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Gujarati JEE Mathematics : દ્વિઘાત સમીકરણ

Multiple Choice Questions

31. જો દ્વિઘાત સમીકરણ ax² + bx + c = 0 નાં બીજ α અને β હોય, તો $fraction numerator bold 1 over denominator bold aα bold plus bold b end fraction bold plus fraction numerator bold 1 over denominator bold aβ bold plus bold b bold space end fraction bold equals bold space bold. bold. bold. bold. bold.$

$bold a over bold bc$
$straight c over bold ab$
$1 over bold abc$
$bold b over bold ac$

32.

જો બીજનાં વર્ગોનો સરવાળો 40 તથા બીજના ઘનનો સરવાળો હોય તથા બીજ સંમેય હોય તેવું દ્વિઘાત સમીકરણ ........ હોય.

x² + 4x - 12 = 0
x² + 4x + 12 = 0
x² - 4x + 12 = 0
x² - 4x - 12 = 0

33. સમીકરણ (x+1) (x+2) (x+3) (x+4) = 120 નો વાસ્તવિક ઉકેલ ...... મળે.

-6, -1
-6, 1
-1, 6
6, 1

34. જો સમીકરણ ax² + bx + c = 0 નું એક બીજ એ બીજા બીજના n થાત જેટલું થાય તો $bold left parenthesis bold ac to the power of bold n bold right parenthesis to the power of begin inline style fraction numerator bold 1 over denominator bold n bold plus bold 1 end fraction end style end exponent bold space bold plus bold space bold left parenthesis bold a to the power of bold n bold c bold right parenthesis to the power of begin inline style fraction numerator bold 1 over denominator bold n bold plus bold 1 end fraction end style end exponent bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold. bold space bold. bold space$

-ab
-b
-c
-a

35. log₂(x² + 5x + 10) = 2 નાં બીજ ...... મળે.

2, 3
-2,3
-2, -3
-3, 2

36. જો x કોઈ વાસ્તવિક સંખ્યા હોય અને $bold k bold space bold equals bold space fraction numerator bold x to the power of bold 2 bold space bold plus bold space bold x bold space bold plus bold space bold 1 over denominator bold x to the power of bold 2 bold space bold minus bold space bold x bold space bold plus bold space bold 1 end fraction$ હોય તો k ∈ ....... .

$open square brackets bold 1 over bold 3 bold comma bold 3 close square brackets$
$open parentheses bold 1 over bold 3 bold comma bold 3 close parentheses$
$bold k bold less or equal than bold space bold 1 over bold 3$
k ≥ 3

37. જો દ્વિઘાત સમીકરણ x² - m (x + 1) - n = 0 નાં બીજ α અને β હોય, તો $fraction numerator bold alpha to the power of bold 2 bold space bold plus bold space bold 2 bold alpha bold space bold plus bold space bold 1 over denominator bold alpha to the power of bold 2 bold space bold plus bold space bold 2 bold alpha bold space bold plus bold space bold n end fraction bold plus bold space fraction numerator bold beta to the power of bold 2 bold space bold plus bold space bold 2 bold beta bold space bold plus bold space bold 1 over denominator bold beta to the power of bold 2 bold space bold plus bold space bold 2 bold beta bold space bold plus bold space bold n end fraction$ ની કિંમત ...... હોય.

38. સમીકરણ $bold 27 to the power of begin inline style bold 1 over bold x end style end exponent bold space bold plus bold space bold 12 to the power of begin inline style bold 1 over bold x end style end exponent bold space bold equals bold space bold 2 bold. bold 8 to the power of begin inline style bold 1 over bold x end style end exponent$ નાં વાસ્તવિક બીજોની સંખ્યા ....... છે.

