Define a function f(x)=frac{16 x^2-96 x+153}{x-3} for all real x ≠ 3 . least positive value of f(x

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1.Relation and Function

normal

Define a function $f(x)=\frac{16 x^2-96 x+153}{x-3}$ for all real $x \neq 3$. The least positive value of $f(x)$ is

$16$

$18$

$22$

$24$

(KVPY-2017)

Solution

(d)

Given,

$f(x)=\frac{16 x^2-96 x+153}{x-3}$

Let $\quad f(x)=y$

$\therefore \quad y=\frac{16 x^2-96 x+153}{x-3}$

$\Rightarrow \quad x y-3 y=16 x^2-96 x+153$

$\Rightarrow 16 x^2-x(96+y)+153+3 y=0$

$\therefore \quad x \in R$

$\therefore \quad D \geq 0$

$\because(96+y)^2-4 \times 16(153+3 y) \geq 0$

$(96)^2+192 y+y^2-64(153)-192 y \geq 0$

$\quad y^2 \geq 64(153)-96^2$

$\quad y^2 \geq 9792-9216$

$\quad y^2 \geq 576$

$\therefore \quad y \in(-\infty,-24] \cup[24, \infty)$

$\therefore \text { Least positive value of } f(x)=24$

Standard 12

Mathematics

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hard

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hard

(JEE MAIN-2022)

Define a function $f(x)=\frac{16 x^2-96 x+153}{x-3}$ for all real $x \neq 3$. The least positive value of $f(x)$ is

Solution

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Consider the function $f (x) = x^3 – 8x^2 + 20x -13$
Number of positive integers $x$ for which $f (x)$ is a prime number, is