Question
Mathematics Question on Quadratic Equations
Let a>0 be a root of the equation 2x2+x−2=0. If limx→a11−ax216(1−cos(2+x−2x2))=α+β17, where α,β∈Z, then α+β is equal to _____.
Given the quadratic equation:
2x2+x−2=0.
To find the roots, use the quadratic formula:
x=2a−b±b2−4acwith a=2,b=1,c=−2.
Substituting the values:
x=4−1±1+16=4−1±17.
Since a>0, we take the positive root:
a=4−1+17.
Step 1: Simplifying the Given Limit
Consider the expression:
limx→1/a16(1−ax21−cos(2+x−2x2)).
Substitute x=1/a into the expression and expand using trigonometric identities and series expansions near x=1/a. Detailed calculations lead to the simplified form:
limx→1/a16(a22(a1−b1)2).
Step 2: Calculating Constants
From the quadratic equation roots:
a=4−1+17,b=4−1−17.
Using these values:
a1=−1+174,b1=−1−174.
The difference:
a1−b1=17817.
Step 3: Final Expression for the Limit
Substituting these values into the limit expression:
limx→1/a16(a22(17817)2)=α+β17.
Simplifying yields:
α=153,β=17.
Step 4: Calculating α+β
α+β=153+17=170.
Therefore, the correct answer is 170.