If the abscissa of the vertex of the parabola | vedclass

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

Question

(B) The vertex of the parabola $y = ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. Given $- \frac{b}{2a} = 1$,we have $b = -2a$.Since $a, b, c > 0$,and $b

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The vertex of the parabola $y = ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. Given $- \frac{b}{2a} = 1$,we have $b = -2a$.Since $a, b, c > 0$,and $b = -2a$,this implies $a$ must be negative,which contradicts $a > 0$. However,assuming the condition $a, b, c > 0$ refers to the coefficients of the parabola and the vertex abscissa is $1$,we proceed with $b = -2a$.Given $f(x) = \int_0^x (3at^2 + bt + c) dt$,then $f'(x) = 3ax^2 + bx + c$.For $f(x)$ to be strictly increasing,$f'(x) > 0$ for all $x \in R$.This requires $3a > 0$ (which is true as $a > 0$) and the discriminant $D $D = b^2 - 4(3a)(c) = b^2 - 12ac Substituting $b = -2a$,we get $(-2a)^2 - 12ac Dividing by $4a$ (since $a > 0$),we get $a - 3c Thus,$\frac{a}{c} The greatest integer value $[\frac{a}{c}]$ can take is $2$.

If the abscissa of the vertex of the parabola $y = ax^2 + bx + c$ is $1$ $(a, b, c > 0)$ and $f(x) = \int_0^x (3at^2 + bt + c) dt$ is a strictly increasing function $\forall x \in R$,then the maximum possible value of $[\frac{a}{c}]$ is (where $[.]$ denotes the greatest integer function).

Similar Questions

Let ${I_1} = \int\limits_0^1 {\frac{{{e^x}}}{{1 + x}}} \,dx$ and ${I_2} = \int\limits_0^1 {\frac{{{x^2}}}{{{e^{{x^3}}}\left( {2 - {x^3}} \right)}}} \,dx$,then the value of $\frac{{{I_1}}}{{{I_2}}}$ is equal to

Let $r_k = \frac{\int_0^1 (1-x^7)^k dx}{\int_0^1 (1-x^7)^{k+1} dx}$,$k \in N$. Then the value of $\sum_{k=1}^{10} \frac{1}{7(r_k-1)}$ is equal to ...........

If $\int_{0}^{\pi} (\sin^{3} x) e^{-\sin^{2} x} dx = \alpha - \frac{\beta}{e} \int_{0}^{1} \sqrt{t} e^{t} dt$,then $\alpha + \beta$ is equal to $....$

If $f(x) = \text{Max}\{\sin x, \cos x\}$ and $g(x) = \text{Min}\{\sin x, \cos x\}$,then $\int_{0}^{\pi} f(x) dx + \int_{0}^{\pi} g(x) dx = $

The values of $\alpha$ which satisfy $\int_{\pi /2}^{\alpha} \sin x \, dx = \sin 2\alpha$,where $\alpha \in [0, 2\pi]$,are equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module