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উচ্চতর গণিত

১. \sqrt {x - 4}  + 2 = \sqrt {x + 12}

সমাধান: \sqrt {x - 4}  + 2 = \sqrt {x + 12}

\Rightarrow {\left( {\sqrt {x - 4}  + 2} \right)^2} = {\left( {\sqrt {x + 12} } \right)^2}  [উভয় পক্ষকে বর্গ করে]

\Rightarrow x - 4 + 4 + 2\sqrt {x - 4}  \times 2 = x + 12

\Rightarrow x + 4\sqrt {x - 4}  = x + 12

\Rightarrow 4\sqrt {x - 4}  = x + 12 - x

\Rightarrow 4\sqrt {x - 4}  = 12

\Rightarrow \sqrt {x - 4}  = {{12} \over 4}

\Rightarrow \sqrt {x - 4}  = 3

\Rightarrow {\left( {\sqrt {x - 4} } \right)^2} = {\left( 3 \right)^2}  [ পুনরায় উভয় পক্ষকে বর্গ করে]

\Rightarrow x - 4 = 9

\Rightarrow x = 9 + 4

\therefore x = 13

\therefore নির্ণেয় সমাধান: x = 13

 

২. \sqrt {11x - 6}  = \sqrt {4x + 5}  - \sqrt {x - 1}

সমাধান: \sqrt {11x - 6}  = \sqrt {4x + 5}  - \sqrt {x - 1}

\Rightarrow {\left( {\sqrt {11x - 6} } \right)^2} = {\left( {\sqrt {4x + 5}  - \sqrt {x - 1} } \right)^2}  [উভয় পক্ষকে বর্গ করে]

\Rightarrow 11x - 6 = 4x + 5 + x - 1 - 2\sqrt {4x + 5}  \times \sqrt {x - 1}

\Rightarrow 11x - 6 - 5x - 4 =  - 2\sqrt {\left( {4x + 5} \right)\left( {x - 1} \right)}

\Rightarrow 6x - 10 =  - 2\sqrt {4{x^2} + x - 5}

\Rightarrow 2\left( {3x - 5} \right) =  - 2\sqrt {4{x^2} + x - 5}

\Rightarrow 3x - 5 =  - \sqrt {4{x^2} + x - 5}  [উভয় পক্ষকে 2 দ্বারা ভাগ করে]

\Rightarrow {\left( {3x - 5} \right)^2} = {\left( { - \sqrt {4{x^2} + x - 5} } \right)^2}  [ পুনরায় উভয় পক্ষকে বর্গ করে ]

\Rightarrow 9{x^2} - 30x + 25 = 4{x^2} + x - 5

\Rightarrow 9{x^2} - 30x + 25 - 4{x^2} - x + 5 = 0

\Rightarrow 5{x^2} - 31x + 30 = 0

\Rightarrow 5{x^2} - 6x - 25x + 30 = 0

\Rightarrow x\left( {5x - 6} \right) - 5\left( {5x - 6} \right) = 0

\Rightarrow \left( {5x - 6} \right)\left( {x - 5} \right) = 0

হয়, 5x - 6 = 0

\Rightarrow x = {6 \over 5}

অথবা, x - 5 = 0

\Rightarrow x = 5

 

৩. \sqrt {2x + 7}  + \sqrt {3x - 18}  = \sqrt {7x + 1}

সমাধান: \sqrt {2x + 7}  + \sqrt {3x - 18}  = \sqrt {7x + 1}

\Rightarrow {\left( {\sqrt {2x + 7}  + \sqrt {3x - 18} } \right)^2} = {\left( {\sqrt {7x + 1} } \right)^2}  [উভয় পক্ষকে বর্গ করে]

\Rightarrow 2x + 7 + 3x - 18 + 2\sqrt {2x + 7}  \times \sqrt {3x - 18}  = 7x + 1

\Rightarrow 2\sqrt {2x + 7}  \times \sqrt {3x - 18}  = 7x + 1 - 5x + 11

\Rightarrow 2\sqrt {6{x^2} - 15x - 126}  = 2x + 12

\Rightarrow 2\sqrt {6{x^2} - 15x - 126}  = 2\left( {x + 6} \right)

