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Gujarati JEE Mathematics : દ્વિઘાત સમીકરણ

Multiple Choice Questions

21. જો log₂ x + log_x 2 = $begin inline style bold 10 over bold 3 end style$ = log₂y + log_y2 તથા x ≠ y હોય તો x+y = ......

$bold 8 bold minus bold 2 to the power of begin inline style bold 1 over bold 3 end style end exponent$
$bold 2 to the power of begin inline style bold 1 over bold 3 end style end exponent$
$bold 8 bold plus bold 2 to the power of begin inline style bold 1 over bold 3 end style end exponent$
8

22.

જો દ્વિઘાત સમીકરણ (a² + b²)x² - 2b(a + c) x + (b²+c²)=0 નાં બીજ સમાન હોય તોa, b, c, ....... શ્રેણીમાં હોય a, b, c ∈ R.

સમગુણોત્તર
સમાંતર
સ્વરિત
સમાંતર-સમગુણોત્તર

23. જો $bold x bold space bold equals bold space square root of bold 12 bold minus square root of bold 140 end root$ હોય, તો $bold x bold space bold plus bold space bold 1 over bold x bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold.$

$fraction numerator square root of bold 7 bold minus bold 3 square root of bold 5 over denominator bold 2 end fraction$
$fraction numerator square root of bold 7 bold minus square root of bold 5 over denominator bold 2 end fraction$
$square root of bold 7 bold plus square root of bold 5$
$fraction numerator bold 3 square root of bold 7 bold minus square root of bold 5 over denominator bold 2 end fraction$

24.

જો α ≠ β તથા α² = 5α - 3 તેમજ β² = 5β - 3 હોય, તો $bold alpha over bold beta$ અને $bold beta over bold alpha$ બીજ ધરાવતું દ્વિઘાત સમીકરણ ....... હોય.

x² + 19x -3 = 0
3x² - 19x + 3 = 0
3x²-16x + 1 = 0
3x² -- 19x - 3 = 0

25. સમીકરણ $open parentheses bold 5 bold plus bold 2 square root of bold 6 close parentheses to the power of bold x bold minus bold 3 end exponent bold plus open parentheses bold 5 bold minus bold 2 square root of bold 6 close parentheses to the power of bold x bold minus bold 3 end exponent bold space bold equals bold space bold 10$ નું એક બીજ ....... હોય.

Tips: -

અહીં, $bold 5 bold minus bold 2 square root of bold 6 bold space bold equals bold space fraction numerator bold 1 over denominator bold 5 bold plus bold 2 square root of bold 6 end fraction$
$bold therefore bold space bold left parenthesis bold 5 bold plus bold 2 square root of bold 6 bold right parenthesis to the power of bold x bold minus bold 3 end exponent bold space bold plus bold space bold left parenthesis bold 5 bold minus bold 2 square root of bold 6 bold right parenthesis to the power of bold x bold minus bold 3 end exponent bold space bold equals bold space bold 10 bold space bold therefore bold space bold left parenthesis bold 5 bold space bold plus bold space bold 2 square root of bold 6 bold right parenthesis to the power of bold x bold minus bold 3 end exponent bold space bold plus bold space open parentheses fraction numerator bold 1 over denominator bold 5 bold plus bold 2 square root of bold 6 end fraction close parentheses to the power of bold x bold minus bold 3 end exponent bold space bold equals bold space bold 10$

ધારો કે $bold left parenthesis bold 5 bold space bold plus bold space bold 2 square root of bold 6 bold right parenthesis to the power of bold x bold minus bold 3 end exponent bold space bold equals bold space bold m$
$bold therefore bold m bold space bold plus bold space bold 1 over bold m bold space bold space bold equals bold space bold 10 bold therefore bold space bold m to the power of bold 2 bold space bold minus bold space bold 10 bold m bold space bold plus bold space bold 1 bold space bold equals bold space bold 0 bold therefore bold space bold m bold space bold equals bold space fraction numerator bold 10 bold space bold plus-or-minus bold space square root of bold 100 bold minus bold 4 end root over denominator bold 2 end fraction bold therefore bold space bold m bold space bold equals bold space bold 5 bold space bold plus-or-minus bold space bold 2 square root of bold 6$

$bold therefore bold space bold m bold space bold equals bold space bold left parenthesis bold 5 bold plus bold 2 square root of bold 6 bold space bold right parenthesis to the power of bold x bold minus bold 3 end exponent bold space bold equals bold space bold left parenthesis bold 5 bold plus bold 2 square root of bold 6 bold right parenthesis to the power of bold 1$ અથવા $bold left parenthesis bold 5 bold plus bold 2 square root of bold 6 bold right parenthesis to the power of bold minus bold 1 end exponent$

∴ x- 3 = ± 1

∴ x = 4 અથવા 2

26. જો x, y, z ભિન્ન અને વાસ્તવિક હોય, તો x² + 4y² + 9z² - 6yz - 3zx - 2xy હંમેશાં ....... હોય.

0
ઋણ
અનૃણ
સંકર સંખ્યા

27. સમીકરણ $square root of bold x bold minus bold 2 end root bold left parenthesis bold italic x to the power of bold 2 bold space bold minus bold space bold 4 bold italic x bold space bold plus bold space bold 3 bold right parenthesis bold space bold equals bold space bold 0$ નાં વાસ્તવિક બીજ ...... હોય.

1,2,3
2, 3
1, 3
1, 2

28.

જો 3,3 એ દ્વિઘાત સમીકરણ x² + ax + β = 0 નાં બીજ હોય તથા -8, 2 એ દ્વિઘાત સમીકરણ x² + αx + b = 0 નાં બીજ હોય, તો સમીકરણ x² + ax + b = 0 નાં બીજ ...... હોય.

-8, 2
8, -2
-2, -8
2, 8

29.

દ્વિઘાત સમીકરણનો ઉકેલ શોધતી વખતે બે વિદ્યાર્થીઓ પૈકી એક વિદ્યાર્થી સમીકરણનું અચળ પદ ખોટું લખે છે અને સમીકરણનાં સાચાં બીજનો સરવાળો 3 મળે છે જ્યારે બીજો વિદ્યાર્થી x² નો સહગુણક તથા અચળ પદ સાચાં લખે છે જે અનુક્રમે 1 તથા -18 છે તો મળતા દ્વિઘાત સમીકરણનાં સાચાં બીજ ...... હોય.

-6, 3
6, -3
-3, -6
3, 6

30. સમીકરણ $fraction numerator bold log bold space bold 3 bold space bold plus bold space bold log bold space bold left parenthesis bold x to the power of bold 2 bold space bold plus bold space bold 2 bold right parenthesis over denominator bold log bold space bold left parenthesis bold x bold space bold minus bold space bold 2 bold right parenthesis end fraction bold space bold equals bold space bold 2$ ના ....... ઉકેલ મળે.

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Gujarati JEE Mathematics : દ્વિઘાત સમીકરણ

Multiple Choice Questions

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