JEE mathematics

Sponsor Area

Question
CBSEENMA11015519

Let α,  β be real and z be a complex number. If z2 + αz + β = 0 has two distinct roots on the line Re z = 1, then it is necessary that

  • β ∈(0, 1)

  • β ∈(-1, 0)

  • |β| = 1

  • β ∈ (1, ∞)

Solution

D.

β ∈ (1, ∞)

Let roots be p + iq and p - iq p, q ∈ R
root lie on line Re(z) = 1
⇒ p = 1
product of roots = p2 + q2 =  β = 1 + q2
⇒ β∈ (1, ∞), (q ≠ 0, ∵ roots are distinct)

Sponsor Area

Question
CBSEENMA11015520

The coefficient of x7 in the expansion of (1 - x - x2 +x3)6 is :

  • 144

  • -132

  • -144

  • 132

Solution

C.

-144

left parenthesis 1 minus straight x minus straight x squared plus straight x cubed right parenthesis to the power of 6
left parenthesis 1 minus straight x right parenthesis to the power of 6 left parenthesis 1 minus straight x squared right parenthesis to the power of 6
left parenthesis to the power of 6 straight C subscript 0 minus to the power of 6 straight C subscript 1 straight x to the power of 1 plus to the power of 6 straight C subscript 2 straight x squared minus to the power of 6 straight C subscript 3 straight x cubed plus to the power of 6 straight C subscript 4 straight x to the power of 4 minus space to the power of 6 straight C subscript 5 straight x to the power of 5 space plus to the power of 6 straight C subscript 6 straight x to the power of 6 right parenthesis
left parenthesis to the power of 6 straight C subscript 0 minus to the power of 6 straight C subscript 1 straight x squared space plus to the power of 6 straight C subscript 2 straight x to the power of 4 minus to the power of 6 straight C subscript 3 straight x squared space plus to the power of 6 straight C subscript 4 straight x to the power of 8 space plus...... plus to the power of 6 straight C subscript 6 straight x to the power of 12 right parenthesis
Now space coefficient space of space straight x to the power of 7 space equals space to the power of 6 straight C subscript 1 to the power of 6 straight C subscript 3 minus to the power of 6 straight C subscript 3 to the power of 6 straight C subscript 2 space plus to the power of 6 straight C subscript 5 to the power of 6 straight C subscript 1
space equals space 6 space straight x space 20 space minus space 20 space space straight x space 15 space plus 36
space equals space 120 minus 300 plus 36
space equals space 156 minus 300
space equals space minus 144 space space

Question
CBSEENMA11015521

A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases by Rs. 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs. 11040 after

  • 18 Months

  • 19 Months

  • 20 Months

  • 21 Months

Solution

D.

21 Months

a = Rs. 200
d = Rs. 40
savings in first two months = Rs. 400
remained savings = 200 + 240 + 280 + ..... upto n terms =
n/2[400 + (n -1)40] = 11040 - 400
200n + 20n2 - 20n = 10640
20n2 + 180 n - 10640 = 0
n2 + 9n - 532 = 0
(n + 28) (n - 19) = 0
n = 19
∴ no. of months = 19 + 2 = 21

Question
CBSEENMA11015522

Consider the following statements
P: Suman is brilliant
Q: Suman is rich
R: Suman is honest. The negation of the statement ì Suman is brilliant and dishonest if and only if Suman is richî can be ex- pressed as

  •  ~ P ^ (Q ↔ ~ R)

  • ~ (Q ↔ (P ^ ~R)

  • ~ Q ↔ ~ P ^ R

  • ~ (P ^ ~ R)↔ Q

Solution

B.

~ (Q ↔ (P ^ ~R)

Negation of (PΛ~ R) ↔ Q is ~ ↔(PΛ ~ R)↔Q)
It may also be written as ~ (Q ↔ (PΛ ~ R))

Question
CBSEENMA11015523

If ω(≠1) is a cube root of unity, and (1 + ω)7 = A + Bω.Then (A, B) equals

  • (0,1)

  • (1,1)

  • (1,0)

  • (-1,1)

Solution

B.

(1,1)

(1 + ω)7 = A + Bω
(-ω2)7 = A + Bω
- ω14 = A + Bω

2 = A + Bω
1 + ω = A + Bω
therefore,
(A, B) = (1, 1)