Q.1. Let be defined as . Then is:

Q: What is the correct answer for JEE Main Mathematics Q1?

The correct answer is D. Sol. Let $y = \frac{2x^2-3x+2}{3x^2+x+3}$. Then $3yx^2+yx+3y = 2x^2-3x+2$. Rearranging, $(3y - 2)x^2 + (y + 3)x + (3y - 2) = 0$. For $x$ to be real, the discriminant $D \ge 0$. So, $(y+3)^2 - 4(3y-2)(3y-2) \ge 0$. $(y+3)^2 - 4(3y-2)^2 \ge 0$. $(y+3 - 2(3y-2))(y+3 + 2(3y-2)) \ge 0$. $(y+3-6y+4)(y+3+6y-4) \ge 0$. $(-5y+7)(7y-1) \ge 0$. $(7y-1)(5y-7) \le 0$. This implies $y \in [\frac{1}{7}, \frac{7}

Q: Q.1. Let be defined as . Then is:

The correct answer is D. Sol. Let $y = \frac{2x^2-3x+2}{3x^2+x+3}$. Then $3yx^2+yx+3y = 2x^2-3x+2$. Rearranging, $(3y - 2)x^2 + (y + 3)x + (3y - 2) = 0$. For $x$ to be real, the discriminant $D \ge 0$. So, $(y+3)^2 - 4(3y-2)(3y-2) \ge 0$. $(y+3)^2 - 4(3y-2)^2 \ge 0$. $(y+3 - 2(3y-2))(y+3 + 2(3y-2)) \ge 0$. $(y+3-6y+4)(y+3+6y-4) \ge 0$. $(-5y+7)(7y-1) \ge 0$. $(7y-1)(5y-7) \le 0$. This implies $y \in [\frac{1}{7}, \frac{7}

Q: What are the options for this JEE Main Mathematics question?

A. both one-one and onto · B. one-one but not onto · C. onto but not one-one · D. neither one-one nor onto

Question

Q.1. Let  be defined as . Then  is:

FirstInTest · Accepted Answer

Sol. Let $y = \frac{2x^2-3x+2}{3x^2+x+3}$. Then $3yx^2+yx+3y = 2x^2-3x+2$. Rearranging, $(3y - 2)x^2 + (y + 3)x + (3y - 2) = 0$. For $x$ to be real, the discriminant $D \ge 0$. So, $(y+3)^2 - 4(3y-2)(3y-2) \ge 0$. $(y+3)^2 - 4(3y-2)^2 \ge 0$. $(y+3 - 2(3y-2))(y+3 + 2(3y-2)) \ge 0$. $(y+3-6y+4)(y+3+6y-4) \ge 0$. $(-5y+7)(7y-1) \ge 0$. $(7y-1)(5y-7) \le 0$. This implies $y \in [\frac{1}{7}, \frac{7}{5}]$. The range of $f(x)$ is $[\frac{1}{7}, \frac{7}{5}]$, which is not $R$. Hence, $f(x)$ is not onto. To check if it's one-one, we can find $f'(x)$. $f'(x) = \frac{(4x-3)(3x^2+x+3) - (2x^2-3x+2)(6x+1)}{(3x^2+x+3)^2}$. The numerator is $12x^3+4x^2+12x-9x^2-3x-9 - (12x^3+2x^2-18x^2-3x+12x+2) = 12x^3-5x^2+9x-9 - (12x^3-16x^2+9x+2) = 11x^2-11$. $f'(x) = \frac{11(x^2-1)}{(3x^2+x+3)^2}$. $f'(x)=0$ for $x=\pm 1$. Since $f'(x)$ changes sign, $f(x)$ is not monotonic, and thus not one-one. Therefore, $f(x)$ is neither one-one nor onto.

Answer

both one-one and onto

Answer

one-one but not onto

Answer

onto but not one-one

Answer

neither one-one nor onto

Related Questionsin Mathematics

More from JEE Main

Browse Engineering Questions by Subject

Practice More JEE Main Questions