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Gujarati JEE Mathematics : દ્વિઘાત સમીકરણ

Multiple Choice Questions

31. 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.

32. 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

33. 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.

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

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

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

35. 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

36. 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.

Tips: -

x² - m (x + 1) - n = 0

∴ x² - mx - (m + n) = 0 નાં બીજ તથા આપેલ છે.

$bold therefore bold space bold alpha bold space bold plus bold space bold beta bold space bold equals bold space bold m bold space bold semicolon bold space bold alpha bold space bold beta bold space bold equals bold space bold minus bold left parenthesis bold m bold space bold plus bold space bold n bold right parenthesis bold therefore bold space 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 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 bold equals bold space fraction numerator bold alpha to the power of bold 2 bold beta to the power of bold 2 bold plus bold 2 bold alpha to the power of bold 2 bold beta bold plus bold nα to the power of bold 2 bold plus bold 2 bold αβ to the power of bold 2 bold plus bold 4 bold αβ bold plus bold 2 bold nα bold plus bold beta to the power of bold 2 bold plus bold 2 bold beta bold plus bold n bold plus bold alpha to the power of bold 2 bold beta to the power of bold 2 bold plus bold 2 bold αβ to the power of bold 2 bold plus bold nβ to the power of bold 2 bold plus bold 2 bold alpha to the power of bold 2 bold beta bold plus bold 4 bold αβ bold plus bold 2 bold nβ bold plus bold alpha to the power of bold 2 bold plus bold 2 bold alpha bold plus bold n over denominator bold left parenthesis bold alpha to the power of bold 2 bold plus bold 2 bold alpha bold plus bold n bold right parenthesis bold space bold left parenthesis bold beta to the power of bold 2 bold plus bold 2 bold beta bold plus bold n bold right parenthesis end fraction bold equals bold space fraction numerator bold 2 bold alpha to the power of bold 2 bold beta to the power of bold 2 bold plus bold 4 bold alpha to the power of bold 2 bold beta bold plus bold 4 bold αβ to the power of bold 2 bold space bold plus bold 8 bold αβ bold plus bold n to the power of bold 2 bold plus bold nβ to the power of bold 2 bold plus bold alpha to the power of bold 2 bold plus bold beta to the power of bold 2 bold plus bold 2 bold nα bold plus bold 2 bold nβ bold plus bold 2 bold alpha bold plus bold 2 bold beta bold plus bold 2 bold n over denominator bold alpha to the power of bold 2 bold beta to the power of bold 2 bold plus bold 2 bold alpha to the power of bold 2 bold beta bold plus bold 2 bold αβ to the power of bold 2 bold plus bold 4 bold αβ bold plus bold 2 bold nα bold plus bold nβ to the power of bold 2 bold plus bold 2 bold nβ bold plus bold n to the power of bold 2 end fraction bold equals bold space fraction numerator bold 2 bold left parenthesis bold αβ bold right parenthesis to the power of bold 2 bold plus bold 4 bold αβ bold left parenthesis bold alpha bold plus bold beta bold right parenthesis bold plus bold 8 bold αβ bold plus bold left parenthesis bold alpha to the power of bold 2 bold plus bold beta to the power of bold 2 bold right parenthesis bold left parenthesis bold n bold plus bold 1 bold right parenthesis bold plus bold 2 bold n bold left parenthesis bold alpha bold plus bold beta bold right parenthesis bold plus bold 2 bold left parenthesis bold alpha bold plus bold beta bold right parenthesis bold plus bold 2 bold n over denominator bold left parenthesis bold αβ bold right parenthesis to the power of bold 2 bold space bold plus bold 2 bold αβ bold left parenthesis bold alpha bold plus bold beta bold right parenthesis bold space bold plus bold n bold left parenthesis bold alpha to the power of bold 2 bold plus bold beta to the power of bold 2 bold right parenthesis bold plus bold 4 bold αβ bold space bold plus bold space bold 2 bold n bold space bold left parenthesis bold alpha bold plus bold beta bold right parenthesis bold space bold plus bold space bold n to the power of bold 2 end fraction bold