Inequalities Modulus And Logarithms(11th And 12th > Mathematics ) Questions and answers for exam Preparation

11th And 12th > Mathematics

INEQUALITIES MODULUS AND LOGARITHMS MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1.

The value of $\sqrt{(l o g_{0.5}^{2} 4)}$ is

-2
$\sqrt{(- 4)}$
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 2.

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

$3 l o g_{2} 7$
$1 - 3 l o g_{3} 7$
$1 - 3 l o g_{7} 2$
$N o n e o f t h e s e$

Discuss Question

Answer: Option C. ->

1 - 3 l o g_{7} 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 3.

The value of $l o g_{3} 4 l o g_{4} 5 l o g_{5} 6 l o g_{6} 7 l o g_{7} 8 l o g_{8} 9$ is

Discuss Question

Answer: Option B. -> 2
:
B

l o g_{3} 4. l o g_{4} 5. l o g_{5} 6. l o g_{6} 7. l o g_{7} 8. l o g_{8} 9

= \frac{l o g 4}{l o g 3} . \frac{l o g 5}{l o g 4} . \frac{l o g 6}{l o g 5} . \frac{l o g 7}{l o g 6} . \frac{l o g 8}{l o g 7} . \frac{l o g 9}{l o g 8} = \frac{l o g 9}{l o g 3}

= l o g_{3} 9 = l o g_{3} 3^{2} = 2

Question 4.

The value of $81^{(\frac{1}{l o g_{5} 3})} + 27^{l o g_{9} 36} + 3^{\frac{4}{l o g_{7} 9}}$ is equal to

Discuss Question

Answer: Option D. -> 890
:
D

81^{(\frac{1}{l o g_{5} 3})} + 27^{l o g_{9} 36} + 3^{\frac{4}{l o g_{7} 9}}

= 3^{l o g_{3} 5^{4}} + 3^{l o g_{3} 36^{\frac{3}{2}}} + 3^{l o g_{3} 7^{\frac{4}{2}}}

= 5^{4} + 36^{\frac{3}{2}} + 7^{2} = 890

Question 5.

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

Question 6.

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 $\frac{π}{4}$ is

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 7.

The set of real values of x satisfying $l o g_{\frac{1}{2}} (x^{2} - 6 x + 12) \geq - 2$ is

$(- \infty, 2]$
$[2, 4]$
$[4, + \infty]$
$N o n e o f t h e s e$

Discuss Question

Answer: Option B. ->

[2, 4]

:
B

l o g_{\frac{1}{2}} (x^{2} - 6 x + 12) \geq - 2

...(i)
For log to be defined,

x^{2} - 6 x + 12 > 0

\Rightarrow (x - 3)^{2} + 3 > 0

, which is true

\forall x ϵ R .

From (i),

x^{2} - 6 x + 12 \leq (\frac{1}{2})^{- 2}

\Rightarrow x^{2} - 6 x + 12 \leq 4 \Rightarrow x^{2} - 6 x + 8 \leq 0 \Rightarrow (x - 2) (x - 4) \leq 0 \Rightarrow 2 \leq x \leq 4; ∴ x ϵ [2, 4] .

Question 8.

The set of real values of x for which $2^{{l o g}_{\sqrt{2}} (x - 1)} > x + 5$ is

$(- \infty, - 1) \cup (4, + \infty)$
$(4, + \infty)$
$(- 1, 4)$
$N o n e o f t h e s e$

Discuss Question

Answer: Option B. ->

(4, + \infty)

:
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 9.

If $l o g_{0.04} (x - 1) \geq l o g_{0.2} (x - 1)$ then x belongs to the interval

$(1, 2]$
$(- \infty, 2)$
$[2, + \infty]$
$N o n e o f t h e s e$

Discuss Question

Answer: Option C. ->

[2, + \infty]

:
C

l o g_{0.04} (x - 1) \geq l o g_{0.2} (x - 1)

....(i)
For log to be defined

x - 1 > 0 \Rightarrow x > 1

F r o m (i), l o g_{(0.2)^{2}} (x - 1) \geq l o g_{0.2} (x - 1)

\Rightarrow \frac{1}{2} l o g_{0.2} (x - 1) \geq l o g_{0.2} (x - 1) \Rightarrow \sqrt{x - 1} \leq (x - 1) \Rightarrow \sqrt{x - 1} (1 - \sqrt{x - 1}) \leq 0 \Rightarrow 1 - \sqrt{x - 1} \leq 0 \Rightarrow \sqrt{x - 1} \geq 1 \Rightarrow x \geq 2, ∴ x ϵ [2, \infty) .

Question 10.

Which of the following hold good?

$a^{4} + b^{4} + c^{4} > a b c (a + b + c)$
$a^{5} + b^{5} + c^{5} + d^{5} > a b c d (a + b + c + d)$
$a^{5} + b^{5} + c^{5} > a b c (a b + b c + c a)$
$\frac{a^{8} + b^{8} + c^{8}}{a^{3} b^{3} c^{3}} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$
$\frac{b^{2} + c^{2}}{b + c} + \frac{c^{2} + a^{2}}{c + a} + \frac{a^{2} + b^{2}}{a + b} > a + b + c$