BITSAT 2022
Exam BITSAT
Year 2022
Subject Mathematics
Chapter Functions BITSAT 2022 · Mathematics · Functions
If g(x)=x^2+x-2 and (1)/(2) g o f(x)=2 x^2-5 x+2 then f(x) is equal...
If
g ( x ) = x 2 + x − 2 g(x)=x^2+x-2 g ( x ) = x 2 + x − 2 and
1 2 g o f ( x ) = 2 x 2 − 5 x + 2 \frac{1}{2} g o f(x)=2 x^2-5 x+2 2 1 g o f ( x ) = 2 x 2 − 5 x + 2 then
f ( x ) f(x) f ( x ) is equal to
Options A 2 x − 3 2 x-3 2 x − 3 B 2 x + 3 2 x+3 2 x + 3 C 2 x 2 + 3 x + 1 2 x^2+3 x+1 2 x 2 + 3 x + 1 D 2 x 2 − 3 x + 1 2 x^2-3 x+1 2 x 2 − 3 x + 1 Correct answer A. 2 x − 3 2 x-3 2 x − 3
Step-by-step solution
We have,
1 2 g o f ( x ) = 2 x 2 − 5 x + 2 \frac{1}{2} go f(x)=2 x^2-5 x+2 2 1 g o f ( x ) = 2 x 2 − 5 x + 2 ⇒ g ( f ( x ) ) = 4 x 2 − 10 x + 4 ⇒ ( f ( x ) ) 2 + f ( x ) − 2 = 4 x 2 − 10 x + 4 ⇒ ( f ( x ) ) 2 + f ( x ) − ( 4 x 2 − 10 x + 6 ) = 0 ⇒ f ( x ) = − 1 ± 1 + 4 ( 4 x 2 − 10 x + 6 ) 2 ⇒ f ( x ) = − 1 ± 16 x 2 − 40 x + 25 2 ⇒ f ( x ) = − 1 ± ( 4 x − 5 ) 2 = 2 x − 3 , − 2 x + 2 \begin{aligned} & \Rightarrow g(f(x))=4 x^2-10 x+4 \\ & \Rightarrow(f(x))^2+f(x)-2=4 x^2-10 x+4 \\ & \Rightarrow(f(x))^2+f(x)-\left(4 x^2-10 x+6\right)=0 \\ & \Rightarrow f(x)=\frac{-1 \pm \sqrt{1+4\left(4 x^2-10 x+6\right)}}{2} \\ & \Rightarrow f(x)=\frac{-1 \pm \sqrt{16 x^2-40 x+25}}{2} \\ & \Rightarrow f(x)=\frac{-1 \pm(4 x-5)}{2}=2 x-3,-2 x+2 \end{aligned} ⇒ g ( f ( x )) = 4 x 2 − 10 x + 4 ⇒ ( f ( x ) ) 2 + f ( x ) − 2 = 4 x 2 − 10 x + 4 ⇒ ( f ( x ) ) 2 + f ( x ) − ( 4 x 2 − 10 x + 6 ) = 0 ⇒ f ( x ) = 2 − 1 ± 1 + 4 ( 4 x 2 − 10 x + 6 ) ⇒ f ( x ) = 2 − 1 ± 16 x 2 − 40 x + 25 ⇒ f ( x ) = 2 − 1 ± ( 4 x − 5 ) = 2 x − 3 , − 2 x + 2 $
H e n c e , Hence, He n ce , f(x)=2 x-3$
Practice Functions on Quantrex Academy →
← Previous question If f(x)= x 1+x^2 , then (fof of) (x) isNext question → The domain of the function f(x)= 1 x^ 12 -x^ 9 +x^ 4 -x+1 is given by
More from Functions
Let [x] denote the greatest integer ≤ x . If f(x)=[x] and g(x)=|x| , then the value of f… BITSAT 2024 Number of real solution of 5- _2|x| =3- _2|x| is equal to BITSAT 2024 f(x)=( x)/([(2 x)/(π)]+(1)/(2)) , where x is not an integral multiple of π and [.]… BITSAT 2024 The function f: R arrow R defined by f(x)= x 1+x^2 is BITSAT 2024 If f: R arrow R, g: R arrow R are defined by f(x)=5 x-3, g(x)=x^2+3 , then g o f^ -1 (3)… BITSAT 2024 The domain of the real valued function f(x)= (2 x^2-7 x+5)/(3 x^2-5 x-2) is BITSAT 2024 If a function f: R-\ l\ arrow R-\ m\ defined by f(x)=(x+3)/(x-2) is a bijection, then 3… BITSAT 2024 f(x)= x+ x, g(x)=x^2-1 , then g(f(x)) is invertible if BITSAT 2024 Let the function g:(-∞, ∞) arrow (-(π)/(2), (π)/(2)) be given by g(u)=2 ^ -1… BITSAT 2024 The domain of the function f(x)= x- 1-x^2 is BITSAT 2023 Consider the function f in A=R-\ (2)/(3)\ defined as f(x)=(4 x+3)/(6 x-4) , then f^ -1 is… BITSAT 2023 If f(x)= x 1+x^2 , then (fof of) (x) is BITSAT 2023 The domain of the function f(x)= 1 x^ 12 -x^ 9 +x^ 4 -x+1 is given by BITSAT 2021 Let [x] denote the greatest integer ≤ x . If f(x)=[x] and g(x)=|x|, then the value of f… BITSAT 2020 The domain of the function f(x)= x- 1-x^ 2 is BITSAT 2020 Number of solution of the equation | x | = 2 [ x ] are (where | x | , [ x ] are modulus… BITSAT 2019
Full Functions list
Functions JEE Maths PYQs JEE Mathematics PYQs
All BITSAT PYQs
