If f:[0,5] arrow[7,13] be a surjective linear map such that f(1)f(2) and f (2 α2-α-1)+13= f (α2+1)+ f (0) for some α ∈[0,5], then value of f -1(α+5) equals

Question

If f:[0,5] arrow[7,13] be a surjective linear map such that f(1)>f(2) and f (2 α2-α-1)+13= f (α2+1)+ f (0) for some α ∈[0,5], then value of f -1(α+5) equals (A) 1 (B) 3 (C) 0 (D) 5

Tardigrade Admin · Accepted Answer

Clearly f is one-one decreasing ⇒ f (5)=7, f (0)=13 ∴ f (2 α2-α-1)= f (α2+1) ⇒ α=-1,2 text since α ∈[0,5] ⇒ α=2 ⇒ f -1(α+5)= f -1(7)=5

Q. If $f : [0, 5] \to [7, 13]$ be a surjective linear map such that $f (1) > f (2)$ and $f (2 α^{2} - α - 1) + 13 = f (α^{2} + 1) + f (0)$ for some $α \in [0, 5]$ , then value of $f^{- 1} (α + 5)$ equals

1

3

0

5

Clearly $f$ is one-one & decreasing $\Rightarrow f (5) = 7, f (0) = 13$
$∴ f (2 α^{2} - α - 1) = f (α^{2} + 1) \Rightarrow α = - 1, 2$
$since α \in [0, 5] \Rightarrow α = 2$
$\Rightarrow f^{- 1} (α + 5) = f^{- 1} (7) = 5$

Q. If f:[0,5]→[7,13] be a surjective linear map such that f(1)>f(2) and f(2α2−α−1)+13=f(α2+1)+f(0) for some α∈[0,5], then value of f−1(α+5) equals

1

3

0

5

Clearly f is one-one & decreasing ⇒f(5)=7,f(0)=13 ∴f(2α2−α−1)=f(α2+1)⇒α=−1,2 since α∈[0,5]⇒α=2 ⇒f−1(α+5)=f−1(7)=5

Q. If $f : [0, 5] \to [7, 13]$ be a surjective linear map such that $f (1) > f (2)$ and $f (2 α^{2} - α - 1) + 13 = f (α^{2} + 1) + f (0)$ for some $α \in [0, 5]$ , then value of $f^{- 1} (α + 5)$ equals

Clearly $f$ is one-one & decreasing $\Rightarrow f (5) = 7, f (0) = 13$
$∴ f (2 α^{2} - α - 1) = f (α^{2} + 1) \Rightarrow α = - 1, 2$
$since α \in [0, 5] \Rightarrow α = 2$
$\Rightarrow f^{- 1} (α + 5) = f^{- 1} (7) = 5$