Let a - b - c - R such that a - b - c -1- -text if x -1 1a2textsgnx-1 cos 2 x-2-b x- 2- - text if 1 lt x le 2lend cases -is continuous at x -1- then the value of a 2- b 2-5 Here- sgn k denotes signum function of k

F (1^+) = a + b; f (1) = 1 f (1^ ) =lim_x → 1^ sin(ax^2+bx+c)x^2 1 [00] form nbsp; Applying L' Hopitals' rule, we get f(1^ ) = lim_x → 1^ (2ax+b)cos(ax ...

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Byju's Answer

Standard XII

Mathematics

Continuity in an Interval

Let a , b , c...

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 $\frac{a^{2} + b^{2}}{5}$ is
( Here, $sgn (k)$ denotes signum function of $k$ )

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Solution

$f (1^{+}) = a + b; f (1) = - 1$
$f (1^{-}) = lim x \to 1^{-} \frac{sin (a x^{2} + b x + c)}{x^{2} - 1} [\frac{0}{0}] form$
Applying L' Hopitals' rule, we get
$f (1^{-}) = lim x \to 1^{-} \frac{(2 a x + b) cos (a x^{2} + b x + c)}{2 x}$
$= \frac{(2 a + b) cos (a + b + c)}{2}$
$= - (\frac{2 a + b}{2})$

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

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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+b25 is ( Here, sgn(k) denotes signum function of k )

f(1+)=a+b; f(1)=−1 f(1−)=limx→1−sin(ax2+bx+c)x2−1 [00] form Applying L' Hopitals' rule, we get f(1−)=limx→1−(2ax+b)cos(ax2+bx+c)2x =(2a+b)cos(a+b+c)2 =−(2a+b2) f(x) is continuous at x=1 So, f(1+)=f(1−)=f(1) ⇒a+b=−(2a+b2)=−1 ∴a=3 and b=−4