Let fx-x2-6x-5 and m is the number of points of non-derivability of y-f-x- If -f-x-k- k - has at least m distinct solutions- then the number of integral values of k is

Given : f(x) = x^2 6x +5 = (x 1)(x 5) Vertex of the parabola = ( b2a, D4a) =(3, 4) The graph of y = |f(|x|)| = |(|x| 1)(|x| 5)| is Clearly, y=|f(|x|)| is ...

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Byju's Answer

Standard XIII

Mathematics

Differentiability

Let fx=x2-6x+...

Question

Let $f (x) = x^{2} - 6 x + 5$ and $m$ is the number of points of non-derivability of $y = | f (| x |) | .$ If $| f (| x |) | = k, k \in R$ has at least $m$ distinct solution(s), then the number of integral values of $k$ is

2

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Solution

The correct option is C $4$
Given : $f (x) = x^{2} - 6 x + 5 = (x - 1) (x - 5)$
Vertex of the parabola
$= (- \frac{b}{2 a}, - \frac{D}{4 a}) = (3, - 4)$
The graph of $y = | f (| x |) | = | (| x | - 1) (| x | - 5) |$ is

Clearly, $y = | f (| x |) |$ is non-derivable at $5$ points.
For $| f (| x |) | = k$ to have at least $5$ solutions,
$0 < k \leq 4$

Hence, the number of integral values of $k$ is $4$ .

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Let f(x)=x2−6x+5 and m is the number of points of non-derivability of y=|f(|x|)|. If |f(|x|)|=k,k∈R has at least m distinct solution(s), then the number of integral values of k is

Let $f (x) = x^{2} - 6 x + 5$ and $m$ is the number of points of non-derivability of $y = | f (| x |) | .$ If $| f (| x |) | = k, k \in R$ has at least $m$ distinct solution(s), then the number of integral values of $k$ is