A polynomial function f(x) satisfying the con | vedclass

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

Question

(D) Let $f(x) = ax^2 + bx + c$. Since $f(x) = [f'(x)]^2$,we have $ax^2 + bx + c = (2ax + b)^2 = 4a^2x^2 + 4abx + b^2$. Comparing coefficients: $a = 4a

Vedclass · Accepted Answer

both $(B)$ and $(C)$

Vedclass · Answer

(D) Let $f(x) = ax^2 + bx + c$. Since $f(x) = [f'(x)]^2$,we have $ax^2 + bx + c = (2ax + b)^2 = 4a^2x^2 + 4abx + b^2$. Comparing coefficients: $a = 4a^2 \Rightarrow a = \frac{1}{4}$ (as $a 
eq 0$),$b = 4ab \Rightarrow b = 4(\frac{1}{4})b = b$ (always true),and $c = b^2$. So $f(x) = \frac{1}{4}x^2 + bx + b^2$. Now,$\int_{0}^{1} (\frac{1}{4}x^2 + bx + b^2) dx = [\frac{x^3}{12} + \frac{bx^2}{2} + b^2x]_{0}^{1} = \frac{1}{12} + \frac{b}{2} + b^2$. Given $\frac{1}{12} + \frac{b}{2} + b^2 = \frac{19}{12} \Rightarrow b^2 + \frac{b}{2} - \frac{18}{12} = 0 \Rightarrow b^2 + \frac{b}{2} - \frac{3}{2} = 0$. Multiplying by $2$: $2b^2 + b - 3 = 0 \Rightarrow (2b + 3)(b - 1) = 0$. Thus $b = 1$ or $b = -\frac{3}{2}$. If $b = 1$,$f(x) = \frac{1}{4}x^2 + x + 1$. If $b = -\frac{3}{2}$,$f(x) = \frac{1}{4}x^2 - \frac{3}{2}x + (-\frac{3}{2})^2 = \frac{1}{4}x^2 - \frac{3}{2}x + \frac{9}{4}$. Both options $(B)$ and $(C)$ satisfy the conditions.

$A$ polynomial function $f(x)$ satisfying the conditions $f(x) = [f'(x)]^2$ and $\int_{0}^{1} f(x) dx = \frac{19}{12}$ can be:

Similar Questions

Let $f$ be a real-valued continuous function on $[0, 1]$ and $f(x) = x + \int_{0}^{1} (x - t) f(t) dt$. Then which of the following points $(x, y)$ lies on the curve $y = f(x)$?

If $b_{n} = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{2} nx}{\sin x} dx$,$n \in N$,then

Assertion $(A)$: $\int_0^{\frac{\pi}{2}} (\sin^6 x + \cos^6 x) dx$ lies in the interval $(\frac{\pi}{8}, \frac{\pi}{2})$.
Reason $(R)$: $\sin^6 x + \cos^6 x$ is a periodic function with period $\frac{\pi}{2}$.

If $I = \sum_{k=1}^{98} \int_k^{k+1} \frac{k+1}{x(x+1)} dx$,then which of the following is true?

$\int_{0}^{1/3} (\sum_{r=0}^{101} \{x + \frac{r}{3}\}) dx$ is equal to (where $\{.\}$ represents the fractional part function).

Vedclass Test Series

Exam Paper Generator

Online Exam Module