Free Objective Test 01 Practice Test - 11th and 12th

Question 1

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

A. -2

\sqrt{(- 4)}

C. 2

D. None of these

SOLUTION

Solution : 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

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

A. 1

B. 2

C. 3

D. 4

SOLUTION

Solution : 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 3

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

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

SOLUTION

Solution : 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 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

A. 49

B. 625

C. 216

D. 890

SOLUTION

Solution : 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 -

A. 0

B. 1

C. log 2

D. log 3

SOLUTION

Solution : 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

A. 4

B. 12

C. 3

D. None of these

SOLUTION

Solution : A

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

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

SOLUTION

Solution : 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

SOLUTION

Solution : 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

SOLUTION

Solution : 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

SOLUTION

Solution : A, B, C, and D

(a), (b), (c), (d)
(a) $\frac{a^{4} + b^{4} + c^{4}}{3} > {(\frac{a + b + c}{3})}^{4}$
or $(\frac{a + b + c}{3}) {(\frac{a + b + c}{3})}^{3} > \frac{a + b + c}{3} [(a b c)^{\frac{1}{3}}]^{3}$ ;
$∵ A . M . > G . M .$
or $> (\frac{a + b + c}{3})$
$∴ a^{4} + b^{4} + c^{4} > a b c (a + b + c)$

(b) As above

(c) $(\frac{a^{5} + b^{5} + c^{5}}{3}) > {(\frac{a + b + c}{3})}^{5}$
or $> {(\frac{a + b + c}{3})}^{3} {(\frac{a + b + c}{3})}^{2}$
or $> [(a b c)^{\frac{1}{3}}]^{3} (\frac{a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a}{9})$
But we know that $a^{2} + b^{2} + c^{2} > a b + b c + c a$ by result of two by two rule.
$∴ \frac{a^{5} + b^{5} + c^{5}}{3} > a b c \frac{3 (a b + b c + c a)}{9}$ etc.

(d) $a^{8} + b^{8} + c^{8} > a^{2} b^{2} c^{2} (b c + c a + a b)$
Now $\frac{a^{8} + b^{8} + c^{8}}{3} > {(\frac{a + b + c}{3})}^{8}$
or $> {(\frac{a + b + c}{3})}^{6} {(\frac{a + b + c}{3})}^{2} > [(a b c)^{\frac{1}{3}}]^{6} [\frac{a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a}{9}]$
$∵ A . M . > G . M$
But by two by two rule
$a^{2} + b^{2} + c^{2} > a b + b c + c a$
$∴ \frac{a^{8} + b^{8} + c^{8}}{3} > a^{2} b^{2} c^{2} \frac{(3 a b + 3 b c + 3 c a)}{9}$
$∴ a^{8} + b^{8} + c^{8} > a^{2} b^{2} c^{2} (a b + b c + c a)$
or $\frac{a^{8} b^{8} c^{8}}{a^{3} b^{3} c^{3}} > \frac{a b + b c + c a}{a b c}$ or $> \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$

(e) $\frac{b^{2} + c^{2}}{2} > {(\frac{b + c}{2})}^{2} ∴ \frac{b^{2} + c^{2}}{b + c} > \frac{b + c}{2}$
Write similar inequalities and add.

Question 11

Which of the following hold good? If n is a +ve integer, then

n^{n} > 1.3.5 \dots \dots (2 n - 1)

2.4.6 \dots \dots 2 n < (n + 1)^{n}

(n!)^{3} < n^{n} {(\frac{n + 1}{2})}^{2 n}

[1^{r} + 2^{r} + 3^{r} + \dots \dots + n^{r}]^{n} > n^{n} . (n!)^{r}

SOLUTION

Solution : A, B, C, and D

(a), (b), (c), (d)
(a) Apply A.M. of 1,3,5 .... (2n-1), i.e. of n numbers is > G.M. and remember that 1+3+5+... upto n terms $= \frac{n}{2} [2 a + (n - 1) d] = n^{2}$ as a=1 and d=2

(b) Proceed as in part (a)

(c) Apply A.M. of $1^{3}, 2^{3}, \dots \dots n^{3}$ is greater than their G.M. and remember that
$1^{3} + 2^{3} + \dots \dots n^{3} = {\frac{n (n + 1)}{2}}^{2}$

