Let a, b, c, d be real numbers between -5 and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given equations are:$|a|=\sqrt{4-\sqrt{5-a}}$$|b|=\sqrt{4+\sqrt{5-b}}$$|c|=\sqrt{4-\sqrt{5+c}}$$|d|=\sqrt{4+\sqrt{5+d}}$Squaring the first equatio

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(A) Given equations are:$|a|=\sqrt{4-\sqrt{5-a}}$$|b|=\sqrt{4+\sqrt{5-b}}$$|c|=\sqrt{4-\sqrt{5+c}}$$|d|=\sqrt{4+\sqrt{5+d}}$Squaring the first equation: $a^2 = 4 - \sqrt{5-a} \implies a^2 - 4 = -\sqrt{5-a}$.Squaring again: $(a^2 - 4)^2 = 5 - a \implies a^4 - 8a^2 + 16 = 5 - a \implies a^4 - 8a^2 + a + 11 = 0$.Similarly,for $b, c, d$,we observe that $a, b, -c, -d$ are the roots of the polynomial equation $x^4 - 8x^2 + x + 11 = 0$.By Vieta's formulas,the product of the roots of $x^4 + 0x^3 - 8x^2 + x + 11 = 0$ is given by the constant term,which is $11$.Thus,$a \cdot b \cdot (-c) \cdot (-d) = 11 \implies abcd = 11$.

Let $a, b, c, d$ be real numbers between $-5$ and $5$ such that $|a|=\sqrt{4-\sqrt{5-a}}$,$|b|=\sqrt{4+\sqrt{5-b}}$,$|c|=\sqrt{4-\sqrt{5+c}}$,and $|d|=\sqrt{4+\sqrt{5+d}}$. Then,the product $abcd$ is

Similar Questions

Let $\lambda \in R$ and let the equation $E$ be $|x|^2 - 2|x| + |\lambda - 3| = 0$. Then the largest element in the set $S = \{x + \lambda : x \text{ is an integer solution of } E\}$ is $..........$

The minimum value of $\frac{9 \cdot 3^{2x} + 6 \cdot 3^x + 4}{9 \cdot 3^{2x} - 6 \cdot 3^x + 4}$ is

If $\tan \alpha$ equals the integral solution of the inequality $4x^2 - 16x + 15 < 0$ and $\cos \beta$ equals the slope of the bisector of the first quadrant,then $\sin(\alpha + \beta)\sin(\alpha - \beta)$ is equal to

If $m$ is a root of the equation $(1 - ab)x^2 - (a^2 + b^2)x - (1 + ab) = 0$ and $m$ harmonic means are inserted between $a$ and $b$,then the difference between the last and the first of the means equals

Let $\alpha, \beta$ be the roots of the equation $x^2 - |a|x - |b| = 0$ such that $|\alpha| < |\beta|$. If $|a| < \beta - 1$,then the positive root of $\log_{|\alpha|} \left( \frac{x^2}{\beta^2} \right) - 1 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module