اختبار اختيار من متعدد تفاعلي: هيكل الفصل الدراسي الثالث 2025 - 2026 للصف العاشر متقدم

أ

$[f \circ g](x) = 6x - 2; [g \circ f](x) = 6x - 1$

ب

$[f \circ g](x) = 6x - 1; [g \circ f](x) = 6x - 2$

ج

$[f \circ g](x) = 5x - 1; [g \circ f](x) = 5x - 2$

د

$[f \circ g](x) = 6x; [g \circ f](x) = 6x - 3$

أ

$[f \circ g](x) = 2x^2 + 2; [g \circ f](x) = 2x^2 + 20x + 47$

ب

$[f \circ g](x) = 2x^2 - 3; [g \circ f](x) = 2x^2 + 10x + 22$

ج

$[f \circ g](x) = 2x^2 + 5; [g \circ f](x) = 2x^2 + 20x + 50$

د

$[f \circ g](x) = 2x^2 + 2; [g \circ f](x) = 2x^2 + 10x + 47$

أ

$[f \circ g](x) = x^2 + 2x; [g \circ f](x) = x^2$

ب

$[f \circ g](x) = x^2; [g \circ f](x) = x^2 + 2x$

ج

$[f \circ g](x) = x^2 - 2x; [g \circ f](x) = x^2 - 1$

د

$[f \circ g](x) = x^2 + x; [g \circ f](x) = x^2 + 1$

أ

$[f \circ g](x) = 3x^2 + 4; [g \circ f](x) = 9x^2 - 12x + 6$

ب

$[f \circ g](x) = 3x^2 + 6; [g \circ f](x) = 9x^2 - 12x + 4$

ج

$[f \circ g](x) = 3x^2 - 2; [g \circ f](x) = 9x^2 - 6x + 6$

د

$[f \circ g](x) = 3x^2 + 4; [g \circ f](x) = 3x^2 - 12x + 6$

أ

$[f \circ g](x) = 2x^2 - 8x + 7; [g \circ f](x) = 4x^2 - 20x + 26$

ب

$[f \circ g](x) = 2x^2 - 8x + 10; [g \circ f](x) = 4x^2 - 12x + 14$

ج

$[f \circ g](x) = 2x^2 - 4x + 2; [g \circ f](x) = 4x^2 - 20x + 20$

د

$[f \circ g](x) = 2x^2 - 8x + 7; [g \circ f](x) = 4x^2 - 16x + 26$

أ

$[f \circ g](x) = 9x^2 - 12x + 5; [g \circ f](x) = 3x^2 + 1$

ب

$[f \circ g](x) = 9x^2 - 12x + 4; [g \circ f](x) = 3x^2 + 3$

ج

$[f \circ g](x) = 3x^2 - 12x + 5; [g \circ f](x) = 9x^2 + 1$

د

$[f \circ g](x) = 9x^2 - 6x + 5; [g \circ f](x) = 3x^2 - 1$

أ

$[f \circ g](x) = \frac{1}{2x+5}; [g \circ f](x) = \frac{2}{x} + 5$

ب

$[f \circ g](x) = \frac{1}{x} + 5; [g \circ f](x) = \frac{1}{2x+5}$

ج

$[f \circ g](x) = \frac{2}{2x+5}; [g \circ f](x) = \frac{2x+5}{x}$

د

$[f \circ g](x) = \frac{1}{2x+5}; [g \circ f](x) = \frac{2}{x} + 2$

أ

$[f \circ g](x) = \frac{3}{x+1}; [g \circ f](x) = \frac{3}{x-1} + 2$

ب

$[f \circ g](x) = \frac{3}{x+2}; [g \circ f](x) = \frac{3}{x+1} + 2$

ج

$[f \circ g](x) = \frac{3}{x-1}; [g \circ f](x) = \frac{2x+3}{x-1}$

د

$[f \circ g](x) = \frac{3}{x+1}; [g \circ f](x) = \frac{3}{x} + 2$

أ

$[f \circ g](x) = x; [g \circ f](x) = x$

ب

$[f \circ g](x) = x^2; [g \circ f](x) = x^2$

ج

$[f \circ g](x) = \sqrt{x^2 - 4}; [g \circ f](x) = x + 2$

د

$[f \circ g](x) = x - 2; [g \circ f](x) = x + 2$

أ

$[f \circ g](x) = x; [g \circ f](x) = x$

ب

$[f \circ g](x) = x^2; [g \circ f](x) = x^2$

ج

$[f \circ g](x) = \sqrt{x^2 + 6}; [g \circ f](x) = x - 3$

د

$[f \circ g](x) = x + 3; [g \circ f](x) = x - 3$

أ

$f^{-1}(x) = \frac{x+4}{3}$

ب

$f^{-1}(x) = \frac{x-4}{3}$

ج

f^-1(x) = 3x + 4

د

$f^{-1}(x) = \frac{1}{3}x - 4$

أ

$f^{-1}(x) = \frac{3-x}{2}$

ب

$f^{-1}(x) = \frac{x-3}{2}$

ج

f^-1(x) = -2x - 3

د

f^-1(x) = 2x + 3

أ

f^-1(x) = 2x - 4

ب

f^-1(x) = 2x + 4

ج

$f^{-1}(x) = \frac{1}{2}x - 2$

د

f^-1(x) = -2x + 4

أ

f^-1(x) = -3x - 3

ب

f^-1(x) = -3x + 3

ج

$f^{-1}(x) = -\frac{1}{3}x + 1$

د

f^-1(x) = 3x - 3

أ

$f^{-1}(x) = \sqrt{x+3}, x \ge -3$

ب

$f^{-1}(x) = \sqrt{x-3}, x \ge 3$

ج

$f^{-1}(x) = x^2 + 3, x \ge 0$

د

$f^{-1}(x) = -\sqrt{x+3}, x \ge -3$

أ

$f^{-1}(x) = x^2 - 2, x \ge 0$

ب

$f^{-1}(x) = x^2 + 2, x \ge 0$

ج

$f^{-1}(x) = \sqrt{x-2}, x \ge 2$

د

f^-1(x) = x² - 2

أ

$f^{-1}(x) = \pm \sqrt{x+3}$

ب

$f^{-1}(x) = \pm \sqrt{x-3}$

ج

$f^{-1}(x) = \sqrt{x+3}$

د

f^-1(x) = x² + 3