# Free Objective Test 01 Practice Test - 11th and 12th

The  value of (log20.5 4) is

A. -2
B. (4)
C. 2
D. None of these

#### SOLUTION

Solution : C

(log20.5 4)={log0.5(0.5)2}2=(2)2=2

The value of log3 4log4 5log5 6log6 7log7 8log8 9 is

A. 1
B. 2
C. 3
D. 4

#### SOLUTION

Solution : B

log3 4.log4 5.log5 6.log6 7.log7 8.log8 9
=log 4log 3.log 5log 4.log 6log 5.log 7log 6.log 8log 7.log 9log 8=log 9log 3
=log3 9=log3 32=2

log7 log7 7(77)=

A. 3 log2 7
B. 13 log3 7
C. 13 log7 2
D. None of these

#### SOLUTION

Solution : C

log7 log7 777=log7 log7 778=log7(78)
=log7 7log7 8=1log7 23=13 log7 2

The value of 81(1log5 3)+27log9 36+34log7 9 is equal to

A. 49
B. 625
C. 216
D. 890

#### SOLUTION

Solution : D

81(1log5 3)+27log9 36+34log7 9
=3log3 54+3log3 3632+3log3 742
=54+3632+72=890

7 log (1615)+5 log(2524)+3 log(8180) is equal to -

A. 0
B. 1
C. log 2
D. log 3

#### SOLUTION

Solution : C

7 log (1615)+5 log(2524)+3 log(8180)
a log x=log xa
log (1615)7+log (2524)5+log (8180)3
log x+log y=log (x×y)
log(167157.255245.813803)
=log 2

If log12sin x>0,xϵ[0,4π] then the number of values of x which are integral multiples of π4 is

A. 4
B. 12
C. 3
D. None of these

#### SOLUTION

Solution : A

0<12<1 log12sin x>0,xϵ[0,4π]0<sin x<1
Integral multiple of π4 will be
π4,3π4,9π4,11π4
Number of required values = 4.

The set of real values of x satisfying log12(x26x+12)2 is

A. (,2]
B. [2,4]
C. [4,+]
D. None of these

#### SOLUTION

Solution : B

log12(x26x+12)2      ...(i)
For log to be defined, x26x+12>0
(x3)2+3>0, which is true xϵR.
From (i), x26x+12(12)2
x26x+124x26x+80(x2)(x4)02x4;xϵ[2,4].

The set of real values of x for which 2log2 (x1)>x+5 is

A. (,1)(4,+)
B. (4,+)
C. (1,4)
D. None of these

#### SOLUTION

Solution : B

2log2 (x1)>x+52log2 (x1)2>x+5(x1)2>x+5x23x4>0(x4)(x+1)>0x>4orx<1
But for given log to be defined, x - 1 > 0
i.e.,x>1x>4xϵ(4,).

If log0.04(x1)log0.2(x1) then x belongs to the interval

A. (1,2]
B. (,2)
C. [2,+]
D. None of these

#### SOLUTION

Solution : C

log0.04(x1)log0.2(x1)    ....(i)
For log to be defined x1>0x>1
From(i),log(0.2)2(x1)log0.2(x1)
12log0.2(x1)log0.2(x1)x1(x1)x1(1x1)01x10x11x2,xϵ[2,).

Which of the following hold good?

A. a4+b4+c4>abc(a+b+c)
B. a5+b5+c5+d5>abcd(a+b+c+d)
C. a5+b5+c5>abc(ab+bc+ca)
D. a8+b8+c8a3b3c3>1a+1b+1c
E. b2+c2b+c+c2+a2c+a+a2+b2a+b>a+b+c

#### SOLUTION

Solution : A, B, C, and D

(a), (b), (c), (d)
(a) a4+b4+c43>(a+b+c3)4
or (a+b+c3)(a+b+c3)3>a+b+c3[(abc)13]3;
A.M.>G.M.
or >(a+b+c3)
a4+b4+c4>abc(a+b+c)

(b) As above

(c) (a5+b5+c53)>(a+b+c3)5
or >(a+b+c3)3(a+b+c3)2
or >[(abc)13]3(a2+b2+c2+2ab+2bc+2ca9)
But we know that a2+b2+c2>ab+bc+ca by result of two by two rule.
a5+b5+c53>abc3(ab+bc+ca)9 etc.

