Solution
Step 1: Use the root to find a relationship between p and q
Since x=2 is a root of x2+px+q=0, it must satisfy the equation:
Step 2: Simplify the expression inside the cosine function
Let the argument of the cosine be g(x)=x2−4px+q2+8q+16.
Substitute q=−(2p+4) into the constant part q2+8q+16:
Substituting q=−2p−4:
Now, rewrite g(x):
Step 3: Evaluate the limit of f(x)
The function now becomes:
f(x)=(x−2p)41−cos((x−2p)2)
Let t=(x−2p)2. As x→2p, t→0:
Using the standard limit limθ→0θ21−cosθ=21:
Step 4: Apply the Greatest Integer Function
We need to find limx→2p+[f(x)].
From the Taylor expansion of cost=1−2t2+24t4…, we have:
Since t2 is always positive as x→2p+, the value of f(x) is slightly less than 21:
The greatest integer value of any number in the range 0 \leq \text{value} < 1 is 0:
Final Answer
The limit is 0. Therefore, option (1) is correct.
