Let a function f satisfying the relation f(x+8)=f(x) and f(x)= begincasesx2, 0 ≤ x ≤ 2 (4-x)2, 2

Question

Let a function f satisfying the relation f(x+8)=f(x) and f(x)= begincasesx2, 0 ≤ x ≤ 2 (4-x)2, 2< x ≤ 4 2(x-4), 4< x< 6 -2(x-8), 6 ≤ x ≤ 8 endcases. The value of f(19) + f(63)+f(99) + f(-73) is equal to (A) 1 (B) 2 (C) 4 (D) 6

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/4b90bf04f3708ffb027a2f8f8334e48f-.png" /> f(3) + f(7) + f(3) + f(7) = 2 [f(3) + f(7)] = 2 [1 + 2] = 6.

Q. Let a function $f$ satisfying the relation $f (x + 8) = f (x)$ and $f (x) = ⎩ ⎨ ⎧ x^{2}, (4 - x)^{2}, 2 (x - 4), - 2 (x - 8), 0 \leq x \leq 2 2 < x \leq 4 4 < x < 6 6 \leq x \leq 8$ .
The value of $f (19) + f (63) + f (99) + f (- 73)$ is equal to

1

2

4

6

$f (3) + f (7) + f (3) + f (7)$
$= 2 [f (3) + f (7)]$
$= 2 [1 + 2] = 6.$

Q. Let a function f satisfying the relation f(x+8)=f(x) and f(x)=⎩⎨⎧​x2,(4−x)2,2(x−4),−2(x−8),​0≤x≤22<x≤44<x<66≤x≤8​. The value of f(19)+f(63)+f(99)+f(−73) is equal to

1

2

4

6

f(3)+f(7)+f(3)+f(7) =2[f(3)+f(7)] =2[1+2]=6.

Q. Let a function $f$ satisfying the relation $f (x + 8) = f (x)$ and $f (x) = ⎩ ⎨ ⎧ x^{2}, (4 - x)^{2}, 2 (x - 4), - 2 (x - 8), 0 \leq x \leq 2 2 < x \leq 4 4 < x < 6 6 \leq x \leq 8$ .
The value of $f (19) + f (63) + f (99) + f (- 73)$ is equal to

$f (3) + f (7) + f (3) + f (7)$
$= 2 [f (3) + f (7)]$
$= 2 [1 + 2] = 6.$