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Joint Entrance Examination
Mathematics
Differentiation
Question 61
Question 61 of 67
Hard
MCQ
Let \[y=f(x)=\sin ^{3}\left(\frac{\pi}{3}\left(\cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{\frac{3}{2}}\right)\right)\right)\]. Then, at x = 1,
A
\[2 y^{\prime}+\sqrt{3} \pi^{2} y=0\]
B
\[y^{\prime}+3 \pi^{2} y=0\]
C
\[\sqrt{2} y^{\prime}-3 \pi^{2} y=0\]
D
\[2 y^{\prime}+3 \pi^{2} y=0\]
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Official solution
f
(
x
)
=
sin
3
(
π
3
cos
(
π
3
2
(
−
4
x
3
+
5
x
2
+
1
)
3
/
2
)
)
f(x)=\sin ^{3}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right)
f
(
x
)
=
sin
3
(
3
π
cos
(
3
2
π
(
−
4
x
3
+
5
x
2
+
1
)
3/2
)
)
\[
f
′
(
x
)
=
3
sin
2
(
π
3
cos
(
π
3
2
(
−
4
x
3
+
5
x
2
+
1
)
3
/
2
)
)
cos
(
π
3
cos
(
π
3
2
(
−
4
x
3
+
5
x
2
+
1
)
3
/
2
)
)
π
3
(
−
sin
(
π
3
2
(
−
4
x
3
+
5
x
2
+
1
)
3
/
2
)
)
π
3
2
3
2
(
−
4
x
3
+
5
x
3
+
1
)
1
/
2
(
−
12
x
2
+
10
x
)
\begin{aligned} & f^{\prime}(x)=3 \sin ^{2}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\ & \cos \left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\ & \frac{\pi}{3}\left(-\sin \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\ & \frac{\pi}{3 \sqrt{2}} \frac{3}{2}\left(-4 x^{3}+5 x^{3}+1\right)^{1 / 2}\left(-12 x^{2}+10 x\right) \end{aligned}
f
′
(
x
)
=
3
sin
2
(
3
π
cos
(
3
2
π
(
−
4
x
3
+
5
x
2
+
1
)
3/2
)
)
cos
(
3
π
cos
(
3
2
π
(
−
4
x
3
+
5
x
2
+
1
)
3/2
)
)
3
π
(
−
sin
(
3
2
π
(
−
4
x
3
+
5
x
2
+
1
)
3/2
)
)
3
2
π
2
3
(
−
4
x
3
+
5
x
3
+
1
)
1/2
(
−
12
x
2
+
10
x
)
\]
f
′
(
1
)
=
3
π
2
16
f^{\prime}(1)=\frac{3 \pi^{2}}{16}
f
′
(
1
)
=
16
3
π
2
\[
f
(
1
)
=
sin
3
(
π
3
cos
(
π
3
2
2
2
)
)
=
sin
3
(
−
π
6
)
=
−
1
8
∴
2
f
′
(
1
)
+
3
π
2
f
(
1
)
=
0
\begin{aligned} & f(1)=\sin ^{3}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}} 2 \sqrt{2}\right)\right) \\\\ &=\sin ^{3}\left(-\frac{\pi}{6}\right)=\frac{-1}{8} \\\\ & \therefore 2 f^{\prime}(1)+3 \pi^{2} f(1)=0 \end{aligned}
f
(
1
)
=
sin
3
(
3
π
cos
(
3
2
π
2
2
)
)
=
sin
3
(
−
6
π
)
=
8
−
1
∴
2
f
′
(
1
)
+
3
π
2
f
(
1
)
=
0
\]
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