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Joint Entrance Examination
Mathematics
Differentiation
Question 57
Question 57 of 67
Medium
MCQ
Let \[f(x) = 2x + {\tan ^{ - 1}}x\] and \[g(x) = {\log _e}(\sqrt {1 + {x^2}} + x),x \in [0,3]\]. Then
A
there exists \[\widehat x \in [0,3]\] such that \[f'(\widehat x) < g'(\widehat x)\]
B
there exist \[0 < {x_1} < {x_2} < 3\] such that \[f(x) < g(x),\forall x \in ({x_1},{x_2})\]
C
\[\min f'(x) = 1 + \max g'(x)\]
D
\[\max f(x) > \max g(x)\]
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Solutions · 1
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Official solution
\[
f
′
(
x
)
=
2
+
1
1
+
x
2
,
g
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1
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f
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2
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g
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3
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min
=
f
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3
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=
2
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1
10
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21
10
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g
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ln
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<
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7
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2
\begin{aligned} & f^{\prime}(x)=2+\frac{1}{1+x^2}, g^{\prime}(x)=\frac{1}{\sqrt{x^2+1}} \\\\ & f^{\prime \prime}(x)=-\frac{2 x}{\left(1+x^2\right)^2}<0 \\\\ & g^{\prime \prime}(x)=-\frac{1}{2}\left(x^2+1\right)^{-3 / 2} \cdot 2 x<0 \\\\ & \left.f^{\prime}(x)\right|_{\min }=f^{\prime}(3)=2+\frac{1}{10}=\frac{21}{10} \\\\ & \left.g^{\prime}(x)\right|_{\max }=g^{\prime}(0)=1 \\\\ & \left.f^{\prime}(x)\right|_{\max }=f(3)=2+\tan ^{-1} 3 \\\\ & \left.g(x)\right|_{\max }=g(3)=\ln (3+\sqrt{10})<\ln <7<2 \end{aligned}
f
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=
2
+
1
+
x
2
1
,
g
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x
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1
1
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g
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−
2
1
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3/2
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x
<
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f
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m
i
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=
f
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3
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2
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10
1
=
10
21
g
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a
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1
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m
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g
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3
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ln
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3
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10
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<
ln
<
7
<
2
\]
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