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Gujarati JEE Mathematics : દ્વિઘાત સમીકરણ

Multiple Choice Questions

71. જો 0 ≤ x ≤ $bold pi$ માટે $bold 16 to the power of bold sin to the power of bold 2 bold x end exponent bold space bold plus bold space bold 16 to the power of bold cos to the power of bold 2 bold x end exponent bold space bold equals bold space bold 10$ હોય તો x = .......... .

$bold pi over bold 4$
$fraction numerator bold 3 bold pi over denominator bold 4 end fraction$
$bold pi over bold 3$
$bold pi over 6$

72. જો $bold 7 to the power of bold log subscript bold 7 bold left parenthesis bold x to the power of bold 2 bold minus bold 4 bold x bold plus bold 5 bold right parenthesis end exponent bold space bold equals bold space bold x bold space bold minus bold space bold 1$ હોય તો x ની શક્ય કિંમતો ..... હોય.

-3, -2
7
3
2

73. સમીકરણ x² - 6x - 2 = 0 નાં બીજ α અને β છે. જો a_m = αⁿ - βⁿ ; n ≥ 1 હોય તો $fraction numerator bold a subscript bold 10 bold minus bold 2 bold a subscript bold 8 over denominator bold 2 bold a subscript bold 9 end fraction$

-6
6
3
-3

74.

જો સમીકરણો x² - ax + b = 0 અને x² + bx - a = 0 ને એક બીજ સામાન્ય હોય, તો નીચેનામાંથી શું સત્ય બને ?

a - b = 1
a = - b
a = b
a + b = 1

75. જો a² + b² + c² = 1 હોય તો ab + bc + ca ∈ ......... જોઈ શકે.

$open square brackets bold 1 bold comma bold 1 over bold 2 close square brackets$
$open square brackets bold minus bold 1 over bold 2 bold comma bold 2 close square brackets$
$open square brackets negative bold 1 over bold 2 comma 1 close square brackets$
[-1, 2]

76. જો $bold x bold space bold equals bold space bold 2 bold space bold plus bold space bold 2 to the power of begin inline style bold 1 over bold 3 end style end exponent bold space bold plus bold space bold 2 to the power of begin inline style bold 2 over bold 3 end style end exponent$ હોય તો x³ - 6x² + 6x ની કિંમત ....... હોય.

Tips: -

$bold x bold space bold equals bold space bold 2 bold space bold plus bold space bold 2 to the power of begin inline style bold 1 over bold 3 end style end exponent bold space bold plus bold space bold 2 to the power of begin inline style bold 2 over bold 3 end style end exponent$ આપેલ છે.

$bold therefore bold space bold x bold space bold minus bold space bold 2 bold space bold equals bold space bold 2 to the power of begin inline style bold 1 over bold 3 end style end exponent bold space bold plus bold space bold 2 to the power of begin inline style bold 2 over bold 3 end style end exponent$ ... (1)

bold therefore bold space bold left parenthesis bold x bold space bold minus bold space bold 2 bold right parenthesis to the power of bold 3 bold space bold equals bold space open parentheses bold 2 to the power of begin inline style bold 1 over bold 2 end style end exponent bold plus bold 2 to the power of begin inline style bold 2 over bold 3 end style end exponent close parentheses bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold space bold. bold. bold. bold space bold left parenthesis bold 1 bold right parenthesis bold space bold therefore bold x to the power of bold 3 bold space bold minus bold space bold 8 bold space bold minus bold space bold 6 bold x bold space bold left parenthesis bold x bold space bold minus bold space bold 2 bold right parenthesis bold space bold equals bold space bold 2 bold space bold plus bold space bold 2 to the power of bold 2 bold space bold plus bold space bold 3 bold times bold 2 to the power of begin inline style bold 1 over bold 3 end style end exponent bold times bold 2 to the power of begin inline style bold 2 over bold 3 end style end exponent bold space open parentheses bold 2 to the power of begin inline style bold 1 over bold 3 end style end exponent bold plus bold 2 to the power of begin inline style bold 2 over bold 3 end style end exponent close parentheses bold therefore bold space bold x to the power of bold 3 bold minus bold 6 bold x to the power of bold 2 bold space bold plus bold space bold 12 bold x bold space bold minus bold space bold 8 bold space bold equals bold space bold 6 bold space bold plus bold space bold 3 bold times bold 2 bold left parenthesis bold x bold space bold minus bold space bold 2 bold right parenthesis bold therefore bold space bold x to the power of bold 3 bold space bold minus bold space bold 6 bold x to the power of bold 2 bold space bold plus bold space bold 12 bold x bold space bold minus bold space bold 8 bold space bold equals bold space bold 6 bold space bold plus bold space bold 6 bold x bold space bold minus bold space bold 12 bold therefore bold space bold x to the power of bold 3 bold space bold minus bold space bold 6 bold x to the power of bold 2 bold space bold plus bold space bold 6 bold x bold space bold equals bold space bold 2 bold space

77. $bold 2 to the power of bold cos bold 2 bold x end exponent bold space bold equals bold space bold 3 bold times bold 2 to the power of bold cos to the power of bold 2 bold space bold x end exponent bold space bold 4$ હોય તો x = ....... જ્યાં x ∈ [0, $bold pi$ ]

0
$bold pi$
$bold pi over bold 4$
$bold pi over bold 2$

78. જો સમીકરણ x² + px + q = 0 નાં બીજ p અને q હોય, તો p નીચે કિંમત ....... હોઈ શકે.

-2
1
0
$fraction numerator bold minus bold 1 over denominator bold 2 end fraction$

79.

વાસ્તવિક સહગુણકોવાળું દ્વિઘાત સમીકરણ p(x) = 0 માત્ર શુદ્વ કાલ્પનિક બીજ ધરાવે તો સમીકરણ p(p(x))=0 માટે નીચેનામાંથી કયું સત્ય બને ?

માત્ર એક જ કાલ્પનિક બીજ થાય.
ન કોઈ વાસ્તવિક કે ન હોઈ કાલ્પનિક બીજ હોય.
બધાં જ વાસ્તવિક બીજ હોય.
બે વાસ્તવિક તથા બે કાલ્પનિક બીજ હોય.

80.

જો સમીકરણ px² + qx + r = 0 નાં બીજ α અને β હોય (જ્યાં p ≠ 0) તથા p, q, r સમાંતર શ્રેણીમાં હોય તેમજ $bold 1 over bold alpha bold plus bold 1 over bold beta bold space bold equals bold space bold 4$ હોય તો |α -β| = ........ .

$fraction numerator bold 2 square root of bold 13 over denominator bold 9 end fraction$
$fraction numerator square root of bold 34 over denominator bold 9 end fraction$
$fraction numerator bold 2 square root of bold 17 over denominator bold 9 end fraction$
$fraction numerator square root of bold 61 over denominator bold 9 end fraction$

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Gujarati JEE Mathematics : દ્વિઘાત સમીકરણ

Multiple Choice Questions

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