Free Objective Test 02 Practice Test - 11th and 12th 

Question 1

If log4 5 = a and log5 6 = b, then log3 2 is equal to

A. 12a+1
B. 12b+1
C. 2ab+1
D. 12ab1


Solution : D

ab=log4 5. log5 6=log4 6=12log2 6
ab=12(1+log2 3)2ab1=log2 3
 log3 2=12ab1

Question 2

If logk x.log5 k=logx 5, k1, k>0, then x is equal to

A. k
B. 15
C. 5
D. None of these


Solution : B and C

logk x.log5 k=logx 5log5 x=logx 5
logx 5=1logx 5(logx 5)2=1logx 5=±1

Question 3

log5 a.logax=2, then x is equal to

A. 125
B. a2
C. 25
D. None of these


Solution : C

log5 a.loga x=2log5x=2x=52=25

Question 4

If A=log2 log2 log4 256+2log22, then A is equal to

A. 2
B. 3
C. 5
D. 7


Solution : C

A=log2 log2 log4 256+2log2122
=log2 log2 log4 44+2×1(12)log22
=log2 log24+4=log2 log2 22+4

Question 5

If log10 x=y then log1000x2 is equal to

A. y2
B. 2y
C. 3y2
D. 2y3


Solution : D

log1000 x2
=2 log103x

Question 6

The set of real values of x for which log0.2x+2x1 is

A. (,52](0,+)
B. [52,+)
C. (,2)(0,+)
D. None of these


Solution : A

log0.2x+2x1    ....(i)
For log to be defined, x+2x>0x>0 or
x < -2
Now from(i), log0.2x+2xlog0.20.2
x+2x0.2               .....(ii)
Case (i) x > 0
From (ii),x+20.2x
Case (ii) x < -2
From(ii), x+20.2x0.8x2x52

Question 7

If A, B, C are the angles of a triangle such that C is an obtuse angle, then

A. tan A.tan B<1
B. tan A.tan B>1
C. tan A.tan B=1
D. None


Solution : A

Since C is obtuse therefore A and B both are acute and A + B is also acute.
Hence tan A,tan B and tan(A+B) are all + ive
tan(A+B)=tan A+tan B1tan Atan B=+ive
Above is possible only when tan Atan B<1

Question 8

If y=3x1+3x1 (x real), then the least value of y is

A. 2
B. 6
C. 23
D. None of these


Solution : C

We have
or 3x1+3x123
Hence the least value of 3x1+3x1 is 23

Question 9

The solution set of the inequality ||x|1<1x,2xϵR is equal to

A. (0,)
B. (1,)
C. (1,1)
D. (,0)


Solution : D

Here because of |x| which is equalto either x or -x we shall consider the cases x>0 or x<0
Then we will have |x - 1| = |x + 1| when x < 0
Case 1: x > 0, then |x-1|<(1-x)
or |1-x| < (1-x) i.e. |t|<t
Above is not true for any value of t.
Case 2: x<0,|x1|<1x
or |x+1|<(1x)(1)
If x+10i.ex1 and x<0
i.e. 1x<0 i.e. xϵ[1,0) then (1) reduces to
x+1=1x or 2x<0 or x<0
If x+1<0 i.e. x<1 then (1) reduce to
(x+1)<(1x) or 2<0 is true for all x<1(2)
Hence from (1) and (2) xϵ(,0)

Question 10

Let S1,S2 be squares such that for each n1, the length of a side of Sn equals the length of a diagonal of Sn+1. If the length of a side of S1 is 10 cm, then for which of the following values of n is the area of Sn less than 1 sq. cm?

A. 7
B. 8
C. 9
D. 10


Solution : B, C, and D

(b), (c), (d)
If a be the side of a square, then d=a2
By given condition an=2an+1
or an+1=an2=an1(2)2=an2(2)3=a1(2)n
Replacing n by n-1, we get
Area of Sn<1a2n<1
1002n1<1 or 200<2n or 2n>200
Now 27=128<200<28=256>200
n=8,9,10(b), (c), (d)

Question 11

If a and b are two positive quantities whose sum is λ, then the minimum value of (1+1a)(1+1b) is

A. λ1λ
B. λ2λ
C. 1+1λ
D. 1+2λ


Solution : D

Above will be minimum when ab is maximum.
Now we know that if sum of two quantities is constant then their product is maximum when the quantities are equal.

Question 12

If x=log53+log75+log97, then x

A. 32
B. 1213
C. 3213
D. None


Solution : C

Apply A.M.G.M.

Question 13

 If a, b, c are real numbers such that a + 2b + c = 4 then max. value of ab + bc + ca is

A. 2
B. 4
C. 6
D. 8


Solution : B

Given (a+b)+(b+c)=4
But [(a+b)+(b+c)2]2[(a+b)(b+c)],A.MG.M.
or (42)2ab+bc+ca+b2
4b2ab or ab4b2
Max. value of ab=4(b)

Question 14

The least value of 2log100aloga(0.0001), a>1 is

A. 2
B. 3
C. 4
D. None


Solution : C

E=2 log100aloga(100)2
=[2 log100a+2 loga100]

Question 15

If a, b, c be three real numbers of the same sign then the value of ab+bc+ca lies in the interval

A. [3,)
B. (3,)
C. [2,)
D. (,3)


Solution : A

Since a, b, c are of same sign, ab,bc,ca are all +ive.
Apply A.M. G.M.
therefore E lies in the interval [3,)