Hiya, i've got an A-level question that I am a little stuck with. I was wondering if anyone would know how to do it...
f(x) = 2secx + 2x - 3, where x is in radians.
(a). Evaluate f(0.4) and f(0.5) and deduce the equation f(x) = 0 has a solution in the interval 0.4
^^^^I've done the above by substituting 0.4 and 0.5 into f(x). One value was negative and the other was positive, therefore, I know that the solution lies in the interval 0.4
However, I have tried, but cannot do the next part, part b (below):
(b) Show that the equation f(x) = 0 can be arranged in the form x = p + q/cosx, where p and q are constants, and state the value of p and q.
Does anyone know how to do part (b)? I would be very grateful for any advice and any help would be very much appreciated! Thankyou xx
f(x) = 2secx + 2x - 3, where x is in radians.
(a). Evaluate f(0.4) and f(0.5) and deduce the equation f(x) = 0 has a solution in the interval 0.4
^^^^I've done the above by substituting 0.4 and 0.5 into f(x). One value was negative and the other was positive, therefore, I know that the solution lies in the interval 0.4
However, I have tried, but cannot do the next part, part b (below):
(b) Show that the equation f(x) = 0 can be arranged in the form x = p + q/cosx, where p and q are constants, and state the value of p and q.
Does anyone know how to do part (b)? I would be very grateful for any advice and any help would be very much appreciated! Thankyou xx
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your reasoning in part a is correct.
b) If f(x) = 0
then 2secx + 2x - 3 = 0
2x = -2secx +3
x = - 1/cosx + 3/2 ( secx = 1/cosx )
this is of the required form and p = 3/2 and q = -1
This is the required form and p = -1 and q = 3/2
b) If f(x) = 0
then 2secx + 2x - 3 = 0
2x = -2secx +3
x = - 1/cosx + 3/2 ( secx = 1/cosx )
this is of the required form and p = 3/2 and q = -1
This is the required form and p = -1 and q = 3/2