Mathematics · Straight lines

JEE Main 2024 — 27 January, Shift 2 — Question 4

Let R be the interior region between the lines 3x−y+1=03 x-y+1=0 and x+2y−5=0x+2 y-5=0 containing the origin. The set of all values of aa, for which the points (a2,a+1)\left(a^{2}, a+1\right) lie in RR, is:

  1. Option A:

    (−3,−1)∪(−13,1)(-3,-1) \cup\left(-\frac{1}{3}, 1\right)

  2. Option B:

    (−3,0)∪(13,1)(-3,0) \cup\left(\frac{1}{3}, 1\right)

    Correct
  3. Option C:

    (−3,0)∪(23,1)(-3,0) \cup\left(\frac{2}{3}, 1\right)

  4. Option D:

    (−3,−1)∪(13,1)(-3,-1) \cup\left(\frac{1}{3}, 1\right)

Answer: B

Step-by-step solution

P(a2,a+1)P\left(a^{2}, a+1\right) L1=3x−y+1=0\mathrm{L}_{1}=3 \mathrm{x}-\mathrm{y}+1=0

Origin and P lies same side w.r.t. L1\mathrm{L}_{1} ⇒L1(0).L1(P)>0\Rightarrow \mathrm{L}_{1}(0) . \mathrm{L}_{1}(\mathrm{P})>0

∴3(a2)−(a+1)+1>0\therefore 3\left(a^{2}\right)-(a+1)+1>0

figure

⇒3a2−a>0\Rightarrow 3 \mathrm{a}^{2}-\mathrm{a}>0

a∈(−∞,0)∪(13,∞)\mathrm{a} \in(-\infty, 0) \cup\left(\frac{1}{3}, \infty\right)

Let L2:x+2y−5=0L_{2}: x+2 y-5=0

Origin and P lies same side w.r.t. L2\mathrm{L}_{2}

⇒L2(0)⋅L2(P)>0\Rightarrow \mathrm{L}_{2}(0) \cdot \mathrm{L}_{2}(\mathrm{P})>0

⇒a2+2(a+1)−5<0\Rightarrow \mathrm{a}^{2}+2(\mathrm{a}+1)-5<0

⇒a2+2a−3<0\Rightarrow \mathrm{a}^{2}+2 \mathrm{a}-3<0

⇒(a+3)(a−1)<0\Rightarrow(\mathrm{a}+3)(\mathrm{a}-1)<0

∴a∈(−3,1)\therefore \mathrm{a} \in(-3,1)

Intersection of (1) and (2) a∈(−3,0)∪(13,1)\mathrm{a} \in(-3,0) \cup\left(\frac{1}{3}, 1\right)

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
Straight lines
Topic
Position Of points w.r.t lines.