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1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

If x satisfies the inequality $${\log _{25}}{x^2} + {({\log _5}x)^2} < 2$$, then x belongs to

A
$$\left( {{1 \over 5},5} \right)$$
B
$$\left( {{1 \over {25}},5} \right)$$
C
$$\left( {{1 \over 5},25} \right)$$
D
$$\left( {{1 \over {25}},25} \right)$$

$${\log _{25}}{x^2} + {({\log _5}x)^2} < 2$$ অসমীকরণটিকে সিদ্ধ করে এমন শর্তে x আছে ($$\in$$)

A
$$\left( {{1 \over 5},5} \right)$$
B
$$\left( {{1 \over {25}},5} \right)$$
C
$$\left( {{1 \over 5},25} \right)$$
D
$$\left( {{1 \over {25}},25} \right)$$
2

WB JEE 2020

MCQ (Single Correct Answer)
English
Bengali
If 2 log(x + 1) $$ - $$ log(x2 $$ - $$ 1) = log 2, then x =
A
only 3
B
$$ - $$1 and 3
C
only $$ - $$1
D
1 and 3

Explanation

We have,

2 log (x + 1) $$ - $$ log(x2 $$ - $$ 1) = log 2

$$ \Rightarrow \log {{{{(x + 1)}^2}} \over {{x^2} - 1}} = \log \,2$$

$$ \Rightarrow {(x + 1)^2} = 2({x^2} - 1)$$

$$ \Rightarrow {x^2} + 2x + 1 = 2{x^2} - 2$$

$$ \Rightarrow {x^2} - 2x - 3 = 0$$

$$ \Rightarrow (x - 3)(x + 1) = 0$$

$$ \Rightarrow x = 3,\,x \ne - 1$$

যদি $$2\log (x + 1) - \log ({x^2} - 1) = \log 2$$, তবে x =

A
শুধুমাত্র 3
B
$$-$$1 ও 3
C
শুধুমাত্র $$-$$1
D
1 ও 3

Explanation

$$2\log (x + 1) - \log ({x^2} - 1) = \log 2$$

বা, $$\log {{{{(x + 1)}^2}} \over {{x^2} - 1}} = \log 2$$

বা, $${{x + 1} \over {x - 1}} = 2$$

বা, $$2x - 2 = x + 1 \Rightarrow x = 3$$

3

WB JEE 2019

MCQ (Single Correct Answer)
English
Bengali
If $$\log _2^6 + {1 \over {2x}} = {\log _2}\left( {{2^{{1 \over x}}} + 8} \right)$$, then the value of x are
A
$${1 \over 4},{1 \over 3}$$
B
$${1 \over 4},{1 \over 2}$$
C
$$-$$$${1 \over 4},{1 \over 2}$$
D
$${1 \over 3}, - {1 \over 2}$$

Explanation

We have,

$$\log _2^6 + {1 \over {2x}} = {\log _2}\left( {{2^{{1 \over x}}} + 8} \right)$$

$$ \Rightarrow 1 + \log _2^3 + {1 \over {2x}} = {\log _2}\left( {{2^{{1 \over x}}} + 8} \right)$$

$$ \Rightarrow {{{2^{{1 \over x}}} + 8} \over 3} = {2^{1 + {1 \over {2x}}}}$$

$$ \Rightarrow {2^{{1 \over x}}} + 8 = {3.2.2^{{1 \over {2x}}}}$$

$$ \Rightarrow {2^{{1 \over x}}} + 8 = {6.2^{{1 \over {2x}}}}$$

Let $$y = {2^{{1 \over {2x}}}}$$

$$ \Rightarrow {y^2} + 8 = 6y$$

$$ \Rightarrow {y^2} - 6y + 8 = 0$$

$$ \Rightarrow (y - 4)(y - 2) = 0$$

$$ \Rightarrow y = 4,2$$

$$ \Rightarrow {2^{{1 \over {2x}}}} = 4$$ and $${2^{{1 \over {2x}}}} = 2$$

$$ \Rightarrow {1 \over {2x}} = 2$$ and $${1 \over {2x}} = 1$$

$$ \Rightarrow x = {1 \over 4},{1 \over 2}$$

যদি $${\log _2}6 + {1 \over {2x}} = \log \left( {{2^{{1 \over x}}} + 8} \right)$$ হয়, তবে x-এর মানগুলি হল -

