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Gujarati JEE Mathematics : દ્વિઘાત સમીકરણ

Multiple Choice Questions

31.

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

x² - 4x - 12 = 0

Tips: -

ધારો કે માંગેલ દ્વિઘાત સમીકરણનાં બીજ α અને β છે.

α² + β² = 40 તથા α³ + β³ = 208

∴ (α + β)² - 2αβ = 40 અને (α + β) (α² - αβ + β²) = 208

∴ (α + β) (40 -αβ) = 208

ધારો કે α + β = m તથા α β = n છે.

∴ m² - 2n = 40 તથા m (40 - n) = 208 મળે.

$bold therefore bold space fraction numerator bold m to the power of bold 2 bold minus bold 40 over denominator bold 2 end fraction bold equals bold n$ સમીકરણ m (40 - n) = 208 માં મૂકતાં,
$bold therefore bold m bold space open square brackets bold 40 bold minus open parentheses fraction numerator bold m to the power of bold 2 bold minus bold 40 over denominator bold 2 end fraction close parentheses close square brackets bold space bold equals bold space bold 208 bold therefore bold space bold m bold space bold equals bold space bold left square bracket bold 80 bold space bold minus bold space bold m to the power of bold 2 bold space bold plus bold space bold 40 bold right square bracket bold space bold equals bold space bold 416 bold therefore bold space bold 120 bold space bold m bold space bold minus bold space bold m to the power of bold 3 bold space bold equals bold space bold 416 bold therefore bold space bold m to the power of bold 3 bold space bold minus bold space bold 120 bold space bold m bold space bold plus bold space bold 416 bold space bold equals bold space bold 0 bold therefore bold space bold left parenthesis bold m bold space bold minus bold space bold 4 bold right parenthesis bold space bold left parenthesis bold m to the power of bold 2 bold space bold plus bold space bold 4 bold m bold space bold minus bold space bold 104 bold right parenthesis bold space bold equals bold space bold 0$

∴ m = 4 અથવા m² + 4m - 104 = 0

હવે m² + 4m - 104 = 0 માટે, $bold increment bold space bold equals bold space bold 16 bold space bold minus bold space bold 4 bold left parenthesis bold 1 bold right parenthesis bold space bold left parenthesis bold minus bold 104 bold right parenthesis bold space bold equals bold space bold 432 bold space$

∴ $square root of bold increment bold space bold equals bold space bold 12 square root of bold 3$ અસંમેય સંખ્યા છે. પરંતુ α, β સંમેય છે.

∴ m = 4 શક્ય છે. આથી $bold n bold space bold equals bold space fraction numerator bold 16 bold minus bold 40 over denominator bold 2 end fraction bold space bold equals bold space bold minus bold 12$

∴ α + β = 4 ; α β = -12

∴ માંગેલ દ્વિઘાત સમીકરણ x² - 4x - 12 = 0

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

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

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

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

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

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

35. 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.

36. 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

37. 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

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.

39. 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.

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

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

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

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Gujarati JEE Mathematics : દ્વિઘાત સમીકરણ

Multiple Choice Questions

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