The two quadratics x2 - 2qx - 6 and x2 - 8x + 9q both have the same remainder wh

Question

The two quadratics x2 - 2qx - 6 and x2 - 8x + 9q both have the same remainder when divided by x + 3. Find the value of q. Factorise first and then solve the equation 3x3 - 7x2 - 7x + 3 = 0. Consider the rational function Write out the form of the partial fraction decomposition of the function. Determine the numerical values of the coefficients. Is the function f(x) = 3x3 + 2x2 + 1 even, odd, or neither? Please justify your reasoning (not just with a graph). Find the domain of each of the functions: When a camera flash goes off, the batteries immediately begin to recharge the flash's capacitor, which stores electric charge given by Q(t) =Q0(1 - ) (The maximum charge capacity is Q0 and t is measured in seconds.) Find the inverse of this function and explain its meaning. How long does it take to recharge the capacitor to 90% of capacity if a = 2 Consider two trigonometric functions y = cos 2x and y =1 + sinx. Write down the equation in x that you would solve to find the x coordinate of point(s) of intersection of those graphs on [0, 2 pi] . Solve your equation, and write down the coordinates of the point (s) of intersection . Solve the equation 7x2 = 2 x + l for x.

Explanation / Answer

1)From remainder theorem

R1(-3) =R2(-3)

putting x=-3 in both equation

3+6q=33+9q

=> q=-10 (ans)

2)P(x) =3x^3 - 7x^2 -7x +3=0

P(-1) =0

so from factor theorem

(x+1)is the factor of P(x)

now

P(x) =(x+1) * (3x^2 -10x+3) = (x+1) (3x-1) (x-3) =0

so x= -1,1/3,3 (ans)

3) f(x) =(x^2 - x +6)/(x^3 + 3x ) = A/x + (Bx+C)/(x^2+3) =(A+B)x^2 +Cx +(3A)

equating coefficients

3A =6

A=2

C=-1

B=-1

f(x) =(x^2 - x +6)/(x^3 + 3x ) = A/x + (Bx+C)/(x^2+3) =[2/x - (x+1)/(x^2+3) ] (ans)

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

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

so f(x) is not equal to f(-x) and neither -f(-x)

so f(x) is neither even nor odd.. (ans)

5)a) f(t) =t^2/(t+2)

so t+2 cannot be 0

so t is not equal to -2

domain t= R except -2 or (-infinity ,-2) U (-2,infinity) (ans)

b) g(x) =sqrt(1-x^2) /7x

so x cant be 0 & 1-x^2 >=0 =x^2<=1

or x=[-1,1]

union of both sets

domain =

x = [-1,0) U (0,1] (ans)

6) Q(t) = Qo(1 - e^(-t/a)]

Q(t)/Qo=(1 - e^(-t/a)]

e^(-t/a) = 1 - Q(t)/Qo

e^(t/a) =1/(1 - Q(t)/Qo)

t= a ln [Qo/(1-Q(t)] (ans)

this gives value of t for any Q(t)

b)Q(t) = 0.9 Qo

Q(t) = Qo(1 - e^(-t/a)] =0.9 Qo

(1 - e^(-t/a)] =0.9

e^(-t/a) =0.1

e^(t/a) =10

t=a ln(10) =4.61 sec (ans)

7)cos2x =1+sinx

1 - 2sin^2x =1+sinx

sinx + 2sin^2x =0

sinx (1+2sinx) =0

sinx =0 & sinx=-1/2 (ans)

b) for x=(0 to 2pi)

solution is x={0,pi,7pi/6,11pi/6,2pi} (ans)

8) 7^(x^2) = 2^(x+1)

taking log

x^2 ln7 =(x+1) ln2

solving this

we get

x=0.80 & -0.44 (ans)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.