If f is a function defined on the set of rational numbers such that f(x)=3x2+6x+14x2−3x−3, then f−1(2) is equal to

f−1(2)= the set of values of x such that f(x)=2⇒ 3x2+6x+14x2−3x−3=2⇒ 3x2+6x+14=2x2−3x−3⇒ x2+12x+20=0⇒(x+10)(x+2)=0⇒ x=−2 or x=−10∴ f−1(2)={−2,−10}

If f is a function defined on the set of rational numbers such that f(x)=3x2+6x+14x2−3x−3, then f−1(2) is equal to

f−1(2)= the set of values of x such that f(x)=2⇒ 3x2+6x+14x2−3x−3=2⇒ 3x2+6x+14=2x2−3x−3⇒ x2+12x+20=0⇒(x+10)(x+2)=0⇒ x=−2 or x=−10∴ f−1(2)={−2,−10}

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If f is a function defined on the set of rational numbers such that $f (x) = \frac{3 x^{2} + 6 x + 14}{x^{2} - 3 x - 3}$ , then $f^{- 1} (2)$ is equal to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

{0, 1}

$\{- \frac{1}{2}, \frac{1}{3}\}$

${- 10, - 2}$

$\{- 2, - \frac{1}{7}\}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$f^{- 1} (2) =$ the set of values of x such that $f (x) = 2$

$\begin{array}{l} \Rightarrow \frac{3 x^{2} + 6 x + 14}{x^{2} - 3 x - 3} = 2 \\ \Rightarrow 3 x^{2} + 6 x + 14 = 2 (x^{2} - 3 x - 3) \\ \Rightarrow x^{2} + 12 x + 20 = 0 \Rightarrow (x + 10) (x + 2) = 0 \\ \Rightarrow x = - 2 or x = - 10 \\ ∴ f^{- 1} (2) = {- 2, - 10} \end{array}$