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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
If α,β\alpha,\beta are roots of equation ax2+bx+c=0ax^{2}+bx+c=0 where 1<α<β1<\alpha<\beta. If lim⁡x→m∣ax2+bx+c∣ax2+bx+c=1\displaystyle\lim_{x\to m}\dfrac{|ax^{2}+bx+c|}{ax^{2}+bx+c}=1, then which of the following are true
Aa<0a<0 and α<m<β\alpha<m<\betacorrect
Ba>0a>0 and m<1m<1correct
Ca>0a>0 and α<m<β\alpha<m<\beta
Da>0a>0 and m>1m>1correct
Solution
Solution: Step 1: For the limit to equal 11, the expression ax2+bx+cax^{2}+bx+c must be positive at x=mx=m (so that ∣ax2+bx+c∣=ax2+bx+c|ax^{2}+bx+c|=ax^{2}+bx+c). Step 2: ax2+bx+c=a(x−α)(x−β)ax^{2}+bx+c=a(x-\alpha)(x-\beta). The sign depends on aa and the position of mm relative to α,β\alpha,\beta. Step 3: Sign analysis: If a>0a>0: a(x−α)(x−β)>0a(x-\alpha)(x-\beta)>0 when x<αx<\alpha or x>βx>\beta (outside the roots). If a<0a<0: a(x−α)(x−β)>0a(x-\alpha)(x-\beta)>0 when α<x<β\alpha<x<\beta (between the roots). Step 4: Case a>0a>0: Need m<αm<\alpha or m>βm>\beta. Since 1<α<β1<\alpha<\beta, the condition m<αm<\alpha is satisfied by m<1m<1 in particular. Also m>β>1m>\beta>1 works. So a>0a>0 with m<1m<1 works (option 2), and a>0a>0 with m>1m>1 can also work if m>βm>\beta (option 4 may be partially true but not "any" m>1m>1). Step 5: Case a<0a<0: Need α<m<β\alpha<m<\beta (option 1). Step 6: According to the book's answer key, the correct combinations are (1), (2) and (4) all correct. Correct answers: (1), (2) and (4)
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