Compute α \alpha α
We need
α = lim x → π / 4 tan 3 x − tan x cos ( x + π 4 ) . \alpha=\lim_{x\to \pi/4}\frac{\tan^3 x-\tan x}{\cos\left(x+\frac\pi4\right)}. α = x → π /4 lim cos ( x + 4 π ) tan 3 x − tan x . First, factor the numerator:
tan 3 x − tan x = tan x ( tan 2 x − 1 ) . \tan^3 x-\tan x=\tan x(\tan^2 x-1). tan 3 x − tan x = tan x ( tan 2 x − 1 ) . Using
tan 2 x − 1 = sin 2 x − cos 2 x cos 2 x = − cos 2 x cos 2 x , \tan^2 x-1=\frac{\sin^2 x-\cos^2 x}{\cos^2 x}=-\frac{\cos 2x}{\cos^2 x}, tan 2 x − 1 = cos 2 x sin 2 x − cos 2 x = − cos 2 x cos 2 x , and tan x = sin x cos x \tan x=\dfrac{\sin x}{\cos x} tan x = cos x sin x
tan 3 x − tan x = sin x cos x ⋅ ( − cos 2 x cos 2 x ) = − sin x cos 2 x cos 3 x . \tan^3 x-\tan x
=\frac{\sin x}{\cos x}\cdot \left(-\frac{\cos 2x}{\cos^2 x}\right)
=-\frac{\sin x\cos 2x}{\cos^3 x}. tan 3 x − tan x = cos x sin x ⋅ ( − cos 2 x cos 2 x ) = − cos 3 x sin x cos 2 x . Also,
cos ( x + π 4 ) → cos ( π 4 + π 4 ) = cos π 2 = 0 , \cos\left(x+\frac\pi4\right)\to \cos\left(\frac\pi4+\frac\pi4\right)=\cos\frac\pi2=0, cos ( x + 4 π ) → cos ( 4 π + 4 π ) = cos 2 π = 0 , so this is a 0 / 0 0/0 0/0
A cleaner method is to rewrite near x = π / 4 x=\pi/4 x = π /4
tan 3 x − tan x = tan x ( tan x − 1 ) ( tan x + 1 ) . \tan^3 x-\tan x=\tan x(\tan x-1)(\tan x+1). tan 3 x − tan x = tan x ( tan x − 1 ) ( tan x + 1 ) . Thus
α = lim x → π / 4 tan x ( tan x + 1 ) ⋅ tan x − 1 cos ( x + π / 4 ) . \alpha=\lim_{x\to \pi/4}\tan x(\tan x+1)\cdot \frac{\tan x-1}{\cos(x+\pi/4)}. α = x → π /4 lim tan x ( tan x + 1 ) ⋅ cos ( x + π /4 ) tan x − 1 . As x → π / 4 x\to \pi/4 x → π /4 tan x → 1 \tan x\to 1 tan x → 1
tan x ( tan x + 1 ) → 1 ⋅ 2 = 2. \tan x(\tan x+1)\to 1\cdot 2=2. tan x ( tan x + 1 ) → 1 ⋅ 2 = 2. Hence
α = 2 lim x → π / 4 tan x − 1 cos ( x + π / 4 ) . \alpha=2\lim_{x\to \pi/4}\frac{\tan x-1}{\cos(x+\pi/4)}. α = 2 x → π /4 lim cos ( x + π /4 ) tan x − 1 . Now,
tan x − 1 = sin x − cos x cos x . \tan x-1=\frac{\sin x-\cos x}{\cos x}. tan x − 1 = cos x sin x − cos x . Also,
sin x − cos x = 2 sin ( x − π 4 ) , \sin x-\cos x=\sqrt2\sin\left(x-\frac\pi4\right), sin x − cos x = 2 sin ( x − 4 π ) , and
cos ( x + π 4 ) = − sin ( x − π 4 ) . \cos\left(x+\frac\pi4\right)=-\sin\left(x-\frac\pi4\right). cos ( x + 4 π ) = − sin ( x − 4 π ) . So
\frac{\tan x-1}{\cos(x+\pi/4)}
= rac{\sqrt2\sin(x-\pi/4)}{\cos x\,[-\sin(x-\pi/4)]}
=-\frac{\sqrt2}{\cos x}.
