12th Grade > Mathematics
LOGARITHMS MCQs
Inequalities Modulus And Logarithms (11th And 12th Grade)
Total Questions : 50
| Page 3 of 5 pages
Answer: Option B. -> [2,4]
:
B
log12(x2−6x+12)≥−2 ...(i)
For log to be defined, x2−6x+12>0
⇒(x−3)2+3>0, which is true ∀xϵR.
From (i), x2−6x+12≤(12)−2
⇒x2−6x+12≤4⇒x2−6x+8≤0⇒(x−2)(x−4)≤0⇒2≤x≤4;∴xϵ[2,4].
:
B
log12(x2−6x+12)≥−2 ...(i)
For log to be defined, x2−6x+12>0
⇒(x−3)2+3>0, which is true ∀xϵR.
From (i), x2−6x+12≤(12)−2
⇒x2−6x+12≤4⇒x2−6x+8≤0⇒(x−2)(x−4)≤0⇒2≤x≤4;∴xϵ[2,4].
Answer: Option C. -> [2,+∞]
:
C
log0.04(x−1)≥log0.2(x−1) ....(i)
For log to be defined x−1>0⇒x>1
From(i),log(0.2)2(x−1)≥log0.2(x−1)
⇒12log0.2(x−1)≥log0.2(x−1)⇒√x−1≤(x−1)⇒√x−1(1−√x−1)≤0⇒1−√x−1≤0⇒√x−1≥1⇒x≥2,∴xϵ[2,∞).
:
C
log0.04(x−1)≥log0.2(x−1) ....(i)
For log to be defined x−1>0⇒x>1
From(i),log(0.2)2(x−1)≥log0.2(x−1)
⇒12log0.2(x−1)≥log0.2(x−1)⇒√x−1≤(x−1)⇒√x−1(1−√x−1)≤0⇒1−√x−1≤0⇒√x−1≥1⇒x≥2,∴xϵ[2,∞).
Answer: Option D. -> (−∞,0)
:
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,|−x−1|<1−x
or |x+1|<(1−x)……(1)
If x+1≥0i.ex≥−1 and x<0
i.e. −1≤x<0 i.e. xϵ[−1,0) then (1) reduces to
x+1=1−x or 2x<0 or x<0
If x+1<0 i.e. x<−1 then (1) reduce to
−(x+1)<(1−x) or −2<0 is true for all x<−1……(2)
Hence from (1) and (2) xϵ(−∞,0)
:
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,|−x−1|<1−x
or |x+1|<(1−x)……(1)
If x+1≥0i.ex≥−1 and x<0
i.e. −1≤x<0 i.e. xϵ[−1,0) then (1) reduces to
x+1=1−x or 2x<0 or x<0
If x+1<0 i.e. x<−1 then (1) reduce to
−(x+1)<(1−x) or −2<0 is true for all x<−1……(2)
Hence from (1) and (2) xϵ(−∞,0)
Answer: Option C. -> 25
:
C
log5a.logax=2⇒log5x=2⇒x=52=25
:
C
log5a.logax=2⇒log5x=2⇒x=52=25
Answer: Option C. -> 5
:
C
A=log2log2log4256+2log2122
=log2log2log444+2×1(12)log22
=log2log24+4=log2log222+4
=log22+4=1+4=5
:
C
A=log2log2log4256+2log2122
=log2log2log444+2×1(12)log22
=log2log24+4=log2log222+4
=log22+4=1+4=5
Answer: Option D. -> 2y3
:
D
log1000x2
=log103x2
=2log103x
=23log10x
=23y
:
D
log1000x2
=log103x2
=2log103x
=23log10x
=23y
Answer: Option C. -> 4
:
C
E=2log100a−loga(100)−2
=[2log100a+2loga100]
≥2.2[log100a.loga.100]12
=4[logaa]12=4
:
C
E=2log100a−loga(100)−2
=[2log100a+2loga100]
≥2.2[log100a.loga.100]12
=4[logaa]12=4
Answer: Option A. -> tan A.tan B
:
A
Since C is obtuse therefore A and B both are acute and A + B is also acute.
Hence tanA,tanB and tan(A+B) are all + ive
tan(A+B)=tanA+tanB1−tanAtanB=+ive
Above is possible only when tanAtanB<1
:
A
Since C is obtuse therefore A and B both are acute and A + B is also acute.
Hence tanA,tanB and tan(A+B) are all + ive
tan(A+B)=tanA+tanB1−tanAtanB=+ive
Above is possible only when tanAtanB<1
Answer: Option B. -> 4
:
B
Given (a+b)+(b+c)=4
But [(a+b)+(b+c)2]2≥[(a+b)(b+c)],A.M≥G.M.
or (42)2≥ab+bc+ca+b2
4−b2≥∑ab or ∑ab≤4−b2
∴ Max. value of ∑ab=4⇒(b)
:
B
Given (a+b)+(b+c)=4
But [(a+b)+(b+c)2]2≥[(a+b)(b+c)],A.M≥G.M.
or (42)2≥ab+bc+ca+b2
4−b2≥∑ab or ∑ab≤4−b2
∴ Max. value of ∑ab=4⇒(b)
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