equals bold space fraction numerator bold 2 bold left parenthesis bold m bold plus bold n bold right parenthesis to the power of bold 2 bold minus bold 4 bold m bold left parenthesis bold m bold plus bold n bold right parenthesis bold minus bold 8 bold left parenthesis bold m bold plus bold n bold right parenthesis bold plus bold left parenthesis bold n bold plus bold 1 bold right parenthesis bold left parenthesis bold alpha bold plus bold beta bold right parenthesis to the power of bold 2 bold minus bold 2 bold αβ bold right curly bracket bold plus bold 2 bold nm bold plus bold 2 bold m bold plus bold 2 bold n over denominator bold left parenthesis bold m bold plus bold n bold right parenthesis to the power of bold 2 bold minus bold 2 bold m bold left parenthesis bold m bold plus bold n bold right parenthesis bold plus bold n bold left curly bracket bold left parenthesis bold alpha bold plus bold beta bold right parenthesis to the power of bold 2 bold minus bold 2 bold αβ bold right curly bracket bold minus bold 4 bold left parenthesis bold m bold plus bold n bold right parenthesis bold plus bold 2 bold mn bold plus bold n to the power of bold 2 end fraction bold equals bold space fraction numerator bold 2 open parentheses bold m to the power of bold 2 bold plus bold 2 bold mn bold plus bold n to the power of bold 2 close parentheses bold minus bold 4 bold m to the power of bold 2 bold minus bold 4 bold mn bold minus bold 8 bold m bold minus bold 8 bold n bold plus bold left parenthesis bold n bold plus bold 1 bold right parenthesis bold left curly bracket bold m to the power of bold 2 bold plus bold 2 bold left parenthesis bold m bold plus bold n bold right parenthesis bold right curly bracket bold plus bold 2 bold mn bold plus bold 2 bold m bold plus bold 2 bold n over denominator bold m to the power of bold 2 bold plus bold 2 bold mn bold plus bold n to the power of bold 2 bold minus bold 2 bold m to the power of bold 2 bold minus bold 2 bold mn bold plus bold n bold left curly bracket bold m to the power of bold 2 bold plus bold 2 bold left parenthesis bold m bold plus bold n bold right parenthesis bold right curly bracket bold minus bold 4 bold m bold minus bold 4 bold n bold plus bold 2 bold mn bold plus bold n to the power of bold 2 end fraction bold equals bold space fraction numerator bold 2 bold m to the power of bold 2 bold plus bold 4 bold mn bold plus bold 2 bold n to the power of bold 2 bold minus bold 4 bold m to the power of bold 2 bold minus bold 4 bold mn bold minus bold 8 bold m bold minus bold 8 bold n bold plus bold m to the power of bold 2 bold n bold plus bold m to the power of bold 2 bold plus bold 2 bold mn bold plus bold 2 bold n to the power of bold 2 bold plus bold 2 bold m bold plus bold 2 bold n bold plus bold 2 bold mn bold plus bold 2 bold m bold plus bold 2 bold n over denominator bold m to the power of bold 2 bold plus bold 2 bold mn bold plus bold n to the power of bold 2 bold minus bold 2 bold m to the power of bold 2 bold minus bold 2 bold mn bold plus bold m to the power of bold 2 bold n bold plus bold 2 bold mn bold plus bold 2 bold n to the power of bold 2 bold minus bold 4 bold m bold minus bold 4 bold n bold plus bold 2 bold mn bold plus bold n to the power of bold 2 end fraction bold equals bold space fraction numerator bold minus bold m to the power of bold 2 bold plus bold 4 bold n to the power of bold 2 bold plus bold 4 bold mn bold minus bold 4 bold m bold minus bold 4 bold n bold plus bold m to the power of bold 2 bold n over denominator bold minus bold m to the power of bold 2 bold plus bold 4 bold n to the power of bold 2 bold plus bold 4 bold mn bold plus bold m to the power of bold 2 bold n bold minus bold 4 bold m bold minus bold 4 bold n end fraction bold equals bold 1$

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

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

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

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

39.

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

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

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

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Gujarati JEE Mathematics : દ્વિઘાત સમીકરણ

Multiple Choice Questions

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