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উচ্চতর গণিত

অনুশীলনী-৫.৬

সমাধান কর:

১. {2^x} + {3^y} = 31

{2^x} - {3^y} =  - 23

সমাধান: {2^x} + {3^y} = 31..........(i)

{2^x} - {3^y} =  - 23...........(ii)

এখন (i) নং ও (ii) নং সমীকরণ যোগ করে পাই,

{2.2^x} = 8

\Rightarrow {2^x} = 4

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

\therefore x = 2   [\because {a^m} = {a^n} হলে, m = n]

আবার, (i) নং হতে (ii) নং সমীকরণ বিয়োগ করে পাই,

{2.3^y} = 54

\Rightarrow {3^y} = 27

\Rightarrow {3^y} = {3^3}

\therefore y = 3    [\because {a^m} = {a^n} হলে, m = n]

\therefore নির্ণেয় সমাধান: \left( {x,y} \right) = \left( {2,3} \right)

 

২.  {3^x} = {9^y}

\Rightarrow {5^{x + y + 1}} = {25^{xy}}

সমাধান: {3^x} = {9^y}...........(i)

\Rightarrow {5^{x + y + 1}} = {25^{xy}}...........(ii)

এখন, (i) নং সমীকরণ থেকে পাই, {3^x} = {9^y}

\Rightarrow {3^x} = {\left( {{3^2}} \right)^y}

\Rightarrow {3^x} = {3^{2y}}

\therefore x = 2y........(iii)     [\because {a^m} = {a^n} হলে, m = n]

আবার, (ii) নং সমীকরণ থেকে পাই,

\Rightarrow {5^{x + y + 1}} = {25^{xy}}

\Rightarrow {5^{x + y + 1}} = {\left( {{5^2}} \right)^{xy}}

\Rightarrow {5^{x + y + 1}} = {5^{2xy}}

\Rightarrow x + y + 1 = 2xy     [\because {a^m} = {a^n} হলে, m = n]

\Rightarrow 2y + y + 1 = 4{y^2}  \left[ {\because x = 2y} \right]

\Rightarrow 4{y^2} - 3y - 1 = 0

\Rightarrow 4{y^2} - 4y + y - 1 = 0

\Rightarrow 4y\left( {y - 1} \right) + 1\left( {y - 1} \right) = 0

\Rightarrow \left( {y - 1} \right)\left( {4y + 1} \right) = 0

হয়, y - 1 = 0

\therefore y = 1

অথবা, 4y + 1 = 0

\therefore y =  - \frac{1}{4}

(iii) নং এ y এর মান বসিয়ে পাই,

যখন, y = 1    তখন,   x = 2.1 = 2

যখন, y =  - \frac{1}{4}    তখন, x = 2\left( { - \frac{1}{4}} \right) =  - \frac{1}{2}

\therefore নির্ণেয় সমাধান: \left( {x,y} \right) = \left( {2,1} \right),\left( { - \frac{1}{2}, - \frac{1}{4}} \right)

 

৩. {3^x}{.9^y} = 81

2x - y = 8

সমাধান: {3^x}{.9^y} = 81.......(i)

2x - y = 8........(ii)

এখন, (i) নং সমীকরণ থেকে পাই,

{3^x}{.9^y} = 81

\Rightarrow {3^x}.{\left( {{3^2}} \right)^y} = {3^4}

\Rightarrow {3^x}{.3^{2y}} = {3^4}

\Rightarrow {3^{x + }}^{2y} = {3^4}  \left[ {\because {a^m}.{a^n} = {a^{m + n}}} \right]

\therefore x + 2y = 4    [\because {a^m} = {a^n} হলে, m = n]

\therefore x + 2y - 4 = 0.......(iii)

আবার, (ii) নং সমীকরণ থেকে পাই,

2x - y = 8

\therefore 2x - y - 8 = 0.........(iv)

(iii) ও (iv) নং সমীকরণ জোট থেকে বজ্রগুণন পদ্ধতিতে পাই,

\frac{x}{{ - 16 - 4}} = \frac{y}{{ - 8 + 8}} = \frac{1}{{ - 1 - 4}}

\Rightarrow \frac{x}{{ - 20}} = \frac{y}{0} = \frac{1}{{ - 5}}

\therefore x = 4,y = 0

\therefore নির্ণেয় সমাধান: \left( {x,y} \right) = \left( {4,0} \right)

 

৪. {2^x}{.3^y} = 18

{2^{2x}}{.3^y} = 36

সমাধান: {2^x}{.3^y} = 18..........(i)

{2^{2x}}{.3^y} = 36..........(ii)

এখন (i) নং কে (ii) নং দ্বারা ভাগ করে পাই,

\frac{{{2^{2x}}{{.3}^y}}}{{{2^x}{{.3}^y}}} = \frac{{36}}{{18}}

\Rightarrow {2^x} = 2

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

\therefore x = 1     [\because {a^m} = {a^n} হলে, m = n]

আবার , (iii) নং সমীকরণে  y এর মান বসিয়ে পাই,

{2^1}{.3^y} = 18

\Rightarrow {3^y} = 9

\Rightarrow {3^y} = {3^2}

\therefore y = 2

\therefore নির্ণেয় সমাধান: \left( {x,y} \right) = \left( {1,2} \right)

