If |x + 1||x + 3| – 4|x + 2| + 5 = 0, then sum of squares of solutions is
Correct Answer: 16
Solution and Explanation
Top Questions on General and Particular Solutions of a Differential Equation
- Let \( \alpha, \beta (\alpha \neq \beta) \) be the values of m, for which the equations \(x + y + z = 1\), \(x + 2y + 4z = m\), and \(x + 4y + 10z = m^2\) have infinitely many solutions. Then the value of \(\sum_{n=1}^{10} (n^4 + n^8)\) is equal to:
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Let $[r]$ denote the largest integer not exceeding $r$, and the roots of the equation $ 3z^2 + 6z + 5 + \alpha(x^2 + 2x + 2) = 0 $ are complex numbers whenever $ \alpha > L $ and $ \alpha < M $. If $ (L - M) $ is minimum, then the greatest value of $[r]$ such that $ Ly^2 + My + r < 0 $ for all $ y \in \mathbb{R} $ is:
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Questions Asked in JEE Main exam
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
Let $ a_1, a_2, a_3, \ldots $ be in an A.P. such that $$ \sum_{k=1}^{12} 2a_{2k - 1} = \frac{72}{5}, \quad \text{and} \quad \sum_{k=1}^{n} a_k = 0, $$ then $ n $ is:
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The term independent of $ x $ in the expansion of $$ \left( \frac{x + 1}{x^{3/2} + 1 - \sqrt{x}} \cdot \frac{x + 1}{x - \sqrt{x}} \right)^{10} $$ for $ x>1 $ is:
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Let $ A = \begin{bmatrix} \alpha & -1 \\6 & \beta \end{bmatrix},\ \alpha > 0 $, such that $ \det(A) = 0 $ and $ \alpha + \beta = 1 $. If $ I $ denotes the $ 2 \times 2 $ identity matrix, then the matrix $ (1 + A)^5 $ is:
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