Matematica - Ecuatii si inecuatii. Functii

Ecuatii si inecuatii. Functii - Intrebarea 21

Mergi la:

Să se determine $a element of straight real numbers backslash open curly brackets 1 close curly brackets$ pentru care parabola $y equals left parenthesis a minus 1 right parenthesis x squared plus a x minus 2$ şi dreapta de ecuaţie $y equals x minus 1$ au două puncte distincte comune.

a element of open parentheses negative infinity comma negative 3 close parentheses union open parentheses 1 comma plus infinity close parentheses

a element of open parentheses negative infinity comma negative 1 close parentheses union open parentheses 3 comma plus infinity close parentheses

open parentheses negative 3 semicolon 1 close parentheses

Raspunsul este corect.

Raspunsul este gresit.

Explicatie:

Răspuns corect: A

Punem condiţia ca ecuaţia $left parenthesis a minus 1 right parenthesis x squared plus a x minus 2 equals x minus 1$ să aibă două soluţii reale distincte.

Ecuaţia devine: $left parenthesis a minus 1 right parenthesis x squared plus left parenthesis a minus 1 right parenthesis x minus 1 equals 0$

Punem condiţia: $increment greater than 0$

$rightwards double arrow left parenthesis a minus 1 right parenthesis squared plus 4 left parenthesis a minus 1 right parenthesis greater than 0 left right double arrow left parenthesis a minus 1 right parenthesis left parenthesis a plus 3 right parenthesis greater than 0 left right double arrow a element of open parentheses negative infinity comma negative 3 close parentheses union open parentheses 1 comma plus infinity close parentheses$