Taking limit as x → π / 4 x\to \pi/4 x → π /4
lim x → π / 4 tan x − 1 cos ( x + π / 4 ) = − 2 cos ( π / 4 ) = − 2 1 / 2 = − 2. \lim_{x\to \pi/4}\frac{\tan x-1}{\cos(x+\pi/4)}
=-\frac{\sqrt2}{\cos(\pi/4)}
=-\frac{\sqrt2}{1/\sqrt2}=-2. x → π /4 lim cos ( x + π /4 ) tan x − 1 = − cos ( π /4 ) 2 = − 1/ 2 2 = − 2. Therefore,
α = 2 ( − 2 ) = − 4. \alpha=2(-2)=-4. α = 2 ( − 2 ) = − 4.
Compute β \beta β
We need
β = lim x → 0 ( cos x ) cot x . \beta=\lim_{x\to 0}(\cos x)^{\cot x}. β = x → 0 lim ( cos x ) c o t x . This is of the form 1 ∞ 1^\infty 1 ∞
ln β = lim x → 0 cot x ln ( cos x ) . \ln \beta=\lim_{x\to 0}\cot x\,\ln(\cos x). ln β = x → 0 lim cot x ln ( cos x ) . Use standard expansions near x = 0 x=0 x = 0
ln ( cos x ) ∼ − x 2 2 , cot x ∼ 1 x . \ln(\cos x)\sim -\frac{x^2}{2},
\qquad
\cot x\sim \frac1x. ln ( cos x ) ∼ − 2 x 2 , cot x ∼ x 1 . Hence
ln β ∼ 1 x ( − x 2 2 ) = − x 2 → 0. \ln \beta \sim \frac1x\left(-\frac{x^2}{2}\right)=-\frac x2\to 0. ln β ∼ x 1 ( − 2 x 2 ) = − 2 x → 0. So
ln β = 0 ⇒ β = e 0 = 1. \ln\beta=0 \quad\Rightarrow\quad \beta=e^0=1. ln β = 0 ⇒ β = e 0 = 1. Thus,
β = 1. \beta=1. β = 1.
Form the quadratic using roots α , β \alpha,\beta α , β
The roots of
a x 2 + b x − 4 = 0 ax^2+bx-4=0 a x 2 + b x − 4 = 0 are α = − 4 \alpha=-4 α = − 4 β = 1 \beta=1 β = 1
For a quadratic a x 2 + b x − 4 = 0 ax^2+bx-4=0 a x 2 + b x − 4 = 0 r 1 , r 2 r_1,r_2 r 1 , r 2
r 1 + r 2 = − b a , r 1 r 2 = − 4 a . r_1+r_2=-\frac ba,
\qquad
r_1r_2=\frac{-4}{a}. r 1 + r 2 = − a b , r 1 r 2 = a − 4 . Now,
α + β = − 4 + 1 = − 3 , \alpha+\beta=-4+1=-3, α + β = − 4 + 1 = − 3 , α β = ( − 4 ) ( 1 ) = − 4. \alpha\beta=(-4)(1)=-4. α β = ( − 4 ) ( 1 ) = − 4. From product:
− 4 a = − 4 ⇒ a = 1. \frac{-4}{a}=-4 \Rightarrow a=1. a − 4 = − 4 ⇒ a = 1. Then from sum:
− b a = − 3 ⇒ − b = − 3 ⇒ b = 3. -\frac ba=-3 \Rightarrow -b=-3 \Rightarrow b=3. − a b = − 3 ⇒ − b = − 3 ⇒ b = 3. So
( a , b ) = ( 1 , 3 ) . (a,b)=(1,3). ( a , b ) = ( 1 , 3 ) .
Check options
A: ( 1 , − 3 ) (1,-3) ( 1 , − 3 )
B: ( − 1 , 3 ) (-1,3) ( − 1 , 3 )
C: ( − 1 , − 3 ) (-1,-3) ( − 1 , − 3 )
D: ( 1 , 3 ) (1,3) ( 1 , 3 )
Therefore, the correct option is D .