JEE Main 2024 6 April Morning Shift
Exam JEE Main
Year 2024
Date 6 April
Shift Morning Shift
Subject Mathematics
Chapter Functions JEE Main 2024 · 6 April · Morning Shift · Mathematics · Functions
The function f: R->R, f(x)=(x^2+2 x-15)/(x^2-4 x+9), x ∈ ℝ is
The function
f: R->R, f ( x ) = x 2 + 2 x − 15 x 2 − 4 x + 9 , x ∈ R i s \text{ f: R->R, } f(x)=\frac{x^2+2 x-15}{x^2-4 x+9}, x \in \mathbb{R} is f: R->R, f ( x ) = x 2 − 4 x + 9 x 2 + 2 x − 15 , x ∈ R i s Options A one-one but not onto. B both one-one and onto. C onto but not one-one. D neither one-one nor onto. Correct answer D. neither one-one nor onto.
Step-by-step solution
f ( x ) = ( x + 5 ) ( x − 3 ) x 2 − 4 x + 9 f(x)=\frac{(x+5)(x-3)}{x^2-4 x+9} f ( x ) = x 2 − 4 x + 9 ( x + 5 ) ( x − 3 ) Let g ( x ) = x 2 − 4 x + 9 D < 0 g ( x ) > 0 for x ∈ R \begin{aligned} & \text{ Let } g(x)=x^2-4 x+9 \\ & D < 0 \\ & g(x)>0 \text{\text{ for } } x \in R\end{aligned} Let g ( x ) = x 2 − 4 x + 9 D < 0 g ( x ) > 0 for x ∈ R \therefore\left[ f ( − 5 ) = 0 f ( 3 ) = 0 \begin{array}{l}
\mathrm{f}(-5)=0 \\
\mathrm{f}(3)=0
\end{array} f ( − 5 ) = 0 f ( 3 ) = 0 \right. So,
f ( x ) \mathrm{f}(\mathrm{x}) f ( x ) is many-one.
again,
y x 2 − 4 x y + 9 y = x 2 + 2 x − 15 x 2 ( y − 1 ) − 2 x ( 2 y + 1 ) + ( 9 y + 15 ) = 0 for ∀ x ∈ R ⇒ D ≥ 0 D = 4 ( 2 y + 1 ) 2 − 4 ( y − 1 ) ( 9 y + 15 ) ≥ 0 5 y 2 + 2 y + 16 ≤ 0 ( 5 y − 8 ) ( y + 2 ) ≤ 0 \begin{aligned}
& y x^2-4 x y+9 y=x^2+2 x-15 \\
& x^2(y-1)-2 x(2 y+1)+(9 y+15)=0 \\
& \text{\text{ for } } \forall x \in R \Rightarrow D \geq 0 \\
& D=4(2 y+1)^2-4(y-1)(9 y+15) \geq 0 \\
& 5 y^2+2 y+16 \leq 0 \\
& (5 y-8)(y+2) \leq 0
\end{aligned} y x 2 − 4 x y + 9 y = x 2 + 2 x − 15 x 2 ( y − 1 ) − 2 x ( 2 y + 1 ) + ( 9 y + 15 ) = 0 for ∀ x ∈ R ⇒ D ≥ 0 D = 4 ( 2 y + 1 ) 2 − 4 ( y − 1 ) ( 9 y + 15 ) ≥ 0 5 y 2 + 2 y + 16 ≤ 0 ( 5 y − 8 ) ( y + 2 ) ≤ 0
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