\Rightarrow \sqrt {6{x^2} - 15x - 126}  = \left( {x + 6} \right)   [উভয় পক্ষকে 2 দ্বারা ভাগ করে]

\Rightarrow {\left( {\sqrt {6{x^2} - 15x - 126} } \right)^2} = {\left( {x + 6} \right)^2}   [ পুনরায় উভয় পক্ষকে বর্গ করে ]

\Rightarrow 6{x^2} - 15x - 126 = {x^2} + 12x + 36

\Rightarrow 6{x^2} - 15x - 126 - {x^2} - 12x - 36 = 0

\Rightarrow 5{x^2} - 27x - 162 = 0

\Rightarrow 5{x^2} - 45x + 18x - 162 = 0

\Rightarrow 5x\left( {x - 9} \right) + 18\left( {x - 9} \right) = 0

\Rightarrow \left( {x - 9} \right)\left( {5x + 18} \right) = 0

হয়,  \left( {x - 9} \right) = 0

\therefore x = 9

অথবা,  \left( {5x + 18} \right) = 0

\Rightarrow 5x =  - 18

\therefore x =  - {{18} \over 5}

 

৪. \sqrt {x + 4}  + \sqrt {x + 11}  = \sqrt {8x + 9}

সমাধান:  \sqrt {x + 4}  + \sqrt {x + 11}  = \sqrt {8x + 9}

\Rightarrow {\left( {\sqrt {x + 4}  + \sqrt {x + 11} } \right)^2} = {\left( {\sqrt {8x + 9} } \right)^2}  [উভয় পক্ষকে বর্গ করে]

\Rightarrow x + 4 + 2\sqrt {\left( {x + 4} \right)\left( {x + 11} \right)}  + x + 11 = 8x + 9

\Rightarrow x + 4 + x + 11 - 8x - 9 =  - 2\sqrt {\left( {x + 4} \right)\left( {x + 11} \right)}

\Rightarrow  - 6x + 6 =  - 2\sqrt {{x^2} + 15x + 44}

\Rightarrow  - 2\left( {3x - 3} \right) =  - 2\sqrt {{x^2} + 15x + 44}

\Rightarrow 3x - 3 = \sqrt {{x^2} + 15x + 44}

\Rightarrow {\left( {3x - 3} \right)^2} = {\left( {\sqrt {{x^2} + 15x + 44} } \right)^2}   [ পুনরায় উভয় পক্ষকে বর্গ করে ]

\Rightarrow 9{x^2} - 18x + 9 = {x^2} + 15x + 44

\Rightarrow 9{x^2} - 18x + 9 - {x^2} - 15x - 44 = 0

\Rightarrow 9{x^2} - {x^2} - 18x - 15x + 9 - 44 = 0

\Rightarrow 8{x^2} - 33x - 35 = 0

\Rightarrow 8{x^2} - 40x + 7x - 35 = 0

\Rightarrow 8x\left( {x - 5} \right) + 7\left( {x - 5} \right) = 0

\Rightarrow \left( {x - 5} \right)\left( {8x + 7} \right) = 0

হয়, x - 5 = 0

\therefore x = 5

অথবা,  8x + 7 = 0

\Rightarrow 8x =  - 7

\therefore x =  - {7 \over 8}

 

৫. \sqrt {11x - 6}  = \sqrt {4x + 5}  + \sqrt {x - 1}

সমাধান: \sqrt {11x - 6}  = \sqrt {4x + 5}  + \sqrt {x - 1}

\Rightarrow {\left( {\sqrt {11x - 6} } \right)^2} = {\left( {\sqrt {4x + 5}  + \sqrt {x - 1} } \right)^2}   [উভয় পক্ষকে বর্গ করে]

