If l- m and n are number of points of discontinuity- non-differentiability and local extrema of function fx-max-1-x2-x- in x-1-1- respectively then l-m-n is equal to-where - denotes fractional part function-

f(x) is discontinuous at x=1⇒ l=1 f(x) is non differentiable at x=1,1√(2)⇒ m=2 f(x) is having local extrema at x= 1,0,1√(2),1⇒ n=4 ∴ l+m+n=7 ...

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Byju's Answer

Standard XII

Mathematics

One - One function

If l, m and n...

Question

If $l, m$ and $n$ are number of points of discontinuity, non-differentiability and local extrema of function $f (x) = max [\sqrt{1 - x^{2}}, {x}]$ in $x \in [- 1, 1]$ respectively then $l + m + n$ is equal to
[where ${\cdot}$ denotes fractional part function]

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Solution

$f (x)$ is discontinuous at $x = 1 \Rightarrow l = 1$
$f (x)$ is non-differentiable at $x = 1, \frac{1}{\sqrt{2}} \Rightarrow m = 2$
$f (x)$ is having local extrema at $x = - 1, 0, \frac{1}{\sqrt{2}}, 1 \Rightarrow n = 4$
$∴ l + m + n = 7$

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If l,m and n are number of points of discontinuity, non-differentiability and local extrema of function f(x)=max[√1−x2,{x}] in x∈[−1,1] respectively then l+m+n is equal to [where {⋅} denotes fractional part function]

f(x) is discontinuous at x=1⇒l=1 f(x) is non-differentiable at x=1,1√2⇒m=2 f(x) is having local extrema at x=−1,0,1√2,1⇒n=4 ∴l+m+n=7

If $l, m$ and $n$ are number of points of discontinuity, non-differentiability and local extrema of function $f (x) = max [\sqrt{1 - x^{2}}, {x}]$ in $x \in [- 1, 1]$ respectively then $l + m + n$ is equal to
[where ${\cdot}$ denotes fractional part function]

$f (x)$ is discontinuous at $x = 1 \Rightarrow l = 1$
$f (x)$ is non-differentiable at $x = 1, \frac{1}{\sqrt{2}} \Rightarrow m = 2$
$f (x)$ is having local extrema at $x = - 1, 0, \frac{1}{\sqrt{2}}, 1 \Rightarrow n = 4$
$∴ l + m + n = 7$