(d) $\frac{1^{r} + 2^{r} + 3^{r} + \dots \dots + n^{r}}{n} > [1^{r} . 2^{r} . 3^{r} . \dots \dots . n^{r}]^{\frac{1}{n}}$
or $(1^{r} + 2^{r} + 3^{r} + \dots \dots + n^{r})^{n} > n^{n} (n!)^{r}$

Question 12

Which of the following hold good?

p x^{q - r} + q x^{r - p} + r x^{p - q} > p + q + r

, unless p = q = r, or x = 1

{[\frac{x^{2} + y^{2} + z^{2}}{x + y + z}]}^{x + y + z} > x^{x} y^{y} z^{z} > {[\frac{x + y + z}{3}]}^{x + y + z}

{[\frac{a^{2} + b^{2}}{a + b}]}^{a + b} > a^{a} b^{b}

a^{a} b^{b} > {(\frac{a + b}{2})}^{a + b} > a^{b} . b^{a}

SOLUTION

Solution : A, B, C, and D

(a), (b), (c), (d)
Now because A.M. > G.M.
$(x^{q - r} + x^{q - r} + \dots \dots p t i m e s)$
         $+ (x^{r - p} + x^{r - p} + \dots \dots q t i m e s)$
                      $+ (x^{p - q} + x^{p - q} + \dots \dots r t i m e s) - ------------------------------ -$
                               p+q+r
$> {(x^{q - r} . x^{q - r} \dots \dots p f a c t o r s) . (x^{r - p} . x^{r - p} \dots \dots q f a c t o r s) . (x^{p - q} . x^{p - q} \dots \dots r f a c t o r s)}^{\frac{1}{(p + q + r)}}$
or $\frac{p x^{q - r} + q x^{r - p} + r x^{p - q}}{p + q + r}$
$> {x^{p (q - r) + q (r - p) + r (p - q)}}^{\frac{1}{(p + q + r)}}$
or $\frac{p x^{q - r} + q x^{r - p} + r x^{p - q}}{p + q + r} > [x^{0}]^{\frac{1}{(p + q + r)}}$
or $> 1$
$∴ p x^{q - r} + q x^{r - p} + r x^{p - q} > p + q + r$

(b) In case either p=q=r or x=1, inequality becomes equality
Consider x numbers each equal to x, y numbers each equal to y and z numbers each equal to z.
Apply A.M. > G.M. on these x+y+z numbers
$(x + x + x + \dots \dots x t i m e s) + (y + y + y + \dots \dots y t i m e s) + (z + z + z + \dots \dots z t i m e s) - ---------------------------------------------------------------------------------------------- -$
$∴ x + y + z$
$> [(x, x \dots \dots x f a c t o r s) (y . y \dots \dots f f a c t o r s) (z . z \dots \dots z f a c t o r s)]^{\frac{1}{(x + y + z)}}$
But $x + x + x + \dots \dots x t i m e s = x . x = x^{2}$
and $x . x . x \dots \dots x f a c t o r s = x^{x}$
$∴ \frac{x^{2} + y^{2} + z^{2}}{x + y + z} > (x^{x} y^{y} z^{z})^{\frac{1}{(x + y + z)}}$
or ${(\frac{x^{2} + y^{2} + z^{2}}{x + y + z})}^{x + y + z} > x^{x} y^{y} z^{z}$
Above proves the first part. ......... (A)

