If the function fx-1-cos2pxx2- x0 is continuous at x-0- then the values of 2p-q2 can be

For function to be continuous at x=0 R.H.L.=L.H.L=f(0) Now, R.H.L.=lim_x→0^+1 √(1 x)x×1+√(1 x)1+√(1 x) =lim_x→0^+xx(1+√(1 x)) ∴ R.H.L.=12=q^2 L.H.L.=lim_x→0^ 1 ...

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Standard XII

Mathematics

Differentiability

If the functi...

Question

If the function $f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 2 p x}{x^{2}}, & x < 0 q^{2}, & x = 0 \frac{1 - \sqrt{1 - x}}{x}, & x > 0 \end{matrix}$ is continuous at $x = 0,$ then the value(s) of $2 (p + q^{2})$ can be

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Solution

The correct option is C $2$
For function to be continuous at $x = 0$
$R . H . L . = L . H . L = f (0)$
Now, $R . H . L . = lim x \to 0^{+} \frac{1 - \sqrt{1 - x}}{x} \times \frac{1 + \sqrt{1 - x}}{1 + \sqrt{1 - x}}$
$= lim x \to 0^{+} \frac{x}{x (1 + \sqrt{1 - x})}$
$∴ R . H . L . = \frac{1}{2} = q^{2}$

$L . H . L . = lim x \to 0^{-} \frac{1 - cos 2 p x}{x^{2}}$
$= lim x \to 0^{-} \frac{2 {sin}^{2} p x}{x^{2}} = 2 p^{2}$
$\Rightarrow 2 p^{2} = \frac{1}{2}$
$\Rightarrow p = \pm \frac{1}{2}$
$∴ 2 (p + q^{2}) = \pm 1 + 1 = 0$ or $2$

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Differentiability

Standard XII Mathematics

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If the function f(x)=⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩1−cos2pxx2,x<0q2,x=01−√1−xx,x>0 is continuous at x=0, then the value(s) of 2(p+q2) can be

The correct option is C 2For function to be continuous at x=0 R.H.L.=L.H.L=f(0) Now, R.H.L.=limx→0+1−√1−xx×1+√1−x1+√1−x =limx→0+xx(1+√1−x) ∴R.H.L.=12=q2 L.H.L.=limx→0−1−cos2pxx2 =limx→0−2sin2pxx2=2p2 ⇒2p2=12 ⇒p=±12 ∴2(p+q2)=±1+1=0 or 2

If the function $f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - cos 2 p x}{x^{2}}, & x < 0 q^{2}, & x = 0 \frac{1 - \sqrt{1 - x}}{x}, & x > 0 \end{matrix}$ is continuous at $x = 0,$ then the value(s) of $2 (p + q^{2})$ can be