+pq+mp+mq+nq = a
mnp+mnq+npq+mpq = -b
mnpq = 1
For the objective case,
put m=n=p=q =1
so a = 1+1+1+1+1+1 = 6
-b = 1+1+1+1 =4
so, a = 6 and b = -4

Test: Complex Numbers - Question 25

If a, b are roots of the equation ax2 + bx + c = 0, then the value of a3 + b3 is

Test: Complex Numbers - Question 26

If z1 = 2 + i, z2 = 1 + 3i, then Re ( z1 - z2) =

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The numbers in the questions are not very clear.
z1 = 2 + i
z2 = 1+3i
z1 – z2 = (2 – 1) + i (1 – 3)
= 1 – 2i

Test: Complex Numbers - Question 27

If a, b are roots of the equation ax2 + bx + c = 0 then the equation whose roots are 2a + 3b and 3a + 2b is

Test: Complex Numbers - Question 28

If S is the set of all real x such that  is positive, then S contains

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Test: Complex Numbers - Question 29

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## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests