Let a-b-c- such that a-b-c- If fx-sinax2-bx-cx2-1- if x-1 -1- if x-1 a sgnx-1cos2x-2-bx2- if 1-x-2 - is continuous at x-1- then the value of a2-b2 is -Here- sgnk denotes signum function of k-

F(1^ )=lim_x→ 1^ sin(ax^2+bx+c)(x^2 1) Applying L'Hospital Rule f(1^ )=lim_x→ 1^ cos(ax^2+bx+c)×(2ax+b)2x = (2a+b)2 as a+b+c=π f(1^+)= nbsp;lim_x→ 1^+a sgn(x ...

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Standard XII

Mathematics

Left Hand Limit

Let a,b,c∈ℝ s...

Question

Let $a, b, c \in R$ such that $a + b + c = π$ . If $f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{sin (a x^{2} + b x + c)}{x^{2} - 1}, & if x < 1 - 1, & if x = 1 a sgn (x + 1) cos (2 x - 2) + b x^{2}, & if 1 < x \leq 2 \end{matrix}$ is continuous at $x = 1$ , then the value of $(a^{2} + b^{2})$ is
[Here, $sgn (k)$ denotes signum function of $k$ ]

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Solution

$f (1^{-}) = lim x \to 1^{-} \frac{sin (a x^{2} + b x + c)}{(x^{2} - 1)}$
Applying L'Hospital Rule
$f (1^{-}) = lim x \to 1^{-} \frac{cos (a x^{2} + b x + c) \times (2 a x + b)}{2 x}$
$= \frac{- (2 a + b)}{2}$ as $a + b + c = π$

$f (1^{+}) = lim x \to 1^{+} a sgn (x + 1) cos (2 x - 2) + b x^{2} = a + b$
And $f (1) = - 1$

Since, $f (x)$ is continuous at $x = 1,$
$∴ f (1^{-}) = f (1^{+}) = f (1)$
$\Rightarrow \frac{- (2 a + b)}{2} = a + b = - 1$
$\Rightarrow a = 3, b = - 4$
$∴ a^{2} + b^{2} = (3)^{2} + (- 4)^{2} = 25$

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Similar questions

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Let a,b,c∈R such that a+b+c=π. If f(x)=⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩sin(ax2+bx+c)x2−1,if x<1−1,if x=1a sgn(x+1)cos(2x−2)+bx2,if 1<x≤2 is continuous at x=1, then the value of (a2+b2) is [Here, sgn(k) denotes signum function of k]