Let n be the natural number. Then the range of the function f(n)=8-npn-4,4 le n le 6, is

Question

Let n be the natural number. Then the range of the function f(n)=8-npn-4,4 le n le 6, is (A)  1, 2, 3, 4  (B)  1, 2, 3, 4, 5, 6  (C)  1, 2, 3  (D)  1, 2, 3, 4, 5

Tardigrade Admin · Accepted Answer

Given that, y=2e2x-e-x On differentiating w.r.t. x, we get y1=4e2x+e-x Now, on again differentiating w.r.t. x, we get y2=8e2x-e-x Now, y2-y1-2y =8e2x-e-x-4e2x-e-x-4e2x+2e-x =0

Q. Let $n$ be the natural number. Then the range of the function $f (n) =^{8 - n} p_{n - 4}, 4 \leq n \leq 6,$ is

${1, 2, 3, 4}$

${1, 2, 3, 4, 5, 6}$

${1, 2, 3}$

${1, 2, 3, 4, 5}$

$ϕ$

Given that, $y = 2 e^{2 x} - e^{- x}$
On differentiating w.r.t. $x,$ we get $y_{1} = 4 e^{2 x} + e^{- x}$
Now, on again differentiating w.r.t. $x,$
we get $y_{2} = 8 e^{2 x} - e^{- x}$ Now, $y_{2} - y_{1} - 2 y$
$= 8 e^{2 x} - e^{- x} - 4 e^{2 x} - e^{- x} - 4 e^{2 x} + 2 e^{- x}$
$= 0$

Q. Let n be the natural number. Then the range of the function f(n)=8−npn−4​,4≤n≤6, is

{1,2,3,4}

{1,2,3,4,5,6}

{1,2,3}

{1,2,3,4,5}

ϕ

Given that, y=2e2x−e−x On differentiating w.r.t. x, we get y1​=4e2x+e−x Now, on again differentiating w.r.t. x, we get y2​=8e2x−e−x Now, y2​−y1​−2y =8e2x−e−x−4e2x−e−x−4e2x+2e−x =0