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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

71. જો f(x) = ax² + bx જ્યાં f(1) = 2, f(i²)=0 અને f(z+z¹) = 0 તો |z| = ......

Tips: -

f(x) = ax² + bx ...(1)

f(1) = 2. આથી a + b = 2 ... (2)

(1) અને (2) પરથી, a = 1, b = 1

∴ f(x) = x² + x

f(z+z^-1) = 0, આથી, $open parentheses bold z bold plus bold 1 over bold 2 close parentheses to the power of bold 2 bold space bold plus bold space open parentheses bold z bold plus bold 1 over bold 2 close parentheses bold space bold equals bold space bold 0 bold space$

$bold therefore bold space bold z to the power of bold 4 bold space bold plus bold space bold z to the power of bold 3 bold space bold plus bold space bold 2 bold z to the power of bold 2 bold space bold plus bold space bold z bold space bold plus bold space bold 1 bold space bold equals bold space bold 0 bold space bold therefore bold space bold left parenthesis bold z to the power of bold 2 bold space bold plus bold space bold 1 bold right parenthesis bold space bold left parenthesis bold z to the power of bold 2 bold space bold plus bold space bold z bold space bold plus bold space bold 1 bold space bold right parenthesis bold space bold equals bold space bold 0 bold space bold therefore bold space bold z bold space bold equals bold space bold plus-or-minus bold space bold i bold comma bold space bold w bold comma bold space bold w to the power of bold 2 bold space bold therefore bold space bold vertical line bold z bold vertical line bold space bold equals bold space bold 1 bold space$

72.

P-iv, Q-iii, R-ii, S-i
P-iv, Q-ii, R-iii, S-i
P-i, Q-iv, R-ii, S-iii
P-iv, Q-ii, R-i, S-iii

73. નીચેના ડાબી બાજુના સમીકરણને જમણી બાજુ દર્શાવતા વક્ર સાથે જોડો.

P-iv, Q-ii, R-i, S-i
P-iv, Q-ii, R-iii, S-i
P-iv, Q-iii, R-ii, S-i
P-i, Q-iv, R-ii, S-iii

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Gujarati JEE Mathematics : સંકર સંખ્યાઓ

Multiple Choice Questions

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