If the maximum value of 2x - 7 - ax^2 cannot | vedclass

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

Question

(D) Let $f(x) = -ax^2 + 2x - 7$. For the function to have a maximum value,we must have $a > 0$.Completing the square:$f(x) = -a(x^2 - \frac{2}{a}x) -

Vedclass · Accepted Answer

$\frac{1}{27}$

Vedclass · Answer

(D) Let $f(x) = -ax^2 + 2x - 7$. For the function to have a maximum value,we must have $a > 0$.Completing the square:$f(x) = -a(x^2 - \frac{2}{a}x) - 7$$f(x) = -a(x^2 - \frac{2}{a}x + \frac{1}{a^2} - \frac{1}{a^2}) - 7$$f(x) = -a(x - \frac{1}{a})^2 + \frac{1}{a} - 7$The maximum value of $f(x)$ is $\frac{1}{a} - 7$.Given that the maximum value cannot exceed $20$,we have:$\frac{1}{a} - 7 \leq 20$$\frac{1}{a} \leq 27$Since $a > 0$,we get $a \geq \frac{1}{27}$.Thus,the minimum value of $a$ is $\frac{1}{27}$.

Quadratic expression	The minimum value
i) $x^2 + 4x + 6$	a) $1$
ii) $x^2 - 2x + 5$	b) $2$
iii) $x^2 + 6x + 18$	c) $4$
iv) $x^2 - 4x + 5$	d) $9$

If the maximum value of $2x - 7 - ax^2$ cannot exceed $20$,then the minimum value of $a$ is

Similar Questions

The range of $a$ for which the roots of $x^2 - 2x - a^2 + 1 = 0$ lie between the roots (exclusive) of the equation $x^2 - 2(a + 1)x + a(a - 1) = 0$ is:

Number of integral values of $a$ for which both roots of the quadratic equation $x^2 - (2a + 3)x + a^2 + 3a = 0$ lie in the interval $(0, 4)$ is:

Match the following quadratic expressions with their minimum values:
Quadratic expression The minimum value
i) $x^2 + 4x + 6$ a) $1$
ii) $x^2 - 2x + 5$ b) $2$
iii) $x^2 + 6x + 18$ c) $4$
iv) $x^2 - 4x + 5$ d) $9$

If $b > a$,then the equation $(x - a)(x - b) = 1$ has

The number of integral values of $m$ for which the quadratic expression $(1 + 2m)x^2 - 2(1 + 3m)x + 4(1 + m)$ is always positive for all $x \in R$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module