Free Objective Test 02 Practice Test - 11th and 12th

Question 1

If $l o g_{4}$ 5 = a and $l o g_{5}$ 6 = b, then $l o g_{3}$ 2 is equal to

\frac{1}{2 a + 1}

\frac{1}{2 b + 1}

2 a b + 1

\frac{1}{2 a b - 1}

SOLUTION

Solution : D

$a b = l o g_{4} 5. l o g_{5} 6 = l o g_{4} 6 = \frac{1}{2} l o g_{2} 6$
$a b = \frac{1}{2} (1 + l o g_{2} 3) \Rightarrow 2 a b - 1 = l o g_{2} 3$
$∴ l o g_{3} 2 = \frac{1}{2 a b - 1}$

Question 2

If $l o g_{k} x . l o g_{5} k = l o g_{x} 5, k \neq 1, k > 0$ , then x is equal to

A. k

\frac{1}{5}

C. 5

D. None of these

SOLUTION

Solution : B and C

$l o g_{k} x . l o g_{5} k = l o g_{x} 5 \Rightarrow l o g_{5} x = l o g_{x} 5$
$\Rightarrow l o g_{x} 5 = \frac{1}{l o g_{x} 5} \Rightarrow (l o g_{x} 5)^{2} = 1 \Rightarrow l o g_{x} 5 = \pm 1$
$\Rightarrow x^{\pm 1} = 5 \Rightarrow x = 5, \frac{1}{5} .$

Question 3

$l o g_{5} a . l o g_{a} x = 2$ , then x is equal to

A. 125

a^{2}

C. 25

D. None of these

SOLUTION

Solution : C

$l o g_{5} a . l o g_{a} x = 2 \Rightarrow l o g_{5} x = 2 \Rightarrow x = 5^{2} = 25$

Question 4

If $A = l o g_{2} l o g_{2} l o g_{4} 256 + 2 l o g_{\sqrt{2}} 2$ , then A is equal to

A. 2

B. 3

C. 5

D. 7

SOLUTION

Solution : C

$A = l o g_{2} l o g_{2} l o g_{4} 256 + 2 l o g_{2_{\frac{1}{2}}} 2$
$= l o g_{2} l o g_{2} l o g_{4} 4^{4} + 2 \times \frac{1}{(\frac{1}{2})} l o g_{2} 2$
$= l o g_{2} l o g_{2} 4 + 4 = l o g_{2} l o g_{2} 2^{2} + 4$
$= l o g_{2} 2 + 4 = 1 + 4 = 5$

Question 5

If $l o g_{10} x = y t h e n l o g_{1000} x^{2}$ is equal to

y^{2}

2 y

\frac{3 y}{2}

\frac{2 y}{3}

SOLUTION

Solution : D

$l o g_{1000} x^{2}$
$= l o g_{10^{3}} x^{2}$
$= 2 l o g_{10^{3}} x$
$= \frac{2}{3} l o g_{10} x$
$= \frac{2}{3} y$

Question 6

The set of real values of x for which $l o g_{0.2} \frac{x + 2}{x} \leq 1$ is

(- \infty, - \frac{5}{2}] \cup (0, + \infty)

[\frac{5}{2}, + \infty)

(- \infty, - 2) \cup (0, + \infty)

N o n e o f t h e s e

SOLUTION

Solution : A

$l o g_{0.2} \frac{x + 2}{x} \leq 1$ ....(i)
For log to be defined, $\frac{x + 2}{x} > 0 \Rightarrow x > 0$ or
x < -2
Now from(i), $l o g_{0.2} \frac{x + 2}{x} \leq l o g_{0.2} 0.2$
$\Rightarrow \frac{x + 2}{x} \geq 0.2$ .....(ii)
Case (i) x > 0
From (ii), $x + 2 \geq 0.2 x$
$\Rightarrow 0.8 x \geq - 2$
$\Rightarrow x \geq - \frac{5}{2} .$
Case (ii) x < -2
From(ii), $x + 2 \leq 0.2 x \Rightarrow 0.8 x \leq - 2 \Rightarrow x \leq - \frac{5}{2}$
$\Rightarrow x ϵ (- \infty, - \frac{5}{2}] \cup (0, \infty)$ .

Question 7

If A, B, C are the angles of a triangle such that C is an obtuse angle, then

t a n A . t a n B < 1

t a n A . t a n B > 1

t a n A . t a n B = 1

D. None

SOLUTION

Solution : A

Since C is obtuse therefore A and B both are acute and A + B is also acute.
Hence $t a n A, t a n B$ and $t a n (A + B)$ are all + ive
$t a n (A + B) = \frac{t a n A + t a n B}{1 - t a n A t a n B} = + i v e$
Above is possible only when $t a n A t a n B < 1$

Question 8

If $y = 3^{x - 1} + 3^{- x - 1}$ (x real), then the least value of y is

A. 2

B. 6

\frac{2}{3}

D. None of these

SOLUTION

Solution : C

We have
$3^{x - 1} + 3^{- x - 1} = \frac{1}{3} (3^{x} + 3^{- x}) \geq \frac{1}{3} . 2 \sqrt{3^{x} . 3^{- x}}$ ,
$[∵ A . M . \geq G . M .]$
or $3^{x - 1} + 3^{- x - 1} \geq \frac{2}{3}$
Hence the least value of $3^{x - 1} + 3^{- x - 1}$ is $\frac{2}{3}$

