The domain of the function f(x) = exp (sqrt { | vedclass

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

Question

(D) For the function $f(x) = \exp (\sqrt {5x - 3 - 2{x^2}} )$ to be defined,the expression under the square root must be non-negative.Thus,$5x - 3 - 2

Vedclass · Accepted Answer

$[1, 3/2]$

Vedclass · Answer

(D) For the function $f(x) = \exp (\sqrt {5x - 3 - 2{x^2}} )$ to be defined,the expression under the square root must be non-negative.Thus,$5x - 3 - 2{x^2} \ge 0$.Multiplying by $-1$,we get $2{x^2} - 5x + 3 \le 0$.Factoring the quadratic expression: $2{x^2} - 2x - 3x + 3 \le 0 \implies 2x(x - 1) - 3(x - 1) \le 0 \implies (2x - 3)(x - 1) \le 0$.This can be written as $2(x - 3/2)(x - 1) \le 0$.Using the wavy curve method (sign scheme),the expression is $\le 0$ for $x \in [1, 3/2]$.Therefore,the domain is $[1, 3/2]$.

The domain of the function $f(x) = \exp (\sqrt {5x - 3 - 2{x^2}} )$ is

Similar Questions

If $A = \{1, 2, 4\}, B = \{2, 4, 5\}, C = \{2, 5\},$ then $(A - B) \times (B - C)$ is

If $A = \{1, 2, 3, 4, 5\}, B = \{2, 4, 6\}, C = \{3, 4, 6\},$ then $(A \cup B) \cap C$ is

If the domain of the function $f(x) = x^2 - 6x + 7$ is $(-\infty, \infty)$,then the range of the function is:

If $A = \{a, d\}, B = \{b, c, e\}$ and $C = \{b, c, f\},$ then $A \times (B \cap C) =$

If $U = \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$,$A = \{2, 4, 7\}$,$B = \{3, 5, 7, 9, 11\}$,and $C = \{7, 8, 9, 10, 11\}$,compute: $(A \cap U) \cap (B \cup C)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module