Let a-b- b- 0- Define a functionfx-asin-2x-1- for x- 0tan 2x-sin 2xbx3- for x-0-If f is continuous at x - 0- then 10 - ab is equal to

LHL at x = 0 for f(x) lim_x→ 0^ asinπ2(x 1)=asin ( π2 )= a RHL at x = 0 for f(x) lim_x→ 0^+tan 2x sin 2xbx^3=lim_x→ 0^+tan 2xx(1 cos 2x)/x^2·1b =1blim_x→ 0^+ta ...

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Standard XII

Mathematics

Substitution Method to Remove Indeterminate Form

Let a,b,∈ℝ, b...

Question

Let $a, b, \in R, b \neq 0.$ Define a function
$f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} a sin \frac{π}{2} (x - 1), & for x \leq 0 \frac{tan 2 x - sin 2 x}{b x^{3}}, & for x > 0 \end{matrix} .$
If $f$ is continuous at $x = 0,$ then $10 - a b$ is equal to

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Solution

LHL at $x = 0$ for $f (x)$
$lim x \to 0^{-} a sin \frac{π}{2} (x - 1) = a sin (- \frac{π}{2}) = - a$
RHL at $x = 0$ for $f (x)$
$lim x \to 0^{+} \frac{tan 2 x - sin 2 x}{b x^{3}} = lim x \to 0^{+} \frac{tan 2 x}{x} \frac{(1 - cos 2 x)}{x^{2}} \cdot \frac{1}{b}$
$= \frac{1}{b} lim x \to 0^{+} \frac{tan 2 x}{x} \cdot \frac{1 - cos 2 x}{4 x^{2}} \cdot 4$
$= \frac{1}{b} \cdot 2 \cdot \frac{1}{2} \cdot 4 = \frac{4}{b}$
For continuity, $- a = \frac{4}{b} \Rightarrow a b = - 4$
So, $10 - a b = 14$

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Let a,b,∈R, b≠0. Define a function f(x)=⎧⎪ ⎪⎨⎪ ⎪⎩asinπ2(x−1),for x≤0tan2x−sin2xbx3,for x>0. If f is continuous at x=0, then 10−ab is equal to

LHL at x=0 for f(x) limx→0−asinπ2(x−1)=asin(−π2)=−a RHL at x=0 for f(x) limx→0+tan2x−sin2xbx3=limx→0+tan2xx(1−cos2x)x2⋅1b =1blimx→0+tan2xx⋅1−cos2x4x2⋅4 =1b⋅2⋅12⋅4=4b For continuity, −a=4b⇒ab=−4 So, 10−ab=14

Let $a, b, \in R, b \neq 0.$ Define a function
$f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} a sin \frac{π}{2} (x - 1), & for x \leq 0 \frac{tan 2 x - sin 2 x}{b x^{3}}, & for x > 0 \end{matrix} .$
If $f$ is continuous at $x = 0,$ then $10 - a b$ is equal to