 

৫. {a^x}.{a^{y + 1}} = {a^7}

{a^{2y}}.{a^{3x + 5}} = {a^{20}}

সমাধান: {a^x}.{a^{y + 1}} = {a^7}............(i)

{a^{2y}}.{a^{3x + 5}} = {a^{20}}............(ii)

এখন, (i) নং সমীকরণ থেকে পাই,

{a^x}.{a^{y + 1}} = {a^7}

\Rightarrow x + y + 1 = 7   [\because {a^m} = {a^n} হলে, m = n]

\Rightarrow x + y + 1 - 7 = 0

\therefore x + y - 6 = 0.........(iii)

আবার, (ii) নং সমীকরণ থেকে পাই,

{a^{2y}}.{a^{3x + 5}} = {a^{20}}

\Rightarrow 2y + 3x + 5 = 20    [\because {a^m} = {a^n} হলে, m = n]

\Rightarrow 2y + 3x + 5 - 20 = 0

\therefore 3x + 2y - 15 = 0..........(iv)

(iii) ও (iv) নং সমীকরণ জোট থেকে বজ্রগুণন পদ্ধতিতে পাই,

\frac{x}{{ - 15 + 12}} = \frac{y}{{ - 18 + 15}} = \frac{1}{{2 - 3}}

\Rightarrow \frac{x}{{ - 3}} = \frac{y}{{ - 3}} = \frac{1}{{ - 1}}

\Rightarrow \frac{x}{3} = \frac{y}{3} = 1

\therefore x = 3

আবার, y = 3

\therefore নির্ণেয় সমাধান: \left( {x,y} \right) = \left( {3,3} \right)

 

৬. {y^x} = {x^2}

{x^{2x}} = {y^4}

y \ne 1

সমাধান:  {y^x} = {x^2}.............(i)

{x^{2x}} = {y^4}............(ii)

(ii) নং সমীকরণ থেকে পাই,

{x^{2x}} = {y^4}

\Rightarrow {\left( {{x^2}} \right)^x} = {y^4}

\Rightarrow {\left( {{y^x}} \right)^x} = {y^4}   [ (i) নং সমীকরণ থেকে {x^2} এর মান বসিয়ে ]

\Rightarrow {y^{x2}} = {y^4}   \left[ {\because {{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right]

\Rightarrow {x^2} = 4   [\because {a^m} = {a^n} হলে, m = n]

\therefore x =  \pm 2

আবার, (i) নং সমীকরণে x এর মান বসিয়ে পাই,

যখন, x = 2       তখন,  {y^2} = {2^2}

\Rightarrow {y^2} = 4

\therefore y =  \pm 2

 

যখন, x =  - 2     তখন,  {y^{ - 2}} = {\left( { - 2} \right)^2}

\Rightarrow \frac{1}{{{y^2}}} = 4

\Rightarrow {y^2} = \frac{1}{4}

\therefore y =  \pm \frac{1}{2}

\therefore নির্ণেয় সমাধান:  \left( {x,y} \right) = \left( {2,2} \right),\left( {2, - 2} \right),\left( { - 2,\frac{1}{2}} \right),\left( { - 2, - \frac{1}{2}} \right)

 

৭. {y^x} = 4

{y^2} = {2^x}

সমাধান: {y^x} = 4.........(i)

{y^2} = {2^x}...........(ii)

এখন, (ii) নং সমীকরণ থেকে পাই,

{y^2} = {2^x}

\Rightarrow {\left( {{y^2}} \right)^x} = {\left( {{2^x}} \right)^x}   [উভয় পক্ষের ঘাত  x –এ উন্নীত করে ]

\Rightarrow {y^{2x}} = {2^{{x^2}}}   \left[ {\because {{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right]

\Rightarrow {\left( {{y^x}} \right)^2} = {2^x}^{^2}

\Rightarrow {\left( 4 \right)^2} = {2^x}^{^2}       [ (i) থেকে  {y^x} এর মান বসিয়ে ]

\Rightarrow {\left( {{2^2}} \right)^2} = {2^{{x^2}}}

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

\Rightarrow {x^2} = 4   [\because {a^m} = {a^n} হলে, m = n]

\therefore x =  \pm 2

আবার, (i) নং সমীকরণে x এর মান বসিয়ে পাই,

যখন, x = 2       তখন, {y^2} = 4

\therefore y =  \pm 2

যখন, x =  - 2          তখন, {y^{ - 2}} = 4

\Rightarrow {y^2} = \frac{1}{4}

\therefore y =  \pm \frac{1}{2}

\therefore নির্ণেয় সমাধান: \left( {x,y} \right) = \left( {2,2} \right),\left( {2, - 2} \right),\left( { - 2,\frac{1}{2}} \right),\left( { - 2, - \frac{1}{2}} \right)

 

৮. {4^x} = {2^y}

{\left( {27} \right)^{xy}} = {9^{y + 1}}

সমাধান: {4^x} = {2^y}..........(i)

