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1.Relation and Function
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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
A
$16$
B
$18$
C
$22$
D
$24$
(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