The number of integral values of k for which hx-sgnx2-2kx -sgn k-2 is continuous for all x- is-Note - sgny denotes the signum function of y-

For the signum function to be continuous for all x ∈ℝ, x^2 2kx +sgn (k+2) 0 D 0 ⇒ 4k^2 4sgn (k+2)0 ⇒ k^2 sgn (k+2)0 Case 1: k+2 0 ⇒ k 2, ⇒ k^2 10 ⇒ nbsp;k ...

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Byju's Answer

Standard XIII

Mathematics

Continuity of a Function

The number of...

Question

The number of integral values of $k$ for which $h (x) = sgn (x^{2} - 2 k x + sgn (k + 2))$ is continuous for all $x \in R$ is
[Note : $sgn (y)$ denotes the signum function of $y$ .]

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Solution

The correct option is B $1$
For the signum function to be continuous for all $x \in R$ ,
$x^{2} - 2 k x + sgn (k + 2) > 0$
$D < 0 \Rightarrow 4 k^{2} - 4 sgn (k + 2) < 0$
$\Rightarrow k^{2} - sgn (k + 2) < 0$

Case $1 : k + 2 > 0 \Rightarrow k > - 2$ ,
$\Rightarrow k^{2} - 1 < 0 \Rightarrow k \in (- 1, 1)$
Which is possible.

Case $2 : k + 2 < 0 \Rightarrow k < - 2$
$\Rightarrow k^{2} + 1 < 0$
Which is not possible.

Case $3 : k + 2 = 0 \Rightarrow k = - 2$
$\Rightarrow k^{2} < 0$
Which is not possible.

$∴ k \in (- 1, 1)$
Hence, the number of integral value of $k$ is $1.$

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The number of integral values of k for which h(x)=sgn(x2−2kx+sgn(k+2)) is continuous for all x∈R is [Note : sgn(y) denotes the signum function of y.]

The number of integral values of $k$ for which $h (x) = sgn (x^{2} - 2 k x + sgn (k + 2))$ is continuous for all $x \in R$ is
[Note : $sgn (y)$ denotes the signum function of $y$ .]