Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

The function $f(x)=\left\{\begin{array}{1 1}x+a\sqrt 2\sin x&0\leq x < \large\frac{\pi}{4}\\2x\cot x+b&\large\frac{\pi}{4}\normalsize \leq x < \frac{\pi}{2}\\a\cos 2x-b\sin x&\large\frac{\pi}{2}\normalsize < x \leq \pi\end{array}\right.$ is continuous for $0 \leq x \leq \pi$ then a,b are

$\begin{array}{1 1}(a)\;\large\frac{\pi}{6},\frac{\pi}{12}&(b)\;\large\frac{\pi}{3},\frac{\pi}{6}\\(c)\;\large\frac{\pi}{6},-\frac{\pi}{12}&(d)\;\text{None of these}\end{array}$

Can you answer this question?

1 Answer

0 votes
We apply the test of continuity at $x=\large\frac{\pi}{4}$ and $x=\large\frac{\pi}{2}$ to get the values of a and b.
At $x=\large\frac{\pi}{4}$ LHL =RHL =value
$\Rightarrow \large\frac{\pi}{4}$$+a\sqrt 2\times \large\frac{1}{\sqrt 2}$
$\Rightarrow 2.\large\frac{\pi}{4}$$.(1)+b$
$\Rightarrow a-b=\large\frac{\pi}{4}$------(1)
At $x=\large\frac{\pi}{2}$$,2\big(\large\frac{\pi}{2}\big)$$(0)+b=a(-1)-b(1)$
$\Rightarrow 2b+a=0$------(2)
Solving (1) and (2) we get
Hence (c) is the correct answer.
answered Jan 6, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App