If $f (x) = s g n (x^{2} - a x + 1)$ has maximum number of points of discontinuity, then

a \in (1, 0)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a \in (- 1, 1)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a \in (- \infty, - 2) \cup (2, \infty)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

a \in (- 2, 2)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $a \in (- \infty, - 2) \cup (2, \infty)$
$f (x) = s g n (x^{2} - a x + 1)$
For $f (x)$ to have maximum number of points of discontinuity.
Roots of quadratic equation $x^{2} - a x + 1 = 0$ should be real and different
$\Rightarrow D > 0 \Rightarrow a^{2} - 4 > 0 \Rightarrow a ϵ (- \infty, - 2) \cup (2, \infty)$

Suggest Corrections

Similar questions

Q. If

f (x) = \frac{1}{(x - 1) (x - 2)} a n d g (x) = \frac{1}{x^{2}}

, then points of discontinuity of f(g(x)) are

Let

{lim}_{x \to a} f (x)

exists but it is not equal to f (a). Then f(x) is discontinuous at x

=

a and a is called a removable discontinuity. If

{lim}_{x \to a^{-}} f (x) = l a n d {lim}_{x \to a^{+}} f (x) = m

exist but

l \neq m .

Then a is called a jump discontinuity. If one of the limits (left hand limit or right hand limit ) does not exist, then a is called an infinite discontinuity.

Let f(x)

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} 2 | x |, & x \leq - 1 2 x, & - 1 \leq x \leq 0 x + 1, & 0 < x \leq 1 2 & x > 1 \end{matrix}

Then

f (x)

Q. Let

f (x) = x - ∣ ∣ x - x^{2} ∣ ∣, x ϵ [- 1, 1]

. Then the number of points at which

f (x)

is discontinuous is

Q. Let

f (x) = x | x - x^{2} |, x \in [- 1, 1]

. Then the number of points at which f(x) is discontinuous is

Let $f (x) = {\begin{matrix} x [\frac{1}{x}] + x [x] & if x \neq 0 0 & if x = 0 \end{matrix}$ where $[\cdot]$ denotes the greatest integer function, then which of the following is true?

Join BYJU'S Learning Program

If f(x)=sgn(x2−ax+1) has maximum number of points of discontinuity, then

The correct option is A a∈(−∞,−2)∪(2,∞)f(x)=sgn(x2−ax+1)For f(x) to have maximum number of points of discontinuity.Roots of quadratic equation x2−ax+1=0 should be real and different ⇒D>0⇒a2−4>0⇒aϵ(−∞,−2)∪(2,∞)

If $f (x) = s g n (x^{2} - a x + 1)$ has maximum number of points of discontinuity, then