2nd part:
For 2nd part consider x numbers each equal to $\frac{1}{x}, y$ numbers each equal to $\frac{1}{y}, z$ numbers each equal to $\frac{1}{z}$ . Apply A.M. > G.M. on these x+y+z numbers.
$(\frac{1}{x} + \frac{1}{x} + \dots \dots x t i m e s) + (\frac{1}{y} + \frac{1}{y} + \dots \dots y t i m e s) + (\frac{1}{z} + \frac{1}{z} + \dots \dots z t i m e s) - ---------------------------------------------------------------------------------------- -$
$∴ x + y + z$
$> {[(\frac{1}{x} . \frac{1}{x} \dots \dots x f a c t o r s) (\frac{1}{y} . \frac{1}{y} \dots \dots y f a c t o r s) (\frac{1}{z} . \frac{1}{z} \dots \dots z f a c t o r s)]}^{\frac{1}{(x + y + z)}}$
or $\frac{\frac{1}{x} . x + \frac{1}{y} . y + \frac{1}{z} . z}{x + y + z} > {[\frac{1}{x^{x}} . \frac{1}{y^{y}} . \frac{1}{z^{z}}]}^{\frac{1}{(x + y + z)}}$
or ${(\frac{3}{x + y + z})}^{x + y + z} > \frac{1}{x^{x} y^{y} z^{z}}$
Taking reciprocal and changing the sign of inequality, we get
${(\frac{x + y + z}{3})}^{x + y + z} < x^{x} y^{y} z^{z}$
or $x^{x} y^{y} z^{z} > {(\frac{x + y + z}{3})}^{x + y + z} \dots \dots (B)$
Both (A) and (B) prove the required result.

(c) Proceed as in part (b)
Consider a quantities each equal to $\frac{1}{a}, b$ quantities each equal to $\frac{1}{b}$ ; then since A.M. > G.M.
$\frac{[\frac{1}{a} + \frac{1}{a} + \frac{1}{a} + \dots \dots a t i m e s] + [\frac{1}{b} + \frac{1}{b} + \dots \dots b t i m e s]}{a + b}$
$> {(\frac{1}{a} . \frac{1}{a} . \frac{1}{a} \dots \dots a f a c t o r s) \times {(\frac{1}{b} . \frac{1}{b} . \frac{1}{b} \dots \dots b f a c t o r s)}^{\frac{1}{(a + b)}}}$
or $\frac{a . \frac{1}{a} + b . \frac{1}{b}}{a + b} > {\frac{1}{a} \frac{1}{b}}^{\frac{1}{(a + b)}}$
or ${\frac{2}{a + b}} > {\frac{1}{a^{a} . b^{b}}}^{\frac{1}{(a + b)}}$
or ${\frac{2}{a + b}}^{a + b} > \frac{1}{a^{a} b^{b}}$
Taking reciprocal and hence changing the sign of inequality, we get
${\frac{a + b}{2}}^{a + b} < a^{a} . b^{b}$
or $a^{a} . b^{b} > {\frac{a + b}{2}}^{a + b} \dots \dots (1)$
Note: a and b can be integers as well as fractions

(d) Consider b quantities each equal to a and a quantities each equal to b. Then since A.M. > G.M
$\frac{(a + a + a + \dots \dots b t i m e s) + (b + b + b + \dots \dots a t i m e s)}{a + b}$
$> [(a . a . a \dots \dots b f a c t o r s) (b . b . b \dots \dots a f a c t o r s)^{\frac{1}{(a + b)}}]$
or $\frac{a b + a b}{a + b} > (a^{b} . b^{a})^{\frac{1}{(a + b)}}$
or $\frac{2 a b}{a + b} > (a^{b} . b^{a})^{\frac{1}{(a + b)}}$
But $\frac{a + b}{2}$ being A.M. of a and b is greater than $\frac{2 a b}{a + b}$ their H.M.
$∴ \frac{a + b}{2} > \frac{2 a b}{a + b} > [a^{b} . b^{a}]^{\frac{1}{(a + b)}}$
or ${[\frac{a + b}{2}]}^{a + b} > a^{b} . b^{a} \dots \dots (2)$
$∴ a^{a} . b^{b} > {\frac{a + b}{2}}^{a + b} > a^{b} . b^{a}$
by (1) of part (c) and (2).

Question 13

If A, B, C be the angles of a triangle, then which of the following hold good?

