2
2
2.а) ∫ (5 x 4 + 6 x 2 )dx = ( x5 + 2 x3 ) =25+2⋅23–(–1)5–2(–1)3=32+16+1+2=51;
б)
−1
−1 π 6
1
∫ cos 3xdx = 3 sin 3x
0
3. S =
π 6 0
1⎛ π ⎞ 1 = ⎜ sin − sin 0 ⎟ = . 3⎝ 2 ⎠ 3
3π 4
3π 4
0
0
∫ sin xdx = − cos x
= − cos
3π 2 2+ 2 . + cos 0 = +1 = 4 2 2
ПС–15 1
log 2 16
1 2
1
1. 9log 3 6 : 2 2 = 32 log 3 6 : 2log 2 16 2 = 3log 3 6 : 16 2 = 62 : 4 = 36 : 4 = 9. 2. а) lg(2x – 3) = lg(3x – 2); ⎧2 x − 3 > 0 ⎧ x > 1,5 ⎪ ⎪ ; ⎨ x > 2 3 — данная система не имеет решений. ⎨3 x − 2 > 0 ⎪2 x − 3 = 3 x − 2 ⎪⎩ x = −1 ⎩ Ответ: ∅ . б) (0,2)3x–4 = 52–5x; (0,2)3x–4 = (0,2)–(2–5x); 3x – 4 = –2 + 5x; 2x = –2; x = –1. 3. log22x – 2log2x2 > –3; log22x – 4log2x + 3 > 0; log2x=t, тогда t2–4t+3 > 0; (t – 1)(t – 3) > 0; t ∈ (–∞; 1) ∪ (3; ∞); – + + если t = 1, то log2x = 1; log2x = log22; x = 2, t 1 3 если t = 3, то log2x = 3; log2x = 3log22; log2x = log28; x = 8, значит, x ∈ (0; 2) ∪ (8; ∞). ПС–16 1. а) 22x+1 – 5 ⋅ 2x + 2 = 0; 2x = t, тогда 2t2 – 5t + 2 = 0; D = 25 – 16 = 32; 5±3 1 1 t1,2 = ; t1=2; 2x=2; x=1 или t2= ; 2x= ; 2x=2–1; x = –1. Ответ: ±1. 4 2 2 б)
x + 17 − x + 1 = 2 ;
x + 17 = 2 + x + 1 ;
⎧ x + 17 ≥ 0 ⎧⎪ x ≥ −1 ⎪ ≥ −1 ; x = 8. ; ⎨ ; xx + ⎨x + 1 ≥ 0 1= 9 ⎪ x 3 = + 1 ⎪ ⎩ x x x + = + + + + 17 4 4 1 1 ⎩ Ответ: x = 8. 2. lg(x2 – x + 8) > 1; 2 – + ⎪⎧ x − x + 8 > 0 ; ⎨ –1 ⎪⎩lg( x 2 − x + 8) > lg10 x2–x+8 > 0 при любом значении x; x2–x+8 > 10; x2 – x – 2 > 0; (x + 1)(x – 2) > 0;x ∈ (–∞; –1) ∪ (2; +∞).
{
+ 2
x
97