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Fast 9 | F9 | Fast and Furious 9| All fast movies download

Fast 9 | F9 | Fast and Furious 9| All fast movies HD download free It is one of the best Hollywood movie in the world. Fast & Furious 9 is an upcoming American action film directed by Justin Lin and written by Daniel Casey. It will be the ninth installment in The Fast and the Furious franchise.Little is known about the next Fast & Furious film but we know it will unite the Toretto family once again with Mia's return to the fold. Also, could Cena be playing the film's antagonist - the silence on his role might suggest he will go toe to toe with Dom - which we would love to see! Download from here!
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. Let f : R → R be defined as f(x) = 3x. Choose the correct answer. (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto.

Class 12, NCERT Chapter 1,  Exercise 1.2, Q12 f : R → R defined as  f ( x ) = 3 x . Let  x ,  y  ∈  R  such that  f ( x ) =  f ( y ). ⇒ 3 x  = 3 y ⇒  x  =  y ∴ f  is one-one. Also, for any real number ( y)  in co-domain  R , there exists  in  R  such that . ∴ f  is onto. Hence, function  f  is one-one and onto. The correct answer is A.

Let f : R → R be defined as f(x) = x 4 . Choose the correct answer. (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto.

Class 12, NCERT Chapter 1,  Exercise 1.2, Q11 f : R → R is defined as  f (x) = x 4 Let  x ,  y  ∈  R  such that  f ( x ) =  f ( y ). ∴ does not imply that x 1 =x 2 . For instance, f (1) =  f (-1) = 1 ∴  f  is not one-one. Consider an element 2 in co-domain  R . It is clear that there does not exist any  x  in domain  R  such that  f ( x ) = 2. ∴  f  is not onto. Hence, function  f  is neither one-one nor onto. The correct answer is D

Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by f(x) = (x-2)/(x-3) Is f one-one and onto? Justify your answer.

Class 12, NCERT Chapter 1,  Exercise 1.2, Q10 A =  R  - {3}, B =  R  - {1} f : A → B is defined as Let x,y∈A such that f(x)=f(y). ⇒ (x-2)/(x-3)=(y-2)/(y-3) ⇒ (x-2)(y-3)=(y-2)(x-3) ⇒ xy-3x-2y+6=xy-3y-2x+6 ⇒ -3x-2y=-3y-2x ⇒ 3x-2x=3y-2y ⇒ x=y ∴  f  is one-one. Let  y  ∈B =  R  - {1}. Then,  y  ≠ 1. The function  f  is onto if there exists  x  ∈A such that  f ( x ) =  y. Now, f(x)=y ⇒(x-2)(x-3)=y ⇒x-2=xy-3y ⇒x(1-y)=-3y+2 ⇒ x=2-3y/1-y  ∈A   Thus, for any  y  ∈ B, there exists  x=2-3y/1-y  ∈A     such that Hence, function  f  is one-one and onto.

Let f : N → N be defined by f (n) ={ (n+1)/2, if n is odd and (n-1)/2, if n is even, for all n ∈ N.State whether the function f is bijective. Justify your answer

Class 12, NCERT Chapter 1,  Exercise 1.2, Q9 Thus it is bijective.

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer. (i) f : R → R defined by f(x) = 3 – 4x (ii) f : R → R defined by f(x) = 1 + x^2

Class 12, NCERT Chapter 1,  Exercise 1.2, Q7 (i)  Here f(x) = 3 – 4x Then f(x 1 ) = 3 – 4x 1 and f (x 2 ) = 3 – 4x 2 Now f(x 1 ) = f (x 2 ) ⇒ 3 – 4x 1  = 3 – 4x 2 ⇒ – 4x 1  = –4x 2 ⇒ x 1  = x 2 ∴ f(x 1 ) = f(x 2 ) & x 1  = x 2 So, f is one-one function. Let f(x) = y ∈ R Then y=3 - 4x ⇒ x= 3-y/4 (ii) Let x 1  = 2 and x 2  = –2 ∈ R Here f(x) = 1 + x 2 Then f(x 1 ) = f (2) = 1 + (2)  2  = 5 and f (x 2 ) = f(–2) = 1 + (–2)  2  = 5 ∴ f (x 1 ) = f(x 2 ) but x 1  ≠ x 2 So, f is not one-one function. Let f(x) = – 2 ⇒R Then 1 + x 2  = –2 ⇒ x 2  = –3 = ± √–3 ∈R So, f is not onto function.