A
$${1 \over 4},{1 \over 3}$$
B
$${1 \over 4},{1 \over 2}$$
C
$$ - {1 \over 4},{1 \over 2}$$
D
$${1 \over 3}, - {1 \over 2}$$

Explanation

$${\log _2}6 + {1 \over {2x}} = \log \left( {{2^{{1 \over x}}} + 8} \right)$$

বা, $${\log _2}\left( {{{6.2}^{{1 \over {2x}}}}} \right) = {\log _2}\left( {{2^{{1 \over x}}} + 8} \right)$$

বা, $$6.\sqrt {{2^{{1 \over x}}}} = {2^{{1 \over x}}} + 8$$

বা, $${2^{{1 \over x}}} - 6.\sqrt {{2^{{1 \over x}}}} + 8 = 0$$

বা, $${2^{{1 \over x}}} - 4.\sqrt {{2^{{1 \over x}}}} - 2\sqrt {{2^{{1 \over x}}}} + 8 = 0$$

বা, $$\sqrt {{2^{{1 \over x}}}} \left( {\sqrt {{2^{{1 \over x}}}} - 4} \right) - 2\left( {\sqrt {{2^{{1 \over x}}}} - 4} \right) = 0$$

বা, $$\left( {\sqrt {{2^{{1 \over x}}}} - 4} \right)\left( {\sqrt {{2^{{1 \over x}}}} - 2} \right) = 0$$

$$\therefore$$ $${2^{{1 \over x}}} = 16$$ অথবা $${2^{{1 \over x}}} = 4$$ $$\therefore$$ $${1 \over x} = 4$$ অথবা $${1 \over x} = 2$$

$$ \Rightarrow x = {1 \over 4}$$ অথবা $${1 \over 2}$$

4

WB JEE 2018

MCQ (Single Correct Answer)
English
Bengali
If $$x + {\log _{10}}(1 + {2^x}) = x{\log _{10}}5 + {\log _{10}}6$$, then the value of x is
A
$${1 \over 2}$$
B
$${1 \over 3}$$
C
1
D
2

Explanation

We have,

$$x + {\log _{10}}(1 + {2^x}) = x{\log _{10}}5 + {\log _{10}}6$$

$$ \Rightarrow {\log _{10}}(1 + {2^x}) = x{\log _{10}}5 + {\log _{10}}6 - x$$

$$ = {\log _{10}}{5^x} + {\log _{10}} + - x{\log _{10}}10$$

$$ = {\log _{10}}({5^x}.6) - {\log _{10}}{10^x}$$

$$ \Rightarrow {\log _{10}}(1 + {2^x}) = {\log _{10}}\left( {{{{5^x}.6} \over {{{10}^x}}}} \right)$$

$$ \Rightarrow 1 + {2^x} = {{{5^x}.6} \over {{{10}^x}}} = {6 \over {{2^x}}}$$

$$ \Rightarrow {2^x}(1 + {2^x}) = 6$$

$$ \Rightarrow t(1 + t) = 6$$ (let 2x = t)

$$ \Rightarrow {t^2} + t - 6 = 0$$

$$ \Rightarrow (t + 3)(t - 2) = 0$$

$$ \Rightarrow t + 3 = 0$$

or t $$-$$ 2 = 0

$$ \Rightarrow t = 2$$ [$$ \because $$ neglect t = $$-$$3]

$$ \Rightarrow {2^x} = 2 \Rightarrow x = 1$$

যদি $$x + {\log _{10}}(1 + {2^x}) = x{\log _{10}}5 + {\log _{10}}6$$ হয়, তবে x-এর মান হবে -

A
$${1 \over 2}$$
B
$${1 \over 3}$$
C
1
D
2

Explanation

$$x + {\log _{10}}(1 + {2^x}) = x\log _{10}^5 + \log _{10}^6$$

বা, $$x\left( {1 - \log _{10}^5} \right) = \log _{10}^6 - {\log _{10}}(1 + {2^x})$$

বা, $$x\,.\,\log _{10}^2 = {\log _{10}}{6 \over {1 + {2^x}}}$$

বা, $${2^x} = {6 \over {1 + {2^x}}}$$

বা, $${({2^x})^2} + {2^x} - 6 = 0$$

বা, $$({2^x} + 3)({2^x} - 2) = 0$$

$$\therefore$$ $${2^x} = 2 \Rightarrow x = 1$$

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