Question

1. Find the largest possible domain and the range of the function f(x) = 210. x+1 2. Reduce the rational function 15x71x71, into partial fractions. (x+4)(3x-1)2 3. Find the general solutions of the trigonometric equation: 5 sin x – 12 cos x = 13.

Answer 1

Given function is Answe! 22-22-2 las - 2+1 2 +1 Domain : It is the set of input value for which the functim is real and defined. Take the domain denominato bf 1²-20-2 and compare to zero 2+1 = 0 a = -1 The functim domain is x < -1 d) x>-1 functim range: It is set of values of dependent values for which functim is defined. 22 - 2x-2 +1 2 - x2 = 22-2 = yeti) -) 22-28-2 = 4x+ = x²_2X-2-y-xy cay-xy = x² (2+4)2-2-2 =0 for a quadratic equation of the form-aa2fbx+c=0 the discriminant is b2-4aC 62-4²_4.1.(-2-4) =) }+84 +12 20

517?+2y +64 +12 70 =) 7 (4+2) + 6(7+2) 20 . =) (3+6) y + 2)20 y <- 6 rol y = -6 21 3-2 (61) y = -2 :. The range is If(x) =-6 col) fon) 2-2 Given that Anjwer- 15*2-21X+14 (2144) (3 x-132 partial 70 fractions : I = 1 2210+14 3ity + 3x-1 8x + 1)² (ac+4) (3x-172 = 32C-11 20 +(30+4)(32+1 Jart (I+ya, 1542-2121 +4 - T+43(3x-1) ? (actuBLE) =(3X-1)ão +(30+4)(3x+Wa+c+4)ą2 Ic2_210 Hlu put x=-4 into the equation 15 (4) ²_ 21(4) +14 = (30-4) -1% ao toto 338 = 16920

put a = 1/2 in the equation 156)=21&tly = a034-1)tails+4/344) tas (1+y) ty = 93(3+4) values are substitute in equation a = 2 ,92 = 2 we get, 1573-2134 + 14 = 2138-134 + a, (x+4) 3X-1) +2 (altu -4.91 15X2_2134414 = 606 30,+18 7 +119,7-10x +10 15222-202 + 14 = 3 (391+18) + X(110,-10+ Gua, tio). compare both sides 39,+18 =15 3a, = -3 [a, = -1] substitute a,,a2 a3 values in Quation 24+4 - 3x-1 1582-21x tly | (altu) (x-1)?

ULIUU Given that 15 sina-12 cosx =13 Asinx=1-605 15 sinxe = 13+12 cosa squase both sides (15 sinx)² = (13+12 cos?) ² (15 sinse) ? (13 + 12 cosse) 2=0 -(+13+2.6052)2 + (1-copy je to 7_1695 312 Cost-luucos²x + (1-cos²x/22550 3-169 - 312 cost-144 cosa +225-225 cos²x=0 3-369 Cos?2-3120052 +56=0 - put cosaar +369 V2 +3126015620 - 312 ± √312) ² - 4 (369656) VE 2(369) -312 EN 180000 = -2 (26 & 2552) - 123 Cosa = 2(2582-26) 123 corr -2(26+2552) - 123 1 arccos 2(216Y 1 * = arccos ( -2626:425.6) 42-1, a reces(2016-20); .-2(26+ 25/ 21 = arccos 123 +2017