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Joint Entrance Examination
Mathematics
Limits, Continuity and Differentiability
Question 224
Question 224 of 401
Medium
MCQ
The value of
k
k
k
for which the function
f
(
x
)
=
{
4
5
tan
4
x
tan
5
x
,
0
<
x
<
π
2
k
+
2
5
,
x
=
π
2
f(x)=\begin{cases}\dfrac45\dfrac{\tan4x}{\tan5x}, & 0<x<\dfrac\pi2\\k+\dfrac25, & x=\dfrac\pi2\end{cases}
f
(
x
)
=
⎩
⎨
⎧
5
4
tan
5
x
tan
4
x
,
k
+
5
2
,
0
<
x
<
2
π
x
=
2
π
is continuous at
x
=
π
2
x=\dfrac\pi2
x
=
2
π
, is:
A
17
20
\dfrac{17}{20}
20
17
B
2
5
\dfrac25
5
2
C
3
5
\dfrac35
5
3
D
−
2
5
-\dfrac25
−
5
2
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Official solution
lim
x
→
π
/
2
f
(
x
)
=
f
(
π
2
)
⇒
k
+
2
5
=
1
⇒
k
=
3
5
\displaystyle\lim_{x\to\pi/2}f(x)=f\left(\dfrac\pi2\right)\Rightarrow k+\dfrac25=1\Rightarrow k=\dfrac35
x
→
π
/2
lim
f
(
x
)
=
f
(
2
π
)
⇒
k
+
5
2
=
1
⇒
k
=
5
3
.
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