Question

If f is twice differentiable function for x ∈ R such that f(2) = 5, f'(2) = 8 and f(x) > 1 , f(x) > 4 then (1) f(5) + f(5) ≤ 26 (2)f{5) + f(5) ≤ 28 (3) f(5) + f(5) ≥ 28 (4) none of these

Answer 1

Answer:

Here, we have the following data:

f(0) = 2

f(1) = 3

And the functional relation is:

\boxed{f(x+2)=2f(x)-f(x+1)}

f(x+2)=2f(x)−f(x+1)

We can see that the equation consists of three consecutive terms, namely f(x), f(x+1) and f(x+2). Since we know two values, we can find the other values as follows:

\begin{gathered}f(x+2) = 2f(x) - f(x+1) \\ \\ \\ \text{Put x=0} \\ \\ \\ \implies f(0+2) = 2f(0) - f(0+1) \\ \\ \\ \implies f(2) = 2f(0) - f(1) \\ \\ \\ \implies f(2) = 2(2) - 3 \\ \\ \\ \implies f(2) = 1\end{gathered}

f(x+2)=2f(x)−f(x+1)

Put x=0

⟹f(0+2)=2f(0)−f(0+1)

⟹f(2)=2f(0)−f(1)

⟹f(2)=2(2)−3

⟹f(2)=1

Thus, we have

f(2) = 1

We again use the equation.

\begin{gathered}f(x+2) = 2f(x) - f(x+1) \\ \\ \\ \text{Put x=1} \\ \\ \\ \implies f(1+2) = 2f(1) - f(1+1) \\ \\ \\ \implies f(3) = 2f(1) - f(2) \\ \\ \\ \implies f(3) = 2(3) - 1 \\ \\ \\ \implies f(3) = 5\end{gathered}

f(x+2)=2f(x)−f(x+1)

Put x=1

⟹f(1+2)=2f(1)−f(1+1)

⟹f(3)=2f(1)−f(2)

⟹f(3)=2(3)−1

⟹f(3)=5

Thus, we have:

f(3) = 5

We use the equation again:

\begin{gathered}f(x+2) = 2f(x) - f(x+1) \\ \\ \\ \text{Put x=2} \\ \\ \\ \implies f(2+2) = 2f(2) - f(2+1) \\ \\ \\ \implies f(4) = 2f(2) - f(3) \\ \\ \\ \implies f(4) = 2(1) - 5 \\ \\ \\ \implies f(4) = -3\end{gathered}

f(x+2)=2f(x)−f(x+1)

Put x=2

⟹f(2+2)=2f(2)−f(2+1)

⟹f(4)=2f(2)−f(3)

⟹f(4)=2(1)−5

⟹f(4)=−3

Thus, we have:

f(4) = -3

We use the equation one final time:

\begin{gathered}f(x+2) = 2f(x) - f(x+1) \\ \\ \\ \text{Put x=3} \\ \\ \\ \implies f(3+2) = 2f(3) - f(3+1) \\ \\ \\ \implies f(5) = 2f(3) - f(4) \\ \\ \\ \implies f(5) = 2(5) - (-3) \\ \\ \\ \implies f(5) = 10+3 \\ \\ \\ \implies \boxed{f(5)=13}\end{gathered}

f(x+2)=2f(x)−f(x+1)

Put x=3

⟹f(3+2)=2f(3)−f(3+1)

⟹f(5)=2f(3)−f(4)

⟹f(5)=2(5)−(−3)

⟹f(5)=10+3

⟹

f(5)=13

Thus, f(5) = 13