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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 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.

33.

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

34. 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.

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

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

38. 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

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

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

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

Tips: -

ધારો કે સમીકરણ x² + x - 1 = 0 નાં બીજ α અને β છે.

અહીં $bold alpha over bold beta bold equals bold m over bold n$ છે. વળી $bold alpha bold space bold plus bold space bold beta bold space bold equals bold space bold minus bold 1 bold semicolon bold space bold alpha bold space bold beta bold space bold equals bold space bold minus bold 1$

$bold therefore bold space fraction numerator bold alpha bold space bold plus bold space bold beta over denominator bold alpha bold space bold minus bold space bold beta end fraction bold space bold equals bold space fraction numerator bold m bold space bold plus bold space bold n over denominator bold m bold space bold minus bold space bold n end fraction bold therefore bold space fraction numerator bold left parenthesis bold alpha bold space bold plus bold space bold beta bold right parenthesis to the power of bold 2 over denominator bold left parenthesis bold alpha bold space bold minus bold space bold beta bold right parenthesis to the power of bold 2 end fraction bold space bold equals bold space fraction numerator bold left parenthesis bold m bold space bold plus bold space bold n bold right parenthesis to the power of bold 2 over denominator bold left parenthesis bold m bold space bold minus bold space bold n bold right parenthesis to the power of bold 2 end fraction bold therefore bold space fraction numerator bold left parenthesis bold alpha bold space bold plus bold space bold beta bold right parenthesis to the power of bold 2 over denominator bold left parenthesis bold alpha bold space bold plus bold space bold beta bold right parenthesis to the power of bold 2 bold space bold minus bold space bold 4 bold αβ end fraction bold space bold equals fraction numerator bold left parenthesis bold m bold plus bold n bold right parenthesis to the power of bold 2 over denominator bold left parenthesis bold m bold minus bold n bold right parenthesis to the power of bold 2 end fraction bold therefore bold space fraction numerator bold 1 over denominator bold 1 bold plus bold 4 end fraction bold space bold equals bold space fraction numerator bold left parenthesis bold m bold plus bold n bold right parenthesis to the power of bold 2 over denominator bold left parenthesis bold m bold minus bold n bold right parenthesis to the power of bold 2 end fraction bold therefore bold space bold 1 over bold 5 bold space bold equals bold space open parentheses fraction numerator bold m bold plus bold n over denominator bold m bold minus bold n end fraction close parentheses to the power of bold 2 bold therefore bold space fraction numerator bold m bold plus bold n over denominator bold m bold minus bold n end fraction bold equals fraction numerator bold 1 over denominator square root of bold 5 end fraction bold therefore bold space fraction numerator bold m bold plus bold n bold plus bold m bold minus bold n over denominator bold m bold plus bold n bold minus bold m bold plus bold n end fraction bold space bold equals bold space fraction numerator bold 1 bold plus square root of bold 5 over denominator bold 1 bold minus square root of bold 5 end fraction bold therefore bold space bold m over bold n bold equals fraction numerator square root of bold 5 bold plus bold 1 over denominator bold 1 bold minus square root of bold 5 end fraction bold cross times fraction numerator bold left parenthesis bold 1 bold plus square root of bold 5 bold right parenthesis over denominator bold left parenthesis bold 1 bold plus square root of bold 5 bold right parenthesis end fraction bold space bold equals bold space fraction numerator bold 6 bold plus bold 2 square root of bold 5 over denominator bold minus bold 4 end fraction bold therefore bold space bold m over bold n bold space bold equals bold space fraction numerator bold minus bold 3 over denominator bold 2 end fraction bold minus fraction numerator square root of bold 5 over denominator bold 2 end fraction bold equals bold space fraction numerator bold minus bold 1 over denominator bold 2 end fraction bold left parenthesis bold 3 bold plus square root of bold 5 bold right parenthesis bold therefore bold space bold 2 bold m bold space bold plus bold space bold n bold space bold left parenthesis bold 3 bold space bold plus bold space square root of bold 5 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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