Question
The equation $|x+1|^{\log_{x+1}(3+2x-x^2)} = (x-3)|x|$ has
This question was previously asked in
WBJEE 2026 Physics and Chemistry Question Paper (24-May-2026)
The correct answer is
no solution
To solve the equation \(|x+1|^{\log_{x+1}(3+2x-x^2)} = (x-3)|x|\), we need to break down the equation and analyze the conditions involved.
- First, consider the domain of the logarithmic and absolute value expressions:
- The expression \(\log_{x+1}(3 + 2x - x^2)\) is defined only if \(x+1 > 0\) and \(3 + 2x - x^2 > 0\).
- The condition \(x+1 > 0\) simplifies to \(x > -1\).
- For \(3 + 2x - x^2 > 0\), rewrite the quadratic as \(-x^2 + 2x + 3\). Solving \(3 + 2x - x^2 = 0\) gives the roots \(x=3\) and \(x=-1\). The inequality holds for the interval \(-1 < x < 3\).
- Analyze the right-hand side of the equation:
- The expression \((x-3)|x|\) is defined for all \(x\).
- However, note that for \(-1 < x \leq 0\), \(|x| = -x\), and for \(x > 0\), \(|x| = x\).
- Examine if there can be any solutions within the domain \(-1 < x < 3\):
- If \(x=0\), both sides are zero, but the logarithm becomes undefined due to division by zero. Hence, \(x \neq 0\).
- For \(0 < x < 3\), evaluate:
- The left-hand side is \((x+1)^{\log_{x+1}(3+2x-x^2)}\), which is positive for \(0 < x < 3\).
- The right-hand side \((x-3)x\) is negative for \(x < 3\).
- Conclusion:
- Since there is no overlap between the signs of the left-hand and right-hand sides, the equation has no solutions within the permissible domain.
Therefore, the equation \(|x+1|^{\log_{x+1}(3+2x-x^2)}=(x-3)|x|\) indeed has no solution.
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