\Rightarrow 11x - 6 = 4x + 5 + x - 1 + 2\sqrt {4x + 5}  \times \sqrt {x - 1}

\Rightarrow 11x - 6 - 4x - 5 - x + 1 = 2\sqrt {4{x^2} + x - 5}

\Rightarrow 11x - 6 - 5x - 4 = 2\sqrt {4{x^2} + x - 5}

\Rightarrow 6x - 10 = 2\sqrt {4{x^2} + x - 5}

\Rightarrow 2\left( {3x - 5} \right) = 2\sqrt {4{x^2} + x - 5}

\Rightarrow 3x - 5 = \sqrt {4{x^2} + x - 5}

\Rightarrow {\left( {3x - 5} \right)^2} = {\left( {\sqrt {4{x^2} + x - 5} } \right)^2}   [ পুনরায় উভয় পক্ষকে বর্গ করে ]

\Rightarrow 9{x^2} - 30x + 25 = 4{x^2} + x - 5

\Rightarrow 9{x^2} - 30x + 25 - 4{x^2} - x + 5 = 0

\Rightarrow 5{x^2}31x + 30 = 0

\Rightarrow 5{x^2} - 6x - 25x + 30 = 0

\Rightarrow x\left( {5x - 6} \right) - 5\left( {5x - 6} \right) = 0

\Rightarrow \left( {5x - 6} \right)\left( {x - 5} \right) = 0

হয়,   5x - 6 = 0

\therefore x = {6 \over 5}

অথবা,   x - 5 = 0

\therefore x = 5

 

৬. \sqrt {{x^2} - 8}  + \sqrt {{x^2} - 14}  = 6

সমাধান: \sqrt {{x^2} - 8}  + \sqrt {{x^2} - 14}  = 6

\Rightarrow \sqrt {{x^2} - 8}  = 6 - \sqrt {{x^2} - 14}

\Rightarrow {\left( {\sqrt {{x^2} - 8} } \right)^2} = {\left( {6 - \sqrt {{x^2} - 14} } \right)^2}      [উভয় পক্ষকে বর্গ করে]

\Rightarrow {x^2} - 8 = 36 - 12\sqrt {{x^2} - 14}  + {x^2} - 14

\Rightarrow 12\sqrt {{x^2} - 14}  = 36 + {x^2} - 14 - {x^2} + 8

\Rightarrow 12\sqrt {{x^2} - 14}  = 30

\Rightarrow 2\sqrt {{x^2} - 14}  = 5

\Rightarrow {\left( {2\sqrt {{x^2} - 14} } \right)^2} = {5^2}   [ পুনরায় উভয় পক্ষকে বর্গ করে ]

\Rightarrow 4\left( {{x^2} - 14} \right) = 25

\Rightarrow 4{x^2} - 56 = 25

\Rightarrow 4{x^2} = 81

\Rightarrow {x^2} = {{81} \over 4}

\Rightarrow x =  \pm {9 \over 2}

\therefore x =  \pm {9 \over 2}

 

৭. \sqrt {{x^2} - 6x + 9}  - \sqrt {{x^2} - 6x + 6}  = 1

সমাধান:  \sqrt {{x^2} - 6x + 9}  - \sqrt {{x^2} - 6x + 6}  = 1

\Rightarrow \sqrt {y + 9}  - \sqrt {y + 6}  = 1      ধরি,  \left[ {{x^2} - 6x = y} \right]

\Rightarrow \sqrt {y + 9}  = 1 + \sqrt {y + 6}

\Rightarrow {\left( {\sqrt {y + 9} } \right)^2} = {\left( {1 + \sqrt {y + 6} } \right)^2}    [উভয় পক্ষকে বর্গ করে]

\Rightarrow y + 9 = 1 + 2 \times 1 \times \sqrt {y + 6}  + y + 6

\Rightarrow y + 9 - 1 - y - 6 = 2\sqrt {y + 6}

\Rightarrow 2 = 2\sqrt {y + 6}

\Rightarrow 2\sqrt {y + 6}  = 2

\Rightarrow \sqrt {y + 6}  = 1    [উভয় পক্ষকে 2 দ্বারা ভাগ করে  ]

