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← JEE Main 2022 · 30 Jul · Shift 2
MathematicsSingle correct+4 / -1Question 61 of 90

JEE Main 2022 · 30 Jul · Shift 2, Mathematics question 61

Let f:R−{5}→Rf:\mathbb R-\{5\}\to\mathbb R be defined as f(x)=2x2+3x−2x−5f(x)=\dfrac{2x^2+3x-2}{x-5}. Then ff is

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Answer and worked solution

Official answer: D

Range. f(x)=yf(x)=y gives 2x2+(3−y)x+(5y−2)=02x^2+(3-y)x+(5y-2)=0, real only when y2−46y+25≥0y^2-46y+25\ge0; values of y between the roots are missed, so f is not onto. A quadratic in x also has two roots for most y, so f is not one-one.

(D)