JEE Advanced 2011 Paper 2
Exam JEE Advanced
Year 2011
Paper Paper 2
Subject Mathematics
Chapter Continuity and Differentiability JEE Advanced 2011 · Paper 2 · Mathematics · Continuity and Differentiability
If f(x)=\ array cc -x-(π)/(2), & x ≤-(π)/(2) - x, & -(π)/(2) 1...
If
f(x)=\left\{ − x − π 2 , x ≤ − π 2 − cos x , − π 2 < x ≤ 0 x − 1 , 0 < x ≤ 1 log x , x > 1 \begin{array}{cc}-x-\frac{\pi}{2}, & x \leq-\frac{\pi}{2} \\ -\cos x, & -\frac{\pi}{2} < x \leq 0 \\ x-1, & 0 < x \leq 1 \\ \log x, & x>1\end{array} − x − 2 π , − cos x , x − 1 , log x , x ≤ − 2 π − 2 π < x ≤ 0 0 < x ≤ 1 x > 1 \right. , then
Options A f ( x ) f(x) f ( x ) x = − π 2 x=-\frac{\pi}{2} x = − 2 π B f ( x ) f(x) f ( x ) x = 0 x=0 x = 0 C f ( x ) f(x) f ( x ) x = 1 x=1 x = 1 D f ( x ) f(x) f ( x ) x = − 3 2 x=-\frac{3}{2} x = − 2 3 Correct answer A. f ( x ) f(x) f ( x ) x = − π 2 x=-\frac{\pi}{2} x = − 2 π
Step-by-step solution
f(x)=\left\{
− x − π 2 , x ≤ − π 2 − cos x , − π 2 < x ≤ 0 x − 1 , 0 < x ≤ 1 log x , x > 1 \begin{array}{cc}-x-\frac{\pi}{2}, & x \leq-\frac{\pi}{2} \\ -\cos x, & -\frac{\pi}{2} \lt x \leq 0 \\ x-1, & 0 \lt x \leq 1 \\ \log x, & x \gt 1\end{array} − x − 2 π , − cos x , x − 1 , log x , x ≤ − 2 π − 2 π < x ≤ 0 0 < x ≤ 1 x > 1 \right.
C o n t i n u i t y a t Continuity at C o n t in u i t y a t x=-\frac{\pi}{2}
, , , f ( − π 2 ) = − ( − π 2 ) − π 2 = 0 R H L ⇒ lim h → 0 − cos ( − π 2 + h ) = 0 \begin{aligned}& f\left(-\frac{\pi}{2}\right)=-\left(-\frac{\pi}{2}\right)-\frac{\pi}{2}=0 \\
& \mathrm{RHL} \Rightarrow \lim _{h \rightarrow 0}-\cos \left(-\frac{\pi}{2}+h\right)=0
\end{aligned} f ( − 2 π ) = − ( − 2 π ) − 2 π = 0 RHL ⇒ h → 0 lim − cos ( − 2 π + h ) = 0 ∴ \therefore ∴ Continuous at
x = 0 x=0 x = 0
Continuity at
x = 0 ⇒ f ( 0 ) = − 1 R H L ⇒ lim h → 0 ( 0 + h ) − 1 = − 1 ∴ x=0 \Rightarrow f(0)=-1\mathrm{RHL} \Rightarrow \lim _{h \rightarrow 0}(0+h)-1=-1\therefore x = 0 ⇒ f ( 0 ) = − 1 RHL ⇒ lim h → 0 ( 0 + h ) − 1 = − 1 ∴ Continuous at
x = 0 x=0 x = 0
Continuity at
x = 1 ; f ( 1 ) = 0 R H L ⇒ lim h → 0 log ( 1 + h ) = 0 ∴ x=1 ; f(1)=0\mathrm{RHL} \Rightarrow \lim _{h \rightarrow 0} \log (1+h)=0\therefore x = 1 ; f ( 1 ) = 0 RHL ⇒ lim h → 0 log ( 1 + h ) = 0 ∴ Continuous at
x = 1 x=1 x = 1
Here,
f^{\prime}(x)=\left\{ − 1 , x ≤ − π 2 sin x , − π 2 < x ≤ 0 1 , 0 < x ≤ 1 1 x , x > 1 \begin{array}{cc}-1, & x \leq-\frac{\pi}{2} \\ \sin x, & -\frac{\pi}{2} \lt x \leq 0 \\ 1, & 0 \lt x \leq 1 \\ \frac{1}{x}, & x \gt 1\end{array} − 1 , sin x , 1 , x 1 , x ≤ − 2 π − 2 π < x ≤ 0 0 < x ≤ 1 x > 1 \right.
Differentiable at
x = 0 x=0 x = 0 ,
L H D = 0 , R H D = 1 ∴ \mathrm{LHD}=0, \mathrm{RHD}=1\therefore \quad LHD = 0 , RHD = 1 ∴ Not differentiable at
x = 0 x=0 x = 0
Differentiable at
x = 1 x=1 x = 1 ,
L H D = 1 , R H D = 1 ∴ \mathrm{LHD}=1, \mathrm{RHD}=1\therefore LHD = 1 , RHD = 1 ∴ Differentiable at
x = 1 x=1 x = 1
Also, for
x = − 3 2 ⇒ f ( x ) = − x − 3 2 ∴ x=-\frac{3}{2} \Rightarrow f(x)=-x-\frac{3}{2}\therefore x = − 2 3 ⇒ f ( x ) = − x − 2 3 ∴ Differentiable at
x = − 3 2 x=-\frac{3}{2} x = − 2 3
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