Tips: -

$bold 27 to the power of bold 1 over bold x end exponent bold space bold plus bold space bold 12 to the power of begin inline style bold 1 over bold x end style end exponent bold space bold equals bold space bold 2 bold. bold 8 to the power of begin inline style bold 1 over bold x end style end exponent bold therefore bold space open parentheses bold 27 over bold 8 close parentheses to the power of begin inline style bold 1 over bold x end style end exponent bold space bold plus bold space open parentheses bold 12 over bold 8 close parentheses to the power of begin inline style bold 1 over bold x end style end exponent bold space bold equals bold space bold 2 bold space bold therefore bold space open parentheses bold 3 over bold 2 close parentheses to the power of begin inline style bold 3 over bold x end style end exponent bold space bold plus bold space open parentheses bold 3 over bold 2 close parentheses to the power of begin inline style bold 1 over bold x end style end exponent bold space bold equals bold space bold 2$

ધારો કે $open parentheses bold 3 over bold 2 close parentheses to the power of begin inline style bold 1 over bold x end style end exponent bold space bold equals bold space bold m$
$bold therefore bold space bold m to the power of bold 3 bold space bold plus bold space bold m bold space bold equals bold space bold 2 bold space bold therefore bold space bold m to the power of bold 3 bold space bold plus bold space bold m bold space bold minus bold space bold 2 bold space bold equals bold space bold 0 bold therefore bold space bold left parenthesis bold m to the power of bold 2 bold space bold plus bold space bold m bold space bold plus bold space bold 2 bold right parenthesis bold space bold left parenthesis bold m bold space bold minus bold space bold 1 bold right parenthesis bold space bold equals bold space bold 0$

∴ m = 1 અથવા m² + m + 2 = 0

હવે m² + m + 2 = 0 માટે, $bold increment bold space bold equals bold space bold 1 bold space bold minus bold 4 bold left parenthesis bold 1 bold right parenthesis bold space bold left parenthesis bold 2 bold right parenthesis bold space bold equals bold space bold minus bold 7 bold thin space bold less than bold 0 bold space$ જેથી વાસ્તવિક બીજ ન મળે.

∴ m = 1

$bold therefore bold space open parentheses bold 3 over bold 2 close parentheses to the power of begin inline style bold 1 over bold x end style end exponent bold space bold equals bold space bold 1 bold space bold equals bold space open parentheses bold 3 over bold 2 close parentheses to the power of bold 0$

$bold therefore bold space bold 1 over bold x bold space bold equals bold space bold 0$ જે અશક્ય છે.
$bold therefore$ વાસ્તવિક બીજની સંખ્યા 0 થાય.

39. સમીકરણ x² + x - 1 = 0 નાં બીજનો ગુણોત્તર m : n હોય તો નીચેનામાંથી કયું સત્ય બને ?

3m = 2n
2m = 3n
$bold 2 bold m bold space bold plus bold space bold n bold left parenthesis bold 3 bold plus square root of bold 5 bold right parenthesis bold space bold equals bold 0$
$bold 3 bold m bold minus bold 2 bold n bold left parenthesis square root of bold 5 bold plus bold 1 bold right parenthesis bold space bold equals bold space bold 0$

40. સમીકરણ $bold 3 bold space open parentheses bold x to the power of bold 2 bold plus bold 1 over bold x to the power of bold 2 close parentheses bold space bold plus bold space bold 16 bold space open parentheses bold x bold plus bold 1 over bold x close parentheses bold space bold plus bold space bold 26 bold space bold equals bold space bold 0$ નો ઉકેલ ગણ ...... હોય.

$open curly brackets bold minus bold 1 bold comma bold space bold 3 bold comma bold space fraction numerator bold minus bold 1 over denominator bold 3 end fraction close curly brackets$
$open curly brackets bold 1 bold comma bold space bold 3 bold comma bold space fraction numerator bold minus bold 1 over denominator bold 3 end fraction close curly brackets$
$open curly brackets bold 1 bold comma bold space bold 3 bold comma bold space bold 1 over bold 3 close curly brackets$
$open curly brackets bold minus bold 1 bold comma bold space bold minus bold 3 bold comma bold space fraction numerator bold minus bold 1 over denominator bold 3 end fraction close curly brackets$

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Gujarati JEE Mathematics : દ્વિઘાત સમીકરણ

Multiple Choice Questions

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