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Gujarati JEE Mathematics : દ્વિઘાત સમીકરણ

Multiple Choice Questions

71.

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| = ........ .

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

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.

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

74. 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.

75. 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

76.

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.

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

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

78. 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 = .......... .

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

bold pi over bold 3

bold pi over 6

Tips: -

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 bold space bold therefore bold space 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 bold times bold 16 to the power of bold minus bold sin to the power of bold 2 bold x end exponent bold space bold equals bold space bold 10 bold therefore bold space open parentheses bold 16 to the power of bold sin to the power of bold 2 bold x end exponent close parentheses to the power of bold 2 bold space bold minus bold space bold 10 bold times 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 bold space bold equals bold space bold 0

ધારો કે $bold 16 to the power of bold sin to the power of bold 2 bold x end exponent bold space bold equals bold space bold m$
∴ m² - 10m + 16 = 0

∴ (m - 8) (m - 2) = 0

∴ m = 8 અથવા m = 2

$bold therefore bold space bold 16 to the power of bold sin to the power of bold 2 bold x end exponent bold space bold equals bold space bold 8$ અથવા $bold 16 to the power of bold sin to the power of bold 2 bold x end exponent bold space bold equals bold space bold 2$

bold therefore bold space bold 2 to the power of bold 4 bold sin to the power of bold 2 bold x end exponent bold space bold equals bold space bold 2 to the power of bold 3

અથવા $bold 2 to the power of bold 4 bold space bold sin to the power of bold 2 bold x end exponent bold space bold equals bold space bold 2$
∴ 4 sin²x = 3 અથવા 4 sin²x = 1. આથી $bold sin bold space bold italic x bold space bold equals bold space bold plus-or-minus bold space fraction numerator square root of bold 3 over denominator bold 2 end fraction$ અથવા $bold sin bold space bold italic x bold space bold plus-or-minus bold space bold 1 over bold 2$
પરંતુ $bold x bold space bold element of bold space bold left square bracket bold 0 bold comma bold space bold pi bold right square bracket$ હોવાથી sin x > 0

$bold therefore bold space bold sin bold space bold x bold space bold equals bold space fraction numerator square root of bold 3 over denominator bold 2 end fraction$ અથવા $bold sin bold space bold italic x bold space bold equals bold space bold 1 over bold 2$

$bold x bold space bold equals bold space bold pi over bold 3$ અથવા $bold x bold space bold equals bold space bold pi over bold 6$

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

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

80. 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]

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Gujarati JEE Mathematics : દ્વિઘાત સમીકરણ

Multiple Choice Questions

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