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
SOLUTION
Solution : D
ab=log4 5. log5 6=log4 6=12log2 6
ab=12(1+log2 3)⇒2ab−1=log2 3
∴ log3 2=12ab−1
Question 2
If logk x.log5 k=logx 5, k≠1, k>0, then x is equal to
SOLUTION
Solution : B and C
logk x.log5 k=logx 5⇒log5 x=logx 5
⇒logx 5=1logx 5⇒(logx 5)2=1⇒logx 5=±1
⇒x±1=5⇒x=5,15.
Question 3
log5 a.logax=2, then x is equal to
SOLUTION
Solution : C
log5 a.loga x=2⇒log5x=2⇒x=52=25
Question 4
If A=log2 log2 log4 256+2log√22, then A is equal to
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
Question 5
If log10 x=y then log1000x2 is equal to
SOLUTION
Solution : D
log1000 x2
=log103x2
=2 log103x
=23log10x
=23y
Question 6
The set of real values of x for which log0.2x+2x≤1 is
SOLUTION
Solution : A
log0.2x+2x≤1 ....(i)
For log to be defined, x+2x>0⇒x>0 or
x < -2
Now from(i), log0.2x+2x≤log0.20.2
⇒x+2x≥0.2 .....(ii)
Case (i) x > 0
From (ii),x+2≥0.2x
⇒0.8x≥−2
⇒x≥−52.
Case (ii) x < -2
From(ii), x+2≤0.2x⇒0.8x≤−2⇒x≤−52
⇒xϵ(−∞,−52]∪(0,∞).
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Question 7
If A, B, C are the angles of a triangle such that C is an obtuse angle, then
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 B1−tan Atan B=+ive
Above is possible only when tan Atan B<1
Question 8
If y=3x−1+3−x−1 (x real), then the least value of y is
SOLUTION
Solution : 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
Question 9
The solution set of the inequality ||x|−1<1−x,2∀xϵR is equal to
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,|−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)
Question 10
Let S1,S2…… be squares such that for each n≥1, 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?
SOLUTION
Solution : B, C, and D
(b), (c), (d)
If a be the side of a square, then d=a√2
By given condition an=√2an+1
or an+1=an√2=an−1(√2)2=an−2(√2)3……=a1(√2)n
Replacing n by n-1, we get
an=a1(√2)n−1=102n−1/2
Area of Sn<1⇒a2n<1
⇒1002n−1<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
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=λ⇒a=b=λ2
∴E2=λ2+4λ+4λ2=(λ+2λ)2
∴E=λ+2λ=1+2λ⇒(d)
Question 12
If x=log53+log75+log97, then x ≥
SOLUTION
Solution : C
Apply A.M.≥G.M.
∴x≥3.(log53.log75.log97)13
=3(log93)13=3(log323)13=3(12)13
Question 13
If a, b, c are real numbers such that a + 2b + c = 4 then max. value of ab + bc + ca is
SOLUTION
Solution : 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)
Question 14
The least value of 2log100a−loga(0.0001), a>1 is
SOLUTION
Solution : C
E=2 log100a−loga(100)−2
=[2 log100a+2 loga100]
≥2.2[log100a.loga.100]12
=4[logaa]12=4
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
SOLUTION
Solution : 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,∞)