{\left( {27} \right)^{xy}} = {9^{y + 1}}..........(ii)

এখন, (i) নং সমীকরণ থেকে পাই,

{\left( {{2^2}} \right)^x} = {2^y}

\Rightarrow {2^{2x}} = {2^y}   \left[ {\because {{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right]

\therefore 2x = y..........(iii)    [\because {a^m} = {a^n} হলে, m = n]

আবার, (ii) নং সমীকরণ থেকে পাই,

{\left( {27} \right)^{xy}} = {9^{y + 1}}

\Rightarrow {\left( {{3^3}} \right)^{xy}} = {\left( {{3^2}} \right)^{y + 1}}

\Rightarrow {3^3}^{xy} = {3^2}^{\left( {y + 1} \right)}    \left[ {\because {{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right]

\Rightarrow 3xy = 2\left( {y + 1} \right)

\Rightarrow 3x.2x = 2\left( {2x + 1} \right)   \left[ {\because y = 2x} \right]

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

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

\Rightarrow 3{x^2} - 2x - 1 = 0

\Rightarrow 3{x^2} - 3x + x - 1 = 0

\Rightarrow 3x\left( {x - 1} \right) + 1\left( {x - 1} \right) = 0

\Rightarrow \left( {x - 1} \right)\left( {3x + 1} \right) = 0

হয়, x - 1 = 0           অথবা, 3x + 1 = 0

\therefore x = 1        \therefore x =  - \frac{1}{3}

(iii) নং সমীকরণে x এর মান বসিয়ে পাই,

যখন, x = 1                               তখন, y = 2.1 = 2

যখন, x =  - \frac{1}{3}            তখন, y = 2.\left( { - \frac{1}{3}} \right) =  - \frac{2}{3}

\therefore নির্ণেয় সমাধান: \left( {x,y} \right) = \left( {1,2} \right),\left( { - \frac{1}{3}, - \frac{2}{3}} \right)

 

৯. 8{y^x} - {y^{2x}} = 16

{2^x} = {y^2}

8{y^x} - {y^{2x}} = 16...........(i)

{2^x} = {y^2}...........(ii)

এখন, (i) নং সমীকরণ থেকে পাই,

{y^{2x}} - 8{y^x} + 16 = 0

\Rightarrow {\left( {{y^x}} \right)^2} - 2.{y^x}.4 + {4^2} = 0

\Rightarrow {\left( {{y^x} - 4} \right)^2} = 0

\therefore {y^x} = 4........(iii)

আবার, (ii) নং সমীকরণ থেকে পাই,

{2^x} = {y^2}

\Rightarrow {\left( {{2^x}} \right)^x} = {\left( {{y^2}} \right)^x}     [উভয় পক্ষের ঘাত  x –এ উন্নীত করে ]

\Rightarrow {2^{{x^2}}} = {y^{2x}}    \left[ {\because {{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right]

\Rightarrow {2^{{x^2}}} = {\left( {{y^x}} \right)^2}   \left[ {\because {a^{mn}} = {{\left( {{a^m}} \right)}^n}} \right]

\Rightarrow {2^{{x^2}}} = {4^2}    [(iii) নং সমীকরণ থেকে  {y^x}  এর মান বসিয়ে পাই ]

\Rightarrow {2^{{x^2}}} = 16

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

\Rightarrow {x^2} = 4    [\because {a^m} = {a^n} হলে, m = n]

\therefore x =  \pm 2

এখন, (ii) নং সমীকরণে x এর মান বসিয়ে পাই,

যখন, x = 2        তখন,  {2^2} = {y^2}

\Rightarrow {y^2} = 4

\therefore y =  \pm 2

যখন,  x =  - 2           তখন,  {2^{ - 2}} = {y^2}

\Rightarrow {y^2} = \frac{1}{4}

\therefore y =  \pm \frac{1}{2}

\therefore নির্ণেয় সমাধান: \left( {x,y} \right) = \left( {2,2} \right),\left( {2, - 2} \right),\left( { - 2,\frac{1}{2}} \right),\left( { - 2, - \frac{1}{2}} \right)

SSC বোর্ড পরীক্ষার প্রশ্নের সমাধান

১. {x^y} = {y^x} এবং y = 2x সমীকরণ দুটির সমাধান নিচের কোনটি? [ব. বো. ১৭]

(a) (2,4)              (b) (4,2)             (c) ( - 4,2)                  (d) (4, - 2)

উত্তর: (a) (2,4)  

২. {16^x} = {64^y} হলে, \frac{y}{x} = কত? [সি.বো. ১৬]

(a) \frac{1}{4}                  (b) \frac{2}{3}

(c) \frac{3}{2}                  (d) 4

উত্তর: (b) \frac{2}{3}   

৩. {x^y} = {y^x} এবং y = 2y হলে, (x,y) = কত? [য. বো. ১৬]

(a) (2,4)              (b) (4,2)             (c) (3,1)                  (d) (1,3)

উত্তর:  (b) (4,2)

৪. {x^y} = {y^x}, x = 2y হলে, y = কত? [রা.বো. ১৫]

(a) x              (b) 2              (c) 3                   (d) 4

উত্তর:  (b) 2