(d) a8+b8+c8>a2b2c2(bc+ca+ab)
Now a8+b8+c83>(a+b+c3)8
or >(a+b+c3)6(a+b+c3)2>[(abc)13]6[a2+b2+c2+2ab+2bc+2ca9]
A.M.>G.M
But by two by two rule
a2+b2+c2>ab+bc+ca
a8+b8+c83>a2b2c2(3ab+3bc+3ca)9
a8+b8+c8>a2b2c2(ab+bc+ca)
or a8b8c8a3b3c3>ab+bc+caabc or >1a+1b+1c

(e) b2+c22>(b+c2)2b2+c2b+c>b+c2
Write similar inequalities and add.

Which of the following hold good? If n is a +ve integer, then

A. nn>1.3.5(2n1)
B. 2.4.62n<(n+1)n
C. (n!)3<nn(n+12)2n
D. [1r+2r+3r++nr]n>nn.(n!)r

#### SOLUTION

Solution : A, B, C, and D

(a), (b), (c), (d)
(a) Apply A.M. of 1,3,5 .... (2n-1), i.e. of n numbers is > G.M. and remember that 1+3+5+... upto n terms =n2[2a+(n1)d]=n2 as a=1 and d=2

(b) Proceed as in part (a)

(c) Apply A.M. of 13,23,n3 is greater than their G.M. and remember that
13+23+n3={n(n+1)2}2

(d) 1r+2r+3r++nrn>[1r.2r.3r..nr]1n
or (1r+2r+3r++nr)n>nn(n!)r

Which of the following hold good?

A.
pxqr+qxrp+rxpq>p+q+r,  unless p = q = r, or x = 1

B. [x2+y2+z2x+y+z]x+y+z>xxyyzz>[x+y+z3]x+y+z
C. [a2+b2a+b]a+b>aabb
D. aabb>(a+b2)a+b>ab.ba

#### SOLUTION

Solution : A, B, C, and D

(a), (b), (c), (d)
Now because A.M. > G.M.
(xqr+xqr+p times)
+(xrp+xrp+q times)
+(xpq+xpq+r times)––––––––––––––––––––––––––––––
p+q+r
>{(xqr.xqrp factors).(xrp.xrpq factors).(xpq.xpqr factors)}1(p+q+r)
or pxqr+qxrp+rxpqp+q+r
>{xp(qr)+q(rp)+r(pq)}1(p+q+r)
or pxqr+qxrp+rxpqp+q+r>[x0]1(p+q+r)
or >1
pxqr+qxrp+rxpq>p+q+r

(b) In case either p=q=r or x=1, inequality becomes equality
Consider x numbers each equal to x, y numbers each equal to y and z numbers each equal to z.
Apply A.M. > G.M. on these x+y+z numbers
(x+x+x+x times)+(y+y+y+y times)+(z+z+z+z times)––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
x+y+z
>[(x,xx factors)(y.yf factors)(z.zz factors)]1(x+y+z)
But x+x+x+x times=x.x=x2
and x.x.xx factors=xx
x2+y2+z2x+y+z>(xxyyzz)1(x+y+z)
or (x2+y2+z2x+y+z)x+y+z>xxyyzz
Above proves the first part.  ......... (A)

2nd part:
For 2nd part consider x numbers each equal to 1x,y numbers each equal to 1y,z numbers each equal to 1z. Apply A.M. > G.M. on these x+y+z numbers.
(1x+1x+x times)+(1y+1y+y times)+(1z+1z+z times)––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
x+y+z
>[(1x.1xx factors)(1y.1yy factors)(1z.1zz factors)]1(x+y+z)
or 1x.x+1y.y+1z.zx+y+z>[1xx.1yy.1zz]1(x+y+z)
or (3x+y+z)x+y+z>1xxyyzz
Taking reciprocal and changing the sign of inequality, we get
(x+y+z3)x+y+z<xxyyzz
or xxyyzz>(x+y+z3)x+y+z(B)
Both (A) and (B) prove the required result.

