alpha is the maximum value of 1-2x-5x^2 and b | vedclass

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

Question

(A) We have $f(x) = 1-2x-5x^2 = -5(x^2+\frac{2}{5}x) + 1 = -5(x+\frac{1}{5})^2 + \frac{1}{5} + 1 = -5(x+\frac{1}{5})^2 + \frac{6}{5}$.Since the coeffi

Vedclass · Accepted Answer

$(-11, 13)$

Vedclass · Answer

(A) We have $f(x) = 1-2x-5x^2 = -5(x^2+\frac{2}{5}x) + 1 = -5(x+\frac{1}{5})^2 + \frac{1}{5} + 1 = -5(x+\frac{1}{5})^2 + \frac{6}{5}$.Since the coefficient of $x^2$ is negative,the maximum value is $\alpha = \frac{6}{5}$.Next,$g(x) = x^2-2x+r = (x-1)^2 + r-1$. The minimum value is $\beta = r-1$.The given inequality is $5\alpha x^2 + \beta x + 6 > 0$,which becomes $5(\frac{6}{5})x^2 + (r-1)x + 6 > 0$,or $6x^2 + (r-1)x + 6 > 0$.For this quadratic to be positive for all real $x$,the discriminant $D$ must be less than $0$.$D = (r-1)^2 - 4(6)(6) $(r-1)^2 - 144 $(r-1-12)(r-1+12) $(r-13)(r+11) Thus,$r \in (-11, 13)$.

$\alpha$ is the maximum value of $1-2x-5x^2$ and $\beta$ is the minimum value of $x^2-2x+r$. If $5\alpha x^2+\beta x+6>0$ for all real values of $x$,then the interval in which $r$ lies is

Similar Questions

Let $a, b, c, d$ be distinct real numbers such that $a, b$ are roots of $x^2-5cx-6d=0$,and $c, d$ are roots of $x^2-5ax-6b=0$. Then $b+d$ 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

Let $a_1, a_2, a_3, a_4$ be real numbers such that $a_1+a_2+a_3+a_4=0$ and $a_1^2+a_2^2+a_3^2+a_4^2=1$. Then,the smallest possible value of the expression $(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2$ lies in the interval

If $x = \sqrt[3]{{\sqrt{2} + 1}} - \sqrt[3]{{\sqrt{2} - 1}}$,then ${x^3} + 3x = $

The polynomial equation of degree $5$ whose roots are the translates of the roots of $x^5-2x^4+3x^3-4x^2+5x-6=0$ by $-2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module