Logarithms(12th Grade > Mathematics ) Questions and answers for exam Preparation

12th Grade > Mathematics

LOGARITHMS MCQs

Inequalities Modulus And Logarithms (11th And 12th Grade)

Total Questions : 50 | Page 2 of 5 pages

y \leq 3 x + 1

x - y > 1

Which of the following ordered pairs (x,y) satisfies the system of inequalities above?

(-2,-1)
(-1, 3)
(1,5)
(2,-1)

Discuss Question

Answer: Option D. -> (2,-1)
:
D
Substitute each of the points in the inequalities to check if the inequality is satisfied.
(-2,-1) does not satisfy either inequality
(-1, 3) does not satisfy either inequality
(1,5) does not satisfy either inequality
(2,-1) satisfies both equalities
Alternatively, plotting the graph of the inequalities,

Y≤3x+1x−y>1Which Of The Following Ordered Pairs (x,y)...

It can be seen that, only (2,-1) belongs to the solution set. Hence D is the correct choice.

Question 12. If

21 b x - 28 > 49

, Where b is a positive constant, which of the following best describes all possible values of 4 - 3bx?

Any value less than -7
Any value greater than -7
Any value less than −113b
Any value greater than 113b

Discuss Question

Answer: Option A. -> Any value less than -7
:
A
Let examine the given inequality

21 b x - 28 > 49

Dividing both side by 7

\frac{21 b x}{7} - \frac{28}{7} > \frac{49}{7}

3 b x - 4 > 7

Multiply -1 on both side [remember to flip the sign]

- 1 (3 b x - 4) < 7 (- 1)

4 - 3 b x < - 7

The correct answer is option A.

Question 13. If

3 < p < 7

and

- 9 < q < - 4

, then what is the range of p - q?

(-16, -7)
(16, 7)
(7, 16)
(-7, -16)

Discuss Question

Answer: Option C. -> (7, 16)
:
C

3 < p < 7

and

- 9 < q < - 4

Multiply the second inequality with - 1. This causes the sign to flip.
So,

3 < p < 7

and

4 < - q < 9

Adding these two inequalities, we get

7 < p - q < 16

Question 14. Which of the following numbers is NOT a solution of the inequality

3 x - 5 \geq 4 x - 3

-1
-2
-3
-5

Discuss Question

Answer: Option A. -> -1
:
A

3 x - 5 \geq 4 x - 3

\Rightarrow - 5 + 3 \geq 4 x - 3 x

\Rightarrow - 2 \geq x

\Rightarrow x \leq - 2

Hence, -1 is not a solution

Question 15.

l o g_{7} l o g_{7} \sqrt{7 (\sqrt{7 \sqrt{7}})} =

3 log2 7
1−3 log3 7
1−3 log7 2
None of these

Discuss Question

Answer: Option C. -> 1−3 log7 2
:
C

l o g_{7} l o g_{7} \sqrt{7 \sqrt{7 \sqrt{7}}} = l o g_{7} l o g_{7} 7^{\frac{7}{8}} = l o g_{7} (\frac{7}{8})

= l o g_{7} 7 - l o g_{7} 8 = 1 - l o g_{7} 2^{3} = 1 - 3 l o g_{7} 2

Question 16. If

l o g_{\frac{1}{\sqrt{2}}} s i n x > 0, x ϵ [0, 4 π]

then the number of values of x which are integral multiples of

\frac{π}{4}

4
12
3
None of these

Discuss Question

Answer: Option A. -> 4
:
A

0 < \frac{1}{\sqrt{2}} < 1

If Log1√2sin x>0,xϵ[0,4π] Then The Number Of Values ...

l o g_{\frac{1}{\sqrt{2}}} s i n x > 0, x ϵ [0, 4 π] \Rightarrow 0 < s i n x < 1

∴

Integral multiple of

\frac{π}{4}

will be

\frac{π}{4}, \frac{3 π}{4}, \frac{9 π}{4}, \frac{11 π}{4}

Number of required values = 4.

Question 17. The set of real values of x for which

2^{{l o g}_{\sqrt{2}} (x - 1)} > x + 5

(−∞,−1)∪(4,+∞)
(4,+∞)
(−1,4)
None of these

Discuss Question

Answer: Option B. -> (4,+∞)
:
B

2^{{l o g}_{\sqrt{2}} (x - 1)} > x + 5 \Rightarrow 2^{l o g_{2} (x - 1)^{2}} > x + 5 \Rightarrow (x - 1)^{2} > x + 5 \Rightarrow x^{2} - 3 x - 4 > 0 \Rightarrow (x - 4) (x + 1) > 0 \Rightarrow x > 4 o r x < - 1

But for given log to be defined, x - 1 > 0

i . e ., x > 1 ∴ x > 4 \Rightarrow x ϵ (4, \infty) .

Question 18. If

x ϵ R

, the solution set of the equation

4^{- x + 0.5} - {7.2}^{- x} - 4 < 0

is equal to

(2,72)
(−2,∞)
(2,∞)
(−∞,∞)
b2+c2b+c+c2+a2c+a+a2+b2a+b>a+b+c

Discuss Question

Answer: Option B. -> (−2,∞)
:
B
Put

2^{- x} = t

and

4^{0.5} = 4^{\frac{1}{2}} = 2

∴ 2 t^{2} - 7 t - 4 < 0 \Rightarrow (t - 4) (2 t + 1) < 0

- \frac{1}{2} < t < 4 \Rightarrow 0 < 2^{- x} < 2^{2}

{(\frac{1}{2})}^{\infty} < {(\frac{1}{2})}^{x} < {(\frac{1}{2})}^{- 2}

Since

\frac{1}{2} < 1 ∴ - 2 < x < \infty

∴ x ϵ (- 2, \infty) \Rightarrow (b)

Question 19. The value of

\sqrt{(l o g_{0.5}^{2} 4)}

-2
√(−4)
2
None of these

Discuss Question

Answer: Option C. -> 2
:
C

\sqrt{(l o g_{0.5}^{2} 4)} = \sqrt{{l o g_{0.5} (0.5)^{- 2}}^{2}} = \sqrt{(- 2)^{2}} = 2

Question 20.

7 l o g (\frac{16}{15}) + 5 l o g (\frac{25}{24}) + 3 l o g (\frac{81}{80})

is equal to -

0
1
log 2
log 3

Discuss Question

Answer: Option C. -> log 2
:
C

7 l o g (\frac{16}{15}) + 5 l o g (\frac{25}{24}) + 3 l o g (\frac{81}{80})

∵ a l o g x = l o g x^{a}

\Rightarrow l o g (\frac{16}{15})^{7} + l o g (\frac{25}{24})^{5} + l o g (\frac{81}{80})^{3}

∵ l o g x + l o g y = l o g (x \times y)

\Rightarrow l o g (\frac{16^{7}}{15^{7}} . \frac{25^{5}}{24^{5}} . \frac{81^{3}}{80^{3}})

= l o g 2

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