\Rightarrow {\left( {\sqrt {y + 6} } \right)^2} = {\left( 1 \right)^2}   [ পুনরায় উভয় পক্ষকে বর্গ করে ]

\Rightarrow y + 6 = 1

\Rightarrow y + 6 - 1 = 0

\Rightarrow y + 5 = 0

\Rightarrow {x^2} - 6x + 5 = 0

\Rightarrow {x^2} - 5x - x + 5 = 0

\Rightarrow x\left( {x - 5} \right) - 1\left( {x - 5} \right) = 0

\Rightarrow \left( {x - 5} \right)\left( {x - 1} \right) = 0

হয়,  x - 5 = 0

\therefore x = 5

অথবা,  x - 1 = 0

\therefore x = 1

 

 

৮.  \sqrt {x - 2}  - \sqrt {x - 9}  = 1

সমাধান:  \sqrt {x - 2}  - \sqrt {x - 9}  = 1

\Rightarrow \sqrt {x - 2}  = 1 + \sqrt {x - 9}

\Rightarrow {\left( {\sqrt {x - 2} } \right)^2} = {\left( {1 + \sqrt {x - 9} } \right)^2}    [উভয় পক্ষকে বর্গ করে]

\Rightarrow x - 2 = 1 + x - 9 + 2\sqrt {x - 9}

\Rightarrow x - 2 - 1 - x + 9 = 2\sqrt {x - 9}

\Rightarrow 6 = 2\sqrt {x - 9}

\Rightarrow 3 = \sqrt {x - 9} [উভয় পক্ষকে 2 দ্বারা ভাগ করে]

\Rightarrow {\left( 3 \right)^2} = {\left( {\sqrt {x - 9} } \right)^2}

\Rightarrow 9 = x - 9

\Rightarrow x - 9 - 9 = 0

\Rightarrow x - 18 = 0

\Rightarrow x = 18

 

৯.  6\sqrt {{{2x} \over {x - 1}}}  + 5\sqrt {{{x - 1} \over {2x}}}  = 13

সমাধান:  6\sqrt {{{2x} \over {x - 1}}}  + 5\sqrt {{{x - 1} \over {2x}}}  = 13

{{2x} \over {x - 1}} = {a^2} ধরা হলে প্রদত্ত সমীকরণ দাঁড়ায়,

6\sqrt {{a^2}}  + 5\sqrt {{1 \over {{a^2}}}}  = 13        \left[ {{{2x} \over {x - 1}} = {a^2} \Rightarrow {{x - 1} \over {2x}} = {1 \over {{a^2}}}} \right]

\Rightarrow 6a + {5 \over a} = 13

\Rightarrow 6{a^2} + 5 = 13a

\Rightarrow 6{a^2} - 13a + 5 = 0

\Rightarrow 6{a^2} - 10a - 3a + 5 = 0

\Rightarrow 2a\left( {3a - 5} \right) - 1\left( {3a - 5} \right) = 0

\Rightarrow \left( {3a - 5} \right)\left( {2a - 1} \right) = 0

হয়, 3a - 5 = 0

\therefore a = {5 \over 3}

 

অথবা, 2a - 1 = 0

\therefore a = {1 \over 2}

a = {5 \over 3} হলে আমরা পাই,

\sqrt {{{2x} \over {x - 1}}}  = {5 \over 3}

\Rightarrow {\left( {\sqrt {{{2x} \over {x - 1}}} } \right)^2} = {\left( {{5 \over 3}} \right)^2}   [উভয় পক্ষকে বর্গ করে]

\Rightarrow {{2x} \over {x - 1}} = {{25} \over 9}

\Rightarrow 25x - 25 = 18x

\Rightarrow 25x - 18x = 25

\Rightarrow 7x = 25

\therefore x = {{25} \over 7}

আবার, a = {1 \over 2}   হলে আমরা পাই,

\sqrt {{{2x} \over {x - 1}}}  = {1 \over 2}

\Rightarrow {\left( {\sqrt {{{2x} \over {x - 1}}} } \right)^2} = {\left( {{1 \over 2}} \right)^2}

