  ## UPTU SEE MCA Entrance Exam Important Questions (Mathematics and Statistics)

Important Question Mathematics And Statistics with Hints

1. If iz3+z2-z+i=0, Then |z| is equal to

(a) -1 (b) 1 (c) i (d) –i

2. If arg (z)<0, then arg(-z) – arg(z) is equal to 3. For any complex number z, the minimum value of |z|+|z+1| is

(a) 1 (b) 0 (c) 1/2 (d) 3/2

4. If z1, z2, z3 are in H.P. they lie on a

(a) circle (b) sphere (c) straight line (d) none of these

5. If z4 = i then z can take the value 6. The least value of |z+(1/z)| if |z| >= 3 is

(a) 10/3            (b) 8/3          (c) 4/3              (d) 2

7. The harmonic mean of the roots of the equation (a) 2           (b) 4               (c) 6                       (d) 8

8. Let the positive number a, b, c ,d be in A.P. then abc, abd, acd, bcd are

(a) Not a A.P./ G.P./ H.P. (b) in A.P. (c) in G.P. (d) in H.P.

9. The sum of all two digit number witch are odd is

(a) 2475 (b) 5049 (c) 2530 (d) 4905

10. The harmonic mean of two is 4, their A.M. is A, and G.M. is G satisfy the relation 2A + G2 = 27. The two number are

(a) 6, 3 (b) 5, 4 (c) 5, -2.5 (d) -3, 1

11. If x2 – 3x + 2 is a factor of x4 – px2 + q = 0, then (p, q) are

(a) (4, 5) (b)(5, 4) (c) (2, 3) (d) (0, 0)

1. (b) 2. (a) 3. (a) 4. (c) 5. (a) 6. (b) 7. (b) 8. (d) 9. (a) 10. (a) 11. (b) 12. (a) 13. (d) 14. (c) 15. (a)

Solution (Hint) of these question:

Ans-1

Dividing by I then z3+(1/i)z2-(1/i)z+1=0

We know that (1/i)=-i

z3-iz2+iz+1=0 or z2 (z-i)+i(z-i)=0 as 1=-i2

or (z-i) (z2+i)=0 therefor z=i or z2=-i

therefor |z| = |i| = 1 and | z2| = |z|2 = |-i| = 1

therefor |z| =1

hence in either case |z|=1

Ans-2 Ans-3

1 = z + (1-z)

|1| = | z + (1-z)| <= |z| + |1-z|

= | z + (1-z)|

Therefor | z + (1-z)| >= 1

Therefor minimum value is 1.

Ans – 4

z1, z2, z3 are in H.P. means that their moduli r1, r2, r3 are in H.P. Ans – 8

a, b, c ,d are in A.P. then dividing by abcd

1/bcd, 1/acd ,1/abd, 1/abc are in also in A.P.

Hence there reciprocals are in H.P.

Ans – 9

11+13+15+……..+99

A=11, d=2, n=45

Ans – 10

H=4 and G2=AH gives G2 =4A

Also 2A+G2=27 or 2A+4A=27

2A=9 or a+b=9 ——-(1)

Also G2=4A=18 or ab=18 —–(2)

By eq (1) and (2)

t2 – 9t + 18 =0

Therefor, t=6, 3

Ans – 11

x2 – 3x + 2 = (x-1)(x-2)

therefor, 1 and 2 are root of 2nd equation

q-p+1 = 0, q-4p+16 =0

p=5, q=4

Ans – 12

b2=ac => 2logb =loga + logc

therefor, logna, lognb, lognc are in A.P.

Their reciprocals

logan, logbn, logcn are in H.P.

Ans – 13

(2n)!/{3!(2n-3)!} *{2!(n-2)!}/n! =11/1

=> 4(2n-1) = 33

Therefor, n = 37/8

Ans – 14

10P99P8 =10! – 9!

= 9!(10-1) = 9*9!

Ans – 15

23n = 8n = (1+7)n

=> (1+7)n – 7n – 1 = nC2 72 + nC3 73 + …..

=> 72

Hence divisible by 49

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