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Let $[x]$ denote the largest integer less than or equal to $x.$ The number of points in the open interval $(1,3)$ in which the function $f(x)=a^{[x^2]},a\gt1$ is not differentiable, is

  1. $0$
  2. $3$
  3. $5$
  4. $7$
in Calculus by Boss
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Correct Answer - Option (D)

Here, the given function is $f(x) = a^{[x^2]}$

Let $g(x) = a^x$ 

and $h(x) = [x]$

hence, h(x) represents Greatest Integer Function. 

Now, we know that $g(x)$ is continuous and differentiable over domain of real numbers,

So differentiability of $f(x)$ is solely dependent on the differentiability of $h(x)$. 

Now, by definition, Greatest Integer Functions are not continuous over integer values.

Hence, they are not differentiable over integer values. Take a look at graph of this function, you can see that left hand limit and right hand limit are not equal for integers.

So, $h(x)$ will not be differentiable when x is an integer. 


$h(x^2) = [x^2]$ will not be differentiable when $x^2$ is an integer.

hence, the given function $f(x)$ will not be differentiable when $x^2$ is an integer.

So, we have to find values of $ \sqrt x$ over set $S = \{x| x \in R, 1<x<3, x^2 \in Z\}$ where $R$ and $S$ represent set of Real and Integer numbers respectively. 

So, $f(x)$ will not be differentiable when, 


$x^2 = 2 $ i.e. $x = 1.41 \in S$

$x^2 = 3 $ i.e. $x = 1.72 \in S$

$x^2 = 4 $ i.e. $x = 2 \in S$

$x^2 = 5 $ i.e. $x = 2.236 \in S$

$x^2 = 6 $ i.e. $x = 2.5 \in S$

$x^2 = 7 $ i.e. $x = 2.64 \in S$

$x^2 = 8 $ i.e. $x = 2.82 \in S$

$x^3 = 9$ i.e. $x =3 \not \in S$ since it's open interval $(1,3)$, similar reason for $x=1$. 

Hence, $f(x)$ will not be differentiable for $7$ values in the interval $(1,3)$.

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