Question

f:[-5,9] R என்ற சார்பானது பின்வருமாறு வரையறுக்கப்படுகிறது.
F(x)= { 6x+1 -5≤x<2
5x^2 -1 2≤x<6
3x-4 6≤x≤9 }
என வரையறுக்கப்படுகிறது எனில் பின்வருவனவற்றை காண்க

i) 2 f(4)+f(8) ii) (2f(-2)-f(6))/(f(4)+f(-2))

Answer 1

Explanation:

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Answer 2

விளக்கம்:

$\begin{array}{rr} f(x)= & \\\text { 6 } x+1 & ,-5 \leq x<2 \\5 x^{2}-1 & , 2 \leq x \leq 9 \\3 x-4, & 6 \leq x \leq 9\end{array}$

$\begin{aligned}&f(x)=6 x+1, x=-5,-4,-3,-2,1,0,1\end{aligned}$

$\begin{aligned}&f(x)=5 x^{2}-1\quad x=2,3,4,5,6\\&f(x)=3 x-4\quad x=6,7,8,9\end{aligned}$

$(i) 2 f(4)+f(8)$

$\begin{aligned}&f(x)=5 x^{2}-1 \\&\Rightarrow f(4)=5\left(4^{2}\right)-1\\&=5(16)-1\\&=80-1\\&=79\end{aligned}$

$\begin{aligned}&f(x)=3 x-4\\ &f(8)=3(8)-4=24\\&4=20\end{aligned}$

$\begin{aligned}&\therefore 2 f(4)+f(8)=2(79)+20=158+20=178\end{aligned}$

$2 f(4)+f(8)=178$

$\text { (ii) } \frac{2 f(-2)-f(6)}{f(4)+f(-2)}$

$\begin{aligned}&f(x)=6 x+1\\&f(-2)=6(-2)+1\\&12+1=-11\end{aligned}$

$\begin{aligned}&f(x)=3 x-4 \\ &f(6)=3(6)-4\\&=18-4=14\end{aligned}$

$\begin{aligned}&f(x)=5 x^{2}-1\\&f(4)=5\left(4^{2}\right)-1\\&=5(16)-1\\&=80-1\\&=79\end{aligned}$

$\therefore \frac{2 f(-2)-f(6)}{f(4)+f(-2)}$

$=\frac{2(-11)-14}{79+(-11)}$

$=\frac{-22-14}{79-11}$

$=\frac{-36}{68}$

$\frac{2 f(-2)-f(6)}{f(4)+f(-2)}=\frac{-9}{17}$