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Joint Entrance Examination
Mathematics
Differentiation
Question 60
Question 60 of 67
Medium
MCQ
Let
f
f
f
and
g
g
g
be the twice differentiable functions on \[\mathbb{R}\] such that
f
′
′
(
x
)
=
g
′
′
(
x
)
+
6
x
f''(x)=g''(x)+6x
f
′′
(
x
)
=
g
′′
(
x
)
+
6
x
f
′
(
1
)
=
4
g
′
(
1
)
−
3
=
9
f'(1)=4g'(1)-3=9
f
′
(
1
)
=
4
g
′
(
1
)
−
3
=
9
f
(
2
)
=
3
g
(
2
)
=
12
f(2)=3g(2)=12
f
(
2
)
=
3
g
(
2
)
=
12
. Then which of the following is NOT true?
A
\[g(-2)-f(-2)=20\]
B
There exists \[x_0\in(1,3/2)\] such that \[f(x_0)=g(x_0)\]
C
\[|f'(x)-g'(x)| < 6\Rightarrow -1 < x < 1\]
D
If \[-1 < x < 2\], then \[|f(x)-g(x)| < 8\]
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Solutions · 1
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Official solution
f
′
′
(
x
)
=
g
′
′
(
x
)
+
6
x
f''(x) = g''(x) + 6x
f
′′
(
x
)
=
g
′′
(
x
)
+
6
x
\[ \Rightarrow f'(x) = g'(x) + 3{x^2} + C\]
f
′
(
1
)
=
g
′
(
1
)
+
3
+
C
f'(1) = g'(1) + 3 + C
f
′
(
1
)
=
g
′
(
1
)
+
3
+
C
\[ \Rightarrow g = 3 + 3 + C \Rightarrow C = 3\] \[ \Rightarrow f'(x) = g'(x) + 3{x^2} + 3\] \[ \Rightarrow f(x) = g(x) + {x^2} + 3x + C'\]
x
=
2
x = 2
x
=
2
f
(
2
)
=
g
(
2
)
+
14
+
C
′
f(2) = g(2) + 14 + C'
f
(
2
)
=
g
(
2
)
+
14
+
C
′
12
=
4
+
14
+
C
′
12 = 4 + 14 + C'
12
=
4
+
14
+
C
′
\[ \Rightarrow C' = - 6\] \[ \Rightarrow f(x) = g(2) + {x^3} + 3x - 6\]
f
(
−
2
)
=
g
(
−
2
)
−
8
−
6
−
6
f( - 2) = g( - 2) - 8 - 6 - 6
f
(
−
2
)
=
g
(
−
2
)
−
8
−
6
−
6
g
(
−
2
)
−
f
(
−
2
)
=
20
g( - 2) - f( - 2) = 20
g
(
−
2
)
−
f
(
−
2
)
=
20
f
′
(
x
)
−
g
′
(
x
)
=
3
x
2
+
3
f'(x) - g'(x) = 3{x^2} + 3
f
′
(
x
)
−
g
′
(
x
)
=
3
x
2
+
3
\[x \in ( - 1,1)\] \[3{x^2} + 3 \in (0,6)\] \[ \Rightarrow f'(x) - g'(x) \in (0,6)\]
f
(
x
)
−
g
(
x
)
=
x
3
+
3
x
−
6
f(x) - g(x) = {x^3} + 3x - 6
f
(
x
)
−
g
(
x
)
=
x
3
+
3
x
−
6
At
x
=
−
1
x = - 1
x
=
−
1
∣
f
(
−
1
)
−
g
(
−
1
)
∣
=
10
|f( - 1) - g( - 1)| = 10
∣
f
(
−
1
)
−
g
(
−
1
)
∣
=
10
\[\therefore\] Option (4) is false.
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