Question

If \(\underset{x\rightarrow 1}{lim}\frac{sin(3x^2-4x+1)-x^2+1}{2x^3-7x^2+ax+b} = -2\), then the quadratic equation having roots \(a\) and \(b\) is

• A. \(x^2+5x-24 = 0\)
• B. \(x^2-5x+24 = 0\)
• C. \(x^2-5x-24 = 0\)
• D. \(x^2+5x+24 = 0\)

Hint:

The denominator must vanish at \(x=1\) twice for the limit to be finite and nonzero.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
At \(x=1\) the numerator \(\sin(3-4+1)-1+1=0\). For a finite limit the denominator must also be 0, giving a first condition on \(a\) and \(b\).

Step 2: Key Formula or Approach
Use L'Hopital repeatedly until the limit is determined.

Step 3: Detailed Explanation
Denominator zero at \(x=1\): \(2-7+a+b=0\), so \(a+b=5\).
First derivatives: numerator \(\cos(3x^2-4x+1)(6x-4)-2x\) equals \(2-2=0\) at \(x=1\). Denominator \(6x^2-14x+a\) equals \(a-8\). For the ratio not to be \(0/(a-8)=0\) (we need \(-2\)), we need \(a=8\), so the denominator derivative also vanishes. Then \(b=-3\).
Second derivatives at \(x=1\): numerator \(= -\sin(0)(2)^2+6\cos0-2=4\). Denominator \(=12-14=-2\). Ratio \(=-2\), which matches the given value.
Quadratic with roots \(8\) and \(-3\): sum 5, product \(-24\):
\[ x^2-5x-24=0 \]

Final Answer:
The roots are \(a=8\) and \(b=-3\), so the equation is \(x^2-5x-24=0\), option (C). \[ \boxed{x^2-5x-24=0\ \text{(C)}} \]