If fx= (a+bcosx+csinx)/(x^2) ; x>0 9 ; x≤0 is continuous at x=0, then the value of (a+b)/5 is

Correct answer: 7.20. For fx to be continuous at x=0 limx→0fx=f0 limx→0(a+bcos x+csin x)/(x^2)=9 limx→0(a+b1-x^2)/(2!)+(x^2)/(4!)-....+cx-(x^3)/(3!)+(x^5)/(5!)-....x^2=9 limx→0(a+b+cx+-b)/2x^2+-c/6x^3.....x^2=9 for the above limit to satisfy a+b=0, c=0 and -b/2=9⇒b=-18, c=0, a=18

Q9 / 93 · NTA Abhyas (JEE Main) 2020 · Mathematics · Continuity and Differentiability

If $f (x) = \{\begin{array}{l} \frac{a + b c o s x + c s i n x}{x^{2}} & ; & x > 0 \\ 9 & ; & x \leq 0 \end{array}$ is continuous at $x = 0$ , then the value of $\frac{|a| + |b|}{5}$ is

Correct answer: 7.20

Solution

For

f (x)

to be continuous at

x = 0

\underset{x \to 0}{l i m} f (x) = f (0)

\underset{x \to 0}{l i m} \frac{a + b c o s x + c s i n x}{x^{2}} = 9

\underset{x \to 0}{l i m} \frac{a + b (1 - \frac{x^{2}}{2!} + \frac{x^{2}}{4!} - . . . .) + c (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - . . . .)}{x^{2}} = 9

\underset{x \to 0}{l i m} \frac{(a + b) + c x + (- \frac{b}{2}) x^{2} + (- \frac{c}{6}) x^{3} . . . . .}{x^{2}} = 9

for the above limit to satisfy

a + b = 0

c = 0

and

(- \frac{b}{2}) = 9 \Rightarrow b = - 18, c = 0, a = 18

Hence, the correct answer is 7.20.

If fx=a+bcosx+csinxx2;x>09;x≤0 is continuous at x=0, then the value of a+b5 is

Solution

More Continuity and Differentiability Questions

If $f (x) = \{\begin{array}{l} \frac{a + b c o s x + c s i n x}{x^{2}} & ; & x > 0 \\ 9 & ; & x \leq 0 \end{array}$ is continuous at $x = 0$ , then the value of $\frac{|a| + |b|}{5}$ is