Verified Solution • 2025–26 Session moderate

QJEEM23Ma644f: JEE Main 2023 Mathematics - Class 11 JEEM23Ma644f Solution: Let be a real number. Let be the roots of the equation and be…

Let λ≠0\lambda \ne 0 be a real number. Let α,β\alpha,\beta be the roots of the equation 14x2−31x+3λ=014{x^2} - 31x + 3\lambda = 0 and α,γ\alpha,\gamma be the roots of the equation 35x2−53x+4λ=035{x^2} - 53x + 4\lambda = 0 . Then 3αβ{{3\alpha } \over \beta } and 4αγ{{4\alpha } \over \gamma } are the roots of the equation

Multiple Choice Options

  • A 7x2−245x+250=07{x^2} - 245x + 250 = 0
  • B 49x2−245x+250=049{x^2} - 245x + 250 = 0
  • C 49x2+245x+250=049{x^2} + 245x + 250 = 0
  • D 7x2+245x−250=07{x^2} + 245x - 250 = 0

Step-by-Step Verified Solution

14x2−31x+3λ=014 x^{2}-31 x+3 \lambda=0

amp;α+β=3114….(1) and αβ=3λ14...(2)amp;35x2−53x+4λ=0amp;α+γ=5335…(3) and αγ=4λ35…(4)amp;(2)(4)⇒βγ=3×354×14=158⇒β=158γ\begin{aligned}& \alpha+\beta=\frac{31}{14} \ldots .(1) \text { and } \alpha \beta=\frac{3 \lambda}{14}\quad...(2) \\\\& 35 x^{2}-53 x+4 \lambda=0 \\\\& \alpha+\gamma=\frac{53}{35} \ldots(3) \text { and } \alpha \gamma=\frac{4 \lambda}{35} \quad\ldots(4) \\\\& \frac{(2)}{(4)} \Rightarrow \frac{\beta}{\gamma}=\frac{3 \times 35}{4 \times 14}=\frac{15}{8} \Rightarrow \beta=\frac{15}{8} \gamma\end{aligned}

(1) −(3)⇒β−γ=3114−5335=155−10670=710-(3) \Rightarrow \beta-\gamma=\frac{31}{14}-\frac{53}{35}=\frac{155-106}{70}=\frac{7}{10}

158γ−γ=710⇒γ=45\frac{15}{8} \gamma-\gamma=\frac{7}{10} \Rightarrow \gamma=\frac{4}{5}

⇒β=158×45=32\Rightarrow \beta=\frac{15}{8} \times \frac{4}{5}=\frac{3}{2}

⇒α=3114−β=3114−32=57\Rightarrow \alpha=\frac{31}{14}-\beta=\frac{31}{14}-\frac{3}{2}=\frac{5}{7}

⇒λ=143αβ=143×57×32=5\Rightarrow \lambda=\frac{14}{3} \alpha \beta=\frac{14}{3} \times \frac{5}{7} \times \frac{3}{2}=5

so, sum of roots 3αβ+4αγ=(3αγ+4αββγ)\frac{3 \alpha}{\beta}+\frac{4 \alpha}{\gamma}=\left(\frac{3 \alpha \gamma+4 \alpha \beta}{\beta \gamma}\right)

amp;=(3×4λ35+4×3λ14)βγ=12λ(14+35)14×35βγamp;=49×12×5490×32×45=5\begin{aligned}& =\frac{\left(3 \times \frac{4 \lambda}{35}+4 \times \frac{3 \lambda}{14}\right)}{\beta \gamma}=\frac{12 \lambda(14+35)}{14 \times 35 \beta \gamma} \\\\& =\frac{49 \times 12 \times 5}{490 \times \frac{3}{2} \times \frac{4}{5}}=5\end{aligned}

Product of roots

=3αβ×4αγ=12α2βγ=12×254932×45=25049=\frac{3 \alpha}{\beta} \times \frac{4 \alpha}{\gamma}=\frac{12 \alpha^{2}}{\beta \gamma}=\frac{12 \times \frac{25}{49}}{\frac{3}{2} \times \frac{4}{5}}=\frac{250}{49}

So, required equation is x2−5x+25049=0x^{2}-5 x+\frac{250}{49}=0

⇒49x2−245x+250=0\Rightarrow 49 \mathrm{x}^{2}-245 \mathrm{x}+250=0

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