Question 9

The solution set of the inequality $| | x | - 1 < 1 - x, 2 \forall x ϵ R$ is equal to

(0, \infty)

(- 1, \infty)

(- 1, 1)

(- \infty, 0)

SOLUTION

Solution : D

$(- \infty, 0)$
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 $| x + 1 | < (1 - x) \dots \dots (1)$
If $x + 1 \geq 0 i . e x \geq - 1$ and $x < 0$
i.e. $- 1 \leq x < 0$ i.e. $x ϵ [- 1, 0)$ then (1) reduces to
$x + 1 = 1 - x$ or $2 x < 0$ 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 $x < - 1 \dots \dots (2)$
Hence from (1) and (2) $x ϵ (- \infty, 0)$

Question 10

Let $S_{1}, S_{2} \dots \dots$ be squares such that for each $n \geq 1$ , the length of a side of $S_{n}$ equals the length of a diagonal of $S_{n + 1}$ . If the length of a side of $S_{1}$ is 10 cm, then for which of the following values of n is the area of $S_{n}$ less than 1 sq. cm?

A. 7

B. 8

C. 9

D. 10

SOLUTION

Solution : B, C, and D

(b), (c), (d)
If a be the side of a square, then $d = a \sqrt{2}$
By given condition $a_{n} = \sqrt{2} a_{n + 1}$
or $a_{n + 1} = \frac{a_{n}}{\sqrt{2}} = \frac{a_{n - 1}}{(\sqrt{2})^{2}} = \frac{a_{n - 2}}{(\sqrt{2})^{3}} \dots \dots = \frac{a_{1}}{(\sqrt{2})^{n}}$
Replacing n by n-1, we get
$a_{n} = \frac{a_{1}}{(\sqrt{2})^{n - 1}} = \frac{10}{2^{n - 1} / 2}$
Area of $S_{n} < 1 \Rightarrow a_{n}^{2} < 1$
$\Rightarrow \frac{100}{2^{n - 1}} < 1 o r 200 < 2^{n} o r 2^{n} > 200$
Now $2^{7} = 128 < 200 < 2^{8} = 256 > 200$
$∴ n = 8, 9, 10 \Rightarrow (b), (c), (d)$

Question 11

If a and b are two positive quantities whose sum is $λ$ , then the minimum value of $\sqrt{(1 + \frac{1}{a}) (1 + \frac{1}{b})}$ is

λ - \frac{1}{λ}

λ - \frac{2}{λ}

1 + \frac{1}{λ}

1 + \frac{2}{λ}

SOLUTION

Solution : D

$E^{2} = 1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{a b} = 1 + \frac{a + b + 1}{a b} = 1 + \frac{λ + 1}{a b}$
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.
$∴ a + b = λ \Rightarrow a = b = \frac{λ}{2}$
$∴ E^{2} = \frac{λ^{2} + 4 λ + 4}{λ^{2}} = {(\frac{λ + 2}{λ})}^{2}$
$∴ E = \frac{λ + 2}{λ} = 1 + \frac{2}{λ} \Rightarrow (d)$

Question 12

If $x = l o g_{5} 3 + l o g_{7} 5 + l o g_{9} 7$ , then x $\geq$

\frac{3}{2}

\frac{1}{2^{\frac{1}{3}}}

\frac{3}{2^{\frac{1}{3}}}

D. None

SOLUTION

Solution : C

Apply $A . M . \geq G . M .$
$∴ x \geq 3. (l o g_{5} 3. l o g_{7} 5. l o g_{9} 7)^{\frac{1}{3}}$
$= 3 (l o g_{9} 3)^{\frac{1}{3}} = 3 (l o g_{3^{2}} 3)^{\frac{1}{3}} = 3 {(\frac{1}{2})}^{\frac{1}{3}}$

Question 13

If a, b, c are real numbers such that a + 2b + c = 4 then max. value of ab + bc + ca is

A. 2

B. 4

C. 6

D. 8

SOLUTION

Solution : B

Given $(a + b) + (b + c) = 4$
But ${[\frac{(a + b) + (b + c)}{2}]}^{2} \geq [(a + b) (b + c)], A . M \geq G . M .$
or ${(\frac{4}{2})}^{2} \geq a b + b c + c a + b^{2}$
$4 - b^{2} \geq \sum a b$ or $\sum a b \leq 4 - b^{2}$
$∴$ Max. value of $\sum a b = 4 \Rightarrow (b)$

Question 14

The least value of $2 l o g_{100} a - l o g_{a} (0.0001), a > 1$ is

A. 2

B. 3

C. 4

D. None

SOLUTION

Solution : C

$E = 2 l o g_{100} a - l o g_{a} (100)^{- 2}$
$= [2 l o g_{100} a + 2 l o g_{a} 100]$
$\geq 2.2 [l o g_{100} a . l o g_{a} .100]^{\frac{1}{2}}$
$= 4 [l o g_{a} a]^{\frac{1}{2}} = 4$

Question 15

If a, b, c be three real numbers of the same sign then the value of $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ lies in the interval

[3, \infty)

(3, \infty)

[2, \infty)

(- \infty, 3)

SOLUTION

Solution : A

Since a, b, c are of same sign, $\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$ are all +ive.
Apply A.M. $\geq$ G.M.
$∴ E \geq 3 {[\frac{a}{b} . \frac{b}{c} . \frac{c}{a}]}^{\frac{1}{3}} = 3$
therefore E lies in the interval $[3, \infty)$

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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