If both roots of the equation x^2 - 5ax + 6a | vedclass

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

Question

(A) Let $f(x) = x^2 - 5ax + 6a$. For both roots to be greater than $1$,the following conditions must be satisfied:$1$. Discriminant $D \ge 0$: $D = (-

Vedclass · Accepted Answer

$[\frac{24}{25}, \infty)$

Vedclass · Answer

(A) Let $f(x) = x^2 - 5ax + 6a$. For both roots to be greater than $1$,the following conditions must be satisfied:$1$. Discriminant $D \ge 0$: $D = (-5a)^2 - 4(1)(6a) = 25a^2 - 24a \ge 0 \implies a(25a - 24) \ge 0$. This gives $a \in (- \infty, 0] \cup [\frac{24}{25}, \infty)$.$2$. Vertex $x_v > 1$: $x_v = -\frac{b}{2a} = \frac{5a}{2} > 1 \implies a > \frac{2}{5}$.$3$. $f(1) > 0$: $f(1) = 1^2 - 5a(1) + 6a = 1 + a > 0 \implies a > -1$.Taking the intersection of all three conditions: $a \in (- \infty, 0] \cup [\frac{24}{25}, \infty)$ $AND$ $a > \frac{2}{5}$ $AND$ $a > -1$.The intersection is $a \in [\frac{24}{25}, \infty)$.

If both roots of the equation $x^2 - 5ax + 6a = 0$ exceed $1$,then the range of '$a$' is

Similar Questions

The symmetry in the curve $x^3 + y^3 = 3axy$ is along:

To convert the equation $2x^2 + 4xy + 5y^2 - 4x - 22y + 29 = 0$ to homogeneous form,the origin is shifted to the point:

If curves $y = ax^2 + bx + c$ and $y = px^2 + qx + r$ do not intersect each other and $a, b, c, p, q, r \in \{1, 2, 3, 4, \dots, 10\}$,then the maximum value of $(aq - bp)^2 + (c - r)^2$ is-

The point to which the origin is to be shifted to remove the first degree terms from the equation $2x^2+4xy-6y^2+2x+8y+1=0$ is

Find the centre of the conic section represented by the equation $14x^2 - 4xy + 11y^2 - 44x - 58y + 71 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module