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

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

Solution : D

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

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

Solution : B and C

logk x.log5 k=logx 5log5 x=logx 5
logx 5=1logx 5(logx 5)2=1logx 5=±1
x±1=5x=5,15.

log5 a.logax=2, then x is equal to

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

#### SOLUTION

Solution : C

log5 a.loga x=2log5x=2x=52=25

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

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

#### SOLUTION

Solution : C

A=log2 log2 log4 256+2log2122
=log2 log2 log4 44+2×1(12)log22
=log2 log24+4=log2 log2 22+4
=log22+4=1+4=5

If log10 x=y then log1000x2 is equal to

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

#### SOLUTION

Solution : D

log1000 x2
=log103x2
=2 log103x
=23log10x
=23y

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

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
0.8x2
x52.
Case (ii) x < -2
From(ii), x+20.2x0.8x2x52
xϵ(,52](0,). 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

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

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

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

#### SOLUTION

Solution : C

We have
3x1+3x1=13(3x+3x)13.23x.3x,
[A.M.G.M.]
or 3x1+3x123
Hence the least value of 3x1+3x1 is 23

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

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

#### SOLUTION

Solution : D

(,0)
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)

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

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
an=a1(2)n1=102n1/2
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)

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

Solution : D

E2=1+1a+1b+1ab=1+a+b+1ab=1+λ+1ab
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.
a+b=λa=b=λ2
E2=λ2+4λ+4λ2=(λ+2λ)2
E=λ+2λ=1+2λ(d)

If x=log53+log75+log97, then x

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

#### SOLUTION

Solution : C

Apply A.M.G.M.
x3.(log53.log75.log97)13
=3(log93)13=3(log323)13=3(12)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

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)

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

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

#### SOLUTION

Solution : C

E=2 log100aloga(100)2
=[2 log100a+2 loga100]
2.2[log100a.loga.100]12
=4[logaa]12=4

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

Solution : A

Since a, b, c are of same sign, ab,bc,ca are all +ive.
Apply A.M. G.M.
E3[ab.bc.ca]13=3
therefore E lies in the interval [3,)