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Both Roots in (-1,1): Greatest Integral Value of |10 lambda| | JEE

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

Question
Let f(x)=2x2−2(2λ+1)x+λ(λ+1)=0f(x) = 2x^{2} - 2(2\lambda+1)x + \lambda(\lambda+1) = 0, where λ\lambda is a parameter. If both roots of f(x)=0f(x) = 0 belong to (−1,1)(-1, 1), then the greatest integral value of ∣10λ∣\left|10\lambda\right| is
A22
B44
C55
D99correct
Solution
Step 1: Recall the three conditions for both roots of an upward parabola ff to lie in (−1,1)(-1, 1):
(i) D≥0,(ii) f(−1)>0 and f(1)>0,(iii) −1<−b2a<1.\text{(i) } D \ge 0, \qquad \text{(ii) } f(-1) > 0 \ \text{and} \ f(1) > 0, \qquad \text{(iii) } -1 < \frac{-b}{2a} < 1 .
 Step 2: Condition (i) - the discriminant. With a=2a = 2, b=−2(2λ+1)b = -2(2\lambda+1), c=λ(λ+1)c = \lambda(\lambda+1),
D4=(2λ+1)2−2λ(λ+1)=4λ2+4λ+1−2λ2−2λ=2λ2+2λ+1.\frac{D}{4} = (2\lambda+1)^{2} - 2\lambda(\lambda+1) = 4\lambda^{2}+4\lambda+1 - 2\lambda^{2}-2\lambda = 2\lambda^{2} + 2\lambda + 1 .
 Its own discriminant is 4−8=−4<04 - 8 = -4 < 0 and its leading coefficient is positive, so 2λ2+2λ+1>02\lambda^{2}+2\lambda+1 > 0 for every real λ\lambda. Condition (i) holds always. Step 3: Condition (ii), first half.
f(−1)=2+2(2λ+1)+λ(λ+1)=λ2+5λ+4=(λ+1)(λ+4)>0f(-1) = 2 + 2(2\lambda+1) + \lambda(\lambda+1) = \lambda^{2} + 5\lambda + 4 = (\lambda+1)(\lambda+4) > 0
  ⟹  λ<−4  or  λ>−1.\implies \lambda < -4 \ \text{ or } \ \lambda > -1 .
 Step 4: Condition (ii), second half.
f(1)=2−2(2λ+1)+λ(λ+1)=λ2−3λ=λ(λ−3)>0f(1) = 2 - 2(2\lambda+1) + \lambda(\lambda+1) = \lambda^{2} - 3\lambda = \lambda(\lambda - 3) > 0
  ⟹  λ<0  or  λ>3.\implies \lambda < 0 \ \text{ or } \ \lambda > 3 .
 Step 5: Condition (iii) - the vertex.
−b2a=2(2λ+1)4=2λ+12,−1<2λ+12<1  ⟹  −32<λ<12.\frac{-b}{2a} = \frac{2(2\lambda+1)}{4} = \frac{2\lambda+1}{2}, \qquad -1 < \frac{2\lambda+1}{2} < 1 \implies -\frac32 < \lambda < \frac12 .
 Step 6: Intersect the three results. Inside (−32,12)\left(-\frac32, \frac12\right) the branch λ<−4\lambda < -4 is impossible, so Step 3 leaves λ>−1\lambda > -1; Step 4 then leaves λ<0\lambda < 0. Hence
λ∈(−1,0).\lambda \in (-1, 0).
 Step 7: Translate to the required quantity.
∣λ∣∈(0,1)  ⟹  ∣10λ∣∈(0,10).|\lambda| \in (0, 1) \implies \left|10\lambda\right| \in (0, 10).
 Step 8: The greatest integer strictly below 1010 is 99. Answer: (4).
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