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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Continuity and Differentiability

NTA Abhyas 2020 · Mathematics · Continuity and Differentiability

If f(x) = \ cases a + bcosx + csinx x ^ 2 & ; & x > 0 9 & ; & x ≤ 0...

If f\left(x\right) = \left\{\begin{cases}\dfrac{a + bcosx + csinx}{{x}^{2}} & ; & x > 0 \\ 9 & ; & x \le 0\end{cases}\right) is continuous atx = 0,thenthevalueof, then the value of\dfrac{\left|a\right| + \left|b\right|}{5}$ is

Correct answer

7.20

Step-by-step solution

For f(x)f\left(x\right) to be continuous at x=0x = 0
lim⁡x→0f(x)=f(0)\lim_{x \to 0}f\left(x\right) = f\left(0\right)
lim⁡x→0a+bcosx+csinxx2=9\lim_{x \to 0}\dfrac{a + bcosx + csinx}{{x}^{2}} = 9
lim⁡x→0a+b(1−x22!+x24!−....)+c(x−x33!+x55!−....)x2=9\lim_{x \to 0}\dfrac{a + b\left(1 - \dfrac{{x}^{2}}{2 !} + \dfrac{{x}^{2}}{4 !} - . . . .\right) + c\left(x - \dfrac{{x}^{3}}{3 !} + \dfrac{{x}^{5}}{5 !} - . . . .\right)}{{x}^{2}} = 9
lim⁡x→0(a+b)+cx+(−b2)x2+(−c6)x3.....x2=9\lim_{x \to 0}\dfrac{\left(a + b\right) + cx + \left(- \dfrac{b}{2}\right){x}^{2} + \left(- \dfrac{c}{6}\right){x}^{3} . . . . .}{{x}^{2}} = 9
for the above limit to satisfy
a+b=0a + b = 0,
c=0c = 0 and
\left(- \dfrac{b}{2}\right) = 9 ⇒ b = - 18, c = 0, a = 18
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