For every twice differentiable function `f: R->[-2,\ 2]` with `(f(0))^2+(f^(prime)(0))^2=85` , which of the following statement(s) is (are) TRUE? There exist `r ,\ s in R` where `roo)f(x)=1` (d) There exists `alpha in (-4,\ 4)` such that `f(alpha)+f"(alpha)=0` and `f^(prime)(alpha)!=0`

Options

A. There exist, `r, s in R," where " r lt s`, such that f is one- one on the open interval (r, s)
B. There exists `x_(0) in (-4, 0)" such that " |f'(x_(0)) | le 1 `
C. `underset( x to infty) lim f(x) =m = 1`
D. There exists `alpha in (-4, 4) " such that " f(alpha) + f'(alpha) = 0 and f'(alpha) ne 0`

Answer

A, B, D

Explanation

We have,
`(f(0))^(2)+(f'(0))^(2)=85`
`and" "f:Rto[-2,2]`
(a) Since, f is twice differentiable funstion, so f is continuous function.
Hence, we can always find `x in(r,s),` where f(x) is one-one.
`:.` This statement is true .
(b) By L.M.V.T
`f'(c) = (f(b)-f(a))/(b-a) rArr |f'(c)|=|(f(b)-f(a))/(b-a)|`
` rArr" " |f'(x_(0))|=|(f(0)-f(-4))/(0+4)|=|(f(0)-f(-4))/4|`
Range of f is `[-2,2]`
`:. 4le f(0) -f(-4) le 4 rArr 0 le|(f(0)-f(-4))/4| le 1`
Hence, ` |f'(x_(0)) | le 1`.
Hence, statement is true.
(c ) As no function is given, then we assume
`f(x) = 2 sin ((sqrt(85)x)/2) `
` :." " f'(x) = sqrt(85) cos ((sqrt(85)x)/2) `
Now, `(f(0))^(2)+(f'(0))^(2) = (2 sin 0)^(2) + (sqrt(85) cos 0)^(2)`
` (f(0))^(2)+(f'(0))^(2) = 85`
and `underset ( x to infty) lim f(x) ` does not exists.
Hence, statements is false.
(d) From option `b,|f'(x_(0))| le and x_(0) in (-4, 0) `
`:." " (f'(x_(0)))^(2) le 1`
`g(x_(0))=(f(x_(0)))^(2) + (f'(x_(0)))^(2) le 4 + 1 `
`" "[:' f(x_(0)) in [-2, 2]]`
` rArr" " g(x_(0)) le 5`
Now, let `p in (-4, 0)` for which g(p) = 5
Similarly, let q be smallest positive number ` q in (0, 4) ` such that g(q) = 5
Hence, by Rolle's theorem is (p,q)
`g'(c ) = 0" for " alpha in (-4, 4) ` and since g(x) is greater than 5 as we move form ` x = p " to " x = q`
and `f(x))^(2) le 4 rArr (f'(x))^(2) ge 1 ` in (p,q)
Thus, `" " g'(c ) = 0`
` rArr " " f'f+f'f'' = 0`
So, `f(alpha) + f''(alpha) = 0 and f'(alpha) ne 0`
Hence, statement is true.