\Rightarrow {{2x} \over {x - 1}} = {1 \over 4}

\Rightarrow 8x = x - 1

\Rightarrow 8x - x =  - 1

\Rightarrow 7x =  - 1

\therefore x =  - {1 \over 7}

 

১০.  \sqrt {{{x - 1} \over {3x + 2}}}  + 2\sqrt {{{3x + 2} \over {x - 1}}}  = 3

সমাধান:  \sqrt {{{x - 1} \over {3x + 2}}}  + 2\sqrt {{{3x + 2} \over {x - 1}}}  = 3

{{x - 1} \over {3x + 2}} = {a^2}   ধরা হলে প্রদত্ত সমীকরণ দাঁড়ায়,

\sqrt {{a^2}}  + 2\sqrt {{1 \over {{a^2}}}}  = 3      \left[ {{{x - 1} \over {3x + 2}} = {a^2} \Rightarrow {{3x + 2} \over {x - 1}} = {1 \over {{a^2}}}} \right]

\Rightarrow a + {2 \over a} = 3

\Rightarrow {a^2} + 2 = 3a

\Rightarrow {a^2} - 3a + 2 = 0

\Rightarrow {a^2} - 2a - a + 2 = 0

\Rightarrow a\left( {a - 2} \right) - 1\left( {a - 2} \right) = 0

\Rightarrow \left( {a - 2} \right)\left( {a - 1} \right) = 0

হয়,  a - 2 = 0

\therefore a = 2

অথবা,  a - 1 = 0

\therefore a = 1

a = 2  হলে আমরা পাই,

\sqrt {{{x - 1} \over {3x + 2}}}  = 2

\Rightarrow {\left( {\sqrt {{{x - 1} \over {3x + 2}}} } \right)^2} = {\left( 2 \right)^2}    [উভয় পক্ষকে বর্গ করে]

\Rightarrow {{x - 1} \over {3x + 2}} = 4

\Rightarrow x - 1 = 12x + 8

\Rightarrow x - 12x = 8 + 1

\Rightarrow  - 11x = 9

\Rightarrow 11x =  - 9

\therefore x =  - {9 \over {11}}

আবার,   a = 1   হলে আমরা পাই,

\sqrt {{{x - 1} \over {3x + 2}}}  = 1

\Rightarrow {\left( {\sqrt {{{x - 1} \over {3x + 2}}} } \right)^2} = {\left( 1 \right)^2}     [উভয় পক্ষকে বর্গ করে]

\Rightarrow {{x - 1} \over {3x + 2}} = 1

\Rightarrow 3x + 2 = x - 1

\Rightarrow 3x - x =  - 1 - 2

\Rightarrow 2x =  - 3

\therefore x =  - {3 \over 2}

SSC বোর্ড পরীক্ষার প্রশ্নের সমাধান

১. \sqrt {(x - 1)(x - 2)}  + \sqrt {(x - 3)(x - 4)}  = \sqrt 2 এর সমাধান- [ব. বো. ১৬]

(a) (0,2)              (b) (0,3)             (c) (2,3)                  (d) (3,5)

উত্তর:  (c) (2,3)  

২. \sqrt {x - 4}  = \sqrt {x + 12}  - 2 সমীকরণটির বীজ কত? [দি. বো. ১৫]

(a) 5               (b) 7              (c) 13                   (d) 15

উত্তর: (c) 13

৩. \sqrt {\frac{{2x}}{{x - 1}}}  = 1 সমীকরণের মূল কোনটি? [ন. প্র. দি. বো.]

(a) – 2               (b) – 1              (c) 1                   (d) 2

উত্তর: (b) – 1

৪. {(2 + a)^{\frac{1}{3}}} = 2 হলে a এর মান নিচের কোনটি? [ন. প্র. চ. বো.

(a) 4               (b) 6              (c) 7                   (d) 8

উত্তর:  (b) 6