Quadratic Equation and Inequalities · Mathematics · JEE Advanced
Numerical
Let $a=3 \sqrt{2}$ and $b=\frac{1}{5^{1 / 6} \sqrt{6}}$. If $x, y \in \mathbb{R}$ are such that
$$ \begin{aligned} & 3 x+2 y=\log _a(18)^{\frac{5}{4}} \quad \text { and } \\ & 2 x-y=\log _b(\sqrt{1080}), \end{aligned} $$
then $4 x+5 y$ is equal to __________.
$$ x^{\left(16\left(\log _{5} x\right)^{3}-68 \log _{5} x\right)}=5^{-16} $$
is __________.
The value of $$6 + {\log _{3/2}}\left( {{1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}\sqrt {4 - {1 \over {3\sqrt 2 }}...} } } } \right)$$ is __________.
MCQ (Single Correct Answer)
ac(a $$-$$ c) + ad(a $$-$$ d) + bc(b $$-$$ c) + bd(b $$-$$ d) is
Let $$\alpha$$(a) and $$\beta$$(a) be the roots of the equation $$(\root 3 \of {1 + a} - 1){x^2} + (\sqrt {1 + a} - 1)x + (\root 6 \of {1 + a} - 1) = 0$$ where $$a > - 1$$. Then $$\mathop {\lim }\limits_{a \to {0^ + }} \alpha (a)$$ and $$\mathop {\lim }\limits_{a \to {0^ + }} \beta (a)$$ are
$$\matrix{ {{{\left( {2x} \right)}^{\ell n2}}\, = {{\left( {3y} \right)}^{\ell n3}}} \cr {{3^{\ell nx}}\, = {2^{\ell ny}}} \cr } $$
Then $${x_0}$$ is
have one root in common is
STATEMENT - 1 : $$\left( {{p^2} - q} \right)\left( {{b^2} - ac} \right) \ge 0$$
and
STATEMENT - 2 : $$b \ne pa$$ or $$c \ne qa$$
If the roots of the equation $${x^2} + 2\left( {a + b + c} \right)x + 3\lambda \left( {ab + bc + ca} \right) = 0$$ are real, then
then $$a,\,b,\,c,\,d$$
$$\left( {b + c - a} \right)\left( {c + a - b} \right)\left( {a + b - c} \right) - abc$$ is
$$(\ell - m)\,{x^2} - 5\,(\ell + m)\,x - 2\,(\ell - m) = 0$$ are
MCQ (More than One Correct Answer)
$${a_n} = {{{\alpha ^n} - {\beta ^n}} \over {\alpha - \beta }},\,n \ge 1$$
$${b_1} = 1\,and\,{b_n} = {a_{n - 1}} + {a_{n + 1}},\,n \ge 2$$
Then which of the following options is/are correct?
Subjective
Find all the real values of $$x,$$ for which $$y$$ takes real values.
has solution satisfying the conditions $$x > 0,\,y > 0.$$
deduce the condition that the equations have a common root.
where $$m$$ and $$n$$ are positive integers $$\left( {n \le m} \right),$$ show that
$$\left( {m,n + 1} \right) = \left( {m - 1,\,n + 1} \right) + {x^{m - n - 1}}\left( {m - 1,n} \right).$$