(c) Proceed as in part (b)
Consider a quantities each equal to 1a,b quantities each equal to 1b; then since A.M. > G.M.
[1a+1a+1a+a times]+[1b+1b+b times]a+b
>{(1a.1a.1aa factors)×(1b.1b.1bb factors)1(a+b)}
or a.1a+b.1ba+b>{1a1b}1(a+b)
or {2a+b}>{1aa.bb}1(a+b)
or {2a+b}a+b>1aabb
Taking reciprocal and hence changing the sign of inequality, we get
{a+b2}a+b<aa.bb
or aa.bb>{a+b2}a+b(1)
Note: a and b can be integers as well as fractions

(d) Consider b quantities each equal to a and a quantities each equal to b. Then since A.M. > G.M
(a+a+a+b times)+(b+b+b+a times)a+b
>[(a.a.ab factors)(b.b.ba factors)1(a+b)]
or ab+aba+b>(ab.ba)1(a+b)
or 2aba+b>(ab.ba)1(a+b)
But a+b2 being A.M. of a and b is greater than 2aba+b their H.M.
a+b2>2aba+b>[ab.ba]1(a+b)
or [a+b2]a+b>ab.ba(2)
aa.bb>{a+b2}a+b>ab.ba
by (1) of part (c) and (2).

If A, B, C be the angles of a triangle, then which of the following hold good?

A. cot A.cot B.cot C133
B. cotA2cotB2cotC23(3)
C. cosA+cosB+cosC<32
D. sin(A2)sin(B2)sin(C2)18

#### SOLUTION

Solution : A, B, C, and D

All the four hold good
(a) We know that when A+B+C=π then
tan(A+B+C)=tanπ=0
or S1S31S2=0S1=S3
or tan A+tan B+tan C=tan Atan Btan C(1)
Now tan A+tan B+tan C3(tan Atan Btan C)13
or (tan Atan Btan C)327tan Atan Btan C
or tan Atan Btan C33
or cot Acot Bcot C133(2)

(b) Again A2+B2+C2=π2
tan(A2+B2+C2)=
=S1S31S2=
1S2=0
or tan(A2)tan(B2)=1
or on dividing by tan(A2)tan(B2)tan(C2),
we get cot(A2)+cot(B2)+cot(C2)
=cot(A2)+cot(B2)+cot(C2)(1)
Now proceed as in part (a) for result (1)

(c) We have cos A+cos B+cos C
=2cos A+B2cosAB2+cos C
=2sin C2cos AB2+cos C2 sin C2+cos C
[0cos A+B21]
=2 sin C2+12 sin2 C2
=2[(sin C212)214]+1
=322(sin C212)232

(d) We know from trigonometry that
cos A+cos B+cos C=1+4 sin A2sin B2sinC232 by (c)
4 sin A2sin B2sin C2321=12
sin A2sin B2sin C218

If xϵR, the solution set of the equation
4x+0.57.2x4<0 is equal to

A. (2,72)
B. (2,)
C. (2,)
D. (,)

#### SOLUTION

Solution : B

Put 2x=t and 40.5=412=2
2t27t4<0(t4)(2t+1)<0
12<t<40<2x<22
or (12)<(12)x<(12)2
Since 12<12<x<
xϵ(2,)(b)

If xϵR and m=x2(x42x2+4), then m lies in the interval

A. [0,14]
B. [0,13]
C. [0,12]
D. [0,15]

#### SOLUTION

Solution : C

m=x2(x21)2+3=+ive i.e.0
Again m=1x2+4x22=1(x2x)2+212
mϵ[0,12]