If the function fx=(1-x)/2tan⁡(πx)/2 is continuous at x=1, then f1 is equal to

Correct answer: (A) 1/(π). f1=limx→1fx=limx→1(1-x)/2tan⁡(πx)/2 Put x=1+h =limh→0⁡(1-1+h)/2tan⁡(π)/21+h =limh→0⁡(-h)/2tan⁡(π)/2+(πh)/2 =limh→0⁡(-h)/2-cot⁡(πh)/2 =limh→0(π)/2⋅htan⁡(πh)/2⋅1/(π)=1/(π)

Q40 / 93 · NTA Abhyas (JEE Main) 2020 · Mathematics · Continuity and Differentiability

If the function $f (x) = \frac{(1 - x)}{2} \tan \frac{π x}{2}$ is continuous at $x = 1,$ then $f (1)$ is equal to

Correct answer
A
$\frac{1}{π}$
B
$\frac{π}{2}$
C
$0$
D
$π$

Solution

f (1) = \underset{x \to 1}{l i m} f (x) = \underset{x \to 1}{l i m} (\frac{1 - x}{2}) \tan (\frac{π x}{2})

Put

x = 1 + h

= \lim_{h \to 0} \frac{(1 - (1 + h))}{2} \tan (\frac{π}{2} (1 + h))

= \lim_{h \to 0} \frac{(- h)}{2} \tan (\frac{π}{2} + \frac{π h}{2})

= \lim_{h \to 0} \frac{(- h)}{2} (- \cot \frac{π h}{2})

= \underset{h \to 0}{l i m} \frac{\frac{π}{2} \cdot h}{\tan (\frac{π h}{2})} \cdot \frac{1}{π} = \frac{1}{π}

Hence, the correct answer is option: 'A'.

If the function fx=1-x2tan⁡πx2 is continuous at x=1, then f1 is equal to

Solution

More Continuity and Differentiability Questions

If the function $f (x) = \frac{(1 - x)}{2} \tan \frac{π x}{2}$ is continuous at $x = 1,$ then $f (1)$ is equal to