12th Grade > Mathematics
LOGARITHMS MCQs
Inequalities Modulus And Logarithms (11th And 12th Grade)
Total Questions : 50
| Page 1 of 5 pages
Answer: Option D. -> 2y3
:
D
log1000x2
=log103x2
=2log103x
=23log10x
=23y
:
D
log1000x2
=log103x2
=2log103x
=23log10x
=23y
Answer: Option C. -> [0,12]
:
C
m=x2(x2−1)2+3=+ivei.e.≥0
Again m=1x2+4x2−2=1(x−2x)2+2≤12
mϵ[0,12]
:
C
m=x2(x2−1)2+3=+ivei.e.≥0
Again m=1x2+4x2−2=1(x−2x)2+2≤12
mϵ[0,12]
Answer: Option D. -> 890
:
D
81(1log53)+27log936+34log79
=3log354+3log33632+3log3742
=54+3632+72=890
:
D
81(1log53)+27log936+34log79
=3log354+3log33632+3log3742
=54+3632+72=890
Answer: Option B. -> 2
:
B
log34.log45.log56.log67.log78.log89
=log4log3.log5log4.log6log5.log7log6.log8log7.log9log8=log9log3
=log39=log332=2
:
B
log34.log45.log56.log67.log78.log89
=log4log3.log5log4.log6log5.log7log6.log8log7.log9log8=log9log3
=log39=log332=2
Answer: Option C. -> 23
:
C
We have
3x−1+3−x−1=13(3x+3−x)≥13.2√3x.3−x,
[∵A.M.≥G.M.]
or 3x−1+3−x−1≥23
Hence the least value of 3x−1+3−x−1 is 23
:
C
We have
3x−1+3−x−1=13(3x+3−x)≥13.2√3x.3−x,
[∵A.M.≥G.M.]
or 3x−1+3−x−1≥23
Hence the least value of 3x−1+3−x−1 is 23
Answer: Option D. -> 1+2λ
:
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=λ⇒a=b=λ2
∴E2=λ2+4λ+4λ2=(λ+2λ)2
∴E=λ+2λ=1+2λ⇒(d)
:
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=λ⇒a=b=λ2
∴E2=λ2+4λ+4λ2=(λ+2λ)2
∴E=λ+2λ=1+2λ⇒(d)
Answer: Option C. -> 3213
:
C
Apply A.M.≥G.M.
∴x≥3.(log53.log75.log97)13
=3(log93)13=3(log323)13=3(12)13
:
C
Apply A.M.≥G.M.
∴x≥3.(log53.log75.log97)13
=3(log93)13=3(log323)13=3(12)13
Answer: Option A. -> [3,∞)
:
A
Since a, b, c are of same sign, ab,bc,ca are all +ive.
Apply A.M. ≥ G.M.
∴E≥3[ab.bc.ca]13=3
therefore E lies in the interval [3,∞)
:
A
Since a, b, c are of same sign, ab,bc,ca are all +ive.
Apply A.M. ≥ G.M.
∴E≥3[ab.bc.ca]13=3
therefore E lies in the interval [3,∞)
Answer: Option D. -> 12ab−1
:
D
ab=log45.log56=log46=12log26
ab=12(1+log23)⇒2ab−1=log23
∴log32=12ab−1
:
D
ab=log45.log56=log46=12log26
ab=12(1+log23)⇒2ab−1=log23
∴log32=12ab−1
Answer: Option A. -> 2n+5≤25
:
A
Cost of entry (dollars) = 5
Cost of playing n games (dollars) = 2n
Total cost (dollars) = 2n + 5
Total cost can be at most $25
Therefore, 2n+5≤25
:
A
Cost of entry (dollars) = 5
Cost of playing n games (dollars) = 2n
Total cost (dollars) = 2n + 5
Total cost can be at most $25
Therefore, 2n+5≤25