Let f(x)=2 x+^{-1} x and g(x)=_{e}({1+x²+x), x[0,3]. Then [1-Feb-2023 Shift 1]

Correct Answer is : max⁡f(x)>max⁡g(x) f(x) > g(x)maxf(x)>maxg(x)

There exists x[0,3] such that f'(x)

Correct Answer is : max⁡f(x)>max⁡g(x) f(x) > g(x)maxf(x)>maxg(x)

Test Index

Differentiation

Section: Mathematics

Question : 28 of 79

Marks: +1, -0

Let

f(x)=2 x+\tan^{-1} x

g(x)=\log_{e}(\sqrt{1+x^2}+x)

x\in[0,3]

There exists $x\in[0,3]$ such that $f'(x) < g'(x)$
$\max f(x) > \max g(x)$
There exist $0 < x_1 < x_2 < 3$ such that $f(x) < g$ (x), $\forall x \in (x_1, x_2)$
$\min f'(x)=1+\max g'(x)$

Solution:

f(x)=2x+\tan^{-1}x

and

g(x)=\ln(\sqrt{1+x^2}+x)

and

x\in[0,3]

g'(x)=\frac{1}{\sqrt{1+x^2}}

Now,

0 \le x \le 3

0 \le x^2 \le 9

1 \le 1+x^2 \le 10

So,

2+\frac{1}{10} \le f'(x) \le 3

\frac{21}{10} \le f'(x) \le 3 \text{ and } \frac{1}{\sqrt{10}} \le g'(x) \le 1

option (4) is incorrect

From above,

g'(x) < f'(x) \forall x \in [0,3]

Option (1) is incorrect.

f'(x) \& g'(x)

both positive so

f(x) \& g(x)

both are increasing

So,

\max(f(x)).

x=3

6+\tan^{-1}3

\max(g(x)

x=3

\ln(3+\sqrt{10})

And

6+\tan^{-1}3 > \ln(3+\sqrt{10})

Option (2) is correct

Go to Question:

Prev Question

Next Question