c o t A . c o t B . c o t C \leq \frac{1}{3 \sqrt{3}}

c o t \frac{A}{2} c o t \frac{B}{2} c o t \frac{C}{2} \geq 3 \sqrt{(3)}

c o s A + c o s B + c o s C < \frac{3}{2}

s i n (\frac{A}{2}) s i n (\frac{B}{2}) s i n (\frac{C}{2}) \leq \frac{1}{8}

SOLUTION

Solution : A, B, C, and D

All the four hold good
(a) We know that when $A + B + C = π$ then
$t a n (A + B + C) = t a n π = 0$
or $\frac{S_{1} - S_{3}}{1 - S_{2}} = 0 ∴ S_{1} = S_{3}$
or $t a n A + t a n B + t a n C = t a n A t a n B t a n C \dots \dots (1)$
Now $t a n A + t a n B + t a n C \geq 3 (t a n A t a n B t a n C)^{\frac{1}{3}}$
or $(t a n A t a n B t a n C)^{3} \geq 27 t a n A t a n B t a n C$
or $t a n A t a n B t a n C \geq 3 \sqrt{3}$
or $c o t A c o t B c o t C \leq \frac{1}{3 \sqrt{3}} \dots \dots (2)$

(b) Again $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{π}{2}$
$∴ t a n (\frac{A}{2} + \frac{B}{2} + \frac{C}{2}) = \infty$
$= \frac{S_{1} - S_{3}}{1 - S_{2}} = \infty$
$∴ 1 - S_{2} = 0$
or $\sum t a n (\frac{A}{2}) t a n (\frac{B}{2}) = 1$
or on dividing by $t a n (\frac{A}{2}) t a n (\frac{B}{2}) t a n (\frac{C}{2})$ ,
we get $c o t (\frac{A}{2}) + c o t (\frac{B}{2}) + c o t (\frac{C}{2})$
$= c o t (\frac{A}{2}) + c o t (\frac{B}{2}) + c o t (\frac{C}{2}) \dots \dots (1)$
Now proceed as in part (a) for result (1)

(c) We have $c o s A + c o s B + c o s C$
$= 2 c o s \frac{A + B}{2} c o s \frac{A - B}{2} + c o s C$
$= 2 s i n \frac{C}{2} c o s \frac{A - B}{2} + c o s C \leq 2 s i n \frac{C}{2} + c o s C$
$[∵ 0 \leq c o s \frac{A + B}{2} \leq 1]$
$= 2 s i n \frac{C}{2} + 1 - 2 s i n^{2} \frac{C}{2}$
$= - 2 [{(s i n \frac{C}{2} - \frac{1}{2})}^{2} - \frac{1}{4}] + 1$
$= \frac{3}{2} - 2 {(s i n \frac{C}{2} - \frac{1}{2})}^{2} \leq \frac{3}{2}$

(d) We know from trigonometry that
$c o s A + c o s B + c o s C = 1 + 4 s i n \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2} \leq \frac{3}{2}$ by (c)
$∴ 4 s i n \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2} \leq \frac{3}{2} - 1 = \frac{1}{2}$
$∴ s i n \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2} \leq \frac{1}{8}$

Question 14

If $x ϵ R$ , the solution set of the equation
$4^{- x + 0.5} - {7.2}^{- x} - 4 < 0$ is equal to

(2, \frac{7}{2})

(- 2, \infty)

(2, \infty)

(- \infty, \infty)

SOLUTION

Solution : B

Put $2^{- x} = t$ and $4^{0.5} = 4^{\frac{1}{2}} = 2$
$∴ 2 t^{2} - 7 t - 4 < 0 \Rightarrow (t - 4) (2 t + 1) < 0$
$- \frac{1}{2} < t < 4 \Rightarrow 0 < 2^{- x} < 2^{2}$
or ${(\frac{1}{2})}^{\infty} < {(\frac{1}{2})}^{x} < {(\frac{1}{2})}^{- 2}$
Since $\frac{1}{2} < 1 ∴ - 2 < x < \infty$
$∴ x ϵ (- 2, \infty) \Rightarrow (b)$

Question 15

If $x ϵ R$ and $m = \frac{x^{2}}{(x^{4} - 2 x^{2} + 4)}$ , then m lies in the interval

[0, \frac{1}{4}]

[0, \frac{1}{3}]

[0, \frac{1}{2}]

[0, \frac{1}{5}]

SOLUTION

Solution : C

$m = \frac{x^{2}}{(x^{2} - 1)^{2} + 3} = + i v e i . e . \geq 0$
Again $m = \frac{1}{x^{2} + \frac{4}{x^{2}} - 2} = \frac{1}{{(x - \frac{2}{x})}^{2} + 2} \leq \frac{1}{2}$
$m ϵ [0, \frac{1}{2}]$

Free Objective Test 01 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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