Given that a, b and c are real numbers such t | vedclass

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

Question

(D) The function is $f(x) = \log \{ax^3 + (a+b)x^2 + (b+c)x + c\}$.Factorizing the expression inside the logarithm:$ax^3 + ax^2 + bx^2 + bx + cx + c =

Vedclass · Accepted Answer

$R - (\{-\frac{b}{2a}\} \cup (-\infty, -1])$

Vedclass · Answer

(D) The function is $f(x) = \log \{ax^3 + (a+b)x^2 + (b+c)x + c\}$.Factorizing the expression inside the logarithm:$ax^3 + ax^2 + bx^2 + bx + cx + c = ax^2(x+1) + bx(x+1) + c(x+1) = (ax^2 + bx + c)(x+1)$.Since $b^2 = 4ac$,the quadratic $ax^2 + bx + c$ can be written as $a(x + \frac{b}{2a})^2$.Thus,$f(x) = \log \{a(x + \frac{b}{2a})^2(x+1)\}$.For $f(x)$ to be defined,the argument of the logarithm must be strictly positive:$a(x + \frac{b}{2a})^2(x+1) > 0$.Given $a > 0$,the term $a(x + \frac{b}{2a})^2$ is always $\geq 0$. It is $0$ when $x = -\frac{b}{2a}$.Therefore,we require $x+1 > 0$ and $x 
eq -\frac{b}{2a}$.This implies $x > -1$ and $x 
eq -\frac{b}{2a}$.The domain $D$ is $(-1, \infty) - \{-\frac{b}{2a}\}$.This is equivalent to $R - (\{-\frac{b}{2a}\} \cup (-\infty, -1])$.Thus,the correct option is $D$.

Given that $a, b$ and $c$ are real numbers such that $b^2 = 4ac$ and $a > 0$. The maximal possible set $D \subseteq R$ on which the function $f: D \rightarrow R$ given by $f(x) = \log \{ax^3 + (a+b)x^2 + (b+c)x + c\}$ is defined,is

Similar Questions

Let $f: R - \{2, 6\} \rightarrow R$ be a real-valued function defined as $f(x) = \frac{x^2+2x+1}{x^2-8x+12}$. Then the range of $f$ is

The range of the function $f(x) = \frac{e^x \ln x \cdot 5^{(x^2 + 2)}(x^2 - 7x + 10)}{2x^2 - 11x + 12}$ is

If $f: R \rightarrow R$ is defined as $f(x) = \frac{x^6}{x^6+2020}$,for all $x \in R$,then the range of $f$ is .......

The range of the function $f(x) = \begin{cases} 4x - 1, & x > 3 \\ x^2 - 2, & -2 \leq x \leq 3 \\ 3x + 4, & x < -2 \end{cases}$ is:

If the domain of the function $f(x) = \log_7(1 - \log_4(x^2 - 9x + 18))$ is $(\alpha, \beta) \cup (\gamma, \delta)$,then $\alpha + \beta + \gamma + \delta$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module