JEE Advanced 2019 Paper 2 Offline | Application of Derivatives Question 4 | Mathematics

JEE Advanced 2019 Paper 2 Offline

MCQ (More than One Correct Answer)

-1

Let f : R $$ \to $$ R be given by

$$f(x) = (x - 1)(x - 2)(x - 5)$$. Define

$$F(x) = \int\limits_0^x {f(t)dt} $$, x > 0

Then which of the following options is/are correct?

F(x) $$ \ne $$ 0 for all x $$ \in $$ (0, 5)

F has a local maximum at x = 2

F has two local maxima and one local minimum in (0, $$\infty $$)

F has a local minimum at x = 1

JEE Advanced 2019 Paper 2 Offline

MCQ (More than One Correct Answer)

-1

Let, $$f(x) = {{\sin \pi x} \over {{x^2}}}$$, x > 0

Let x₁ < x₂ < x₃ < ... < x_n < ... be all the points of local maximum of f and y₁ < y₂ < y₃ < ... < y_n < ... be all the points of local minimum of f.

Then which of the following options is/are correct?

$$|{x_n} - {y_n}|\, > 1$$ for every n

$${x_{n + 1}} - {x_n}\, > 2$$ for every n

x₁ < y₁

$${x_n} \in \left( {2n,\,2n + {1 \over 2}} \right)$$ for every n

JEE Advanced 2017 Paper 2 Offline

MCQ (More than One Correct Answer)

-2

f : R $$ \to $$ R is a differentiable function such that f'(x) > 2f(x) for all x$$ \in $$R, and f(0) = 1 then

f(x) > e^2x in (0, $$\infty $$)

f'(x) < e^2x in (0, $$\infty $$)

f(x) is increasing in (0, $$\infty $$)

f(x) is decreasing in (0, $$\infty $$)

JEE Advanced 2017 Paper 2 Offline

MCQ (More than One Correct Answer)

-2

If $$f(x) = \left| {\matrix{ {\cos 2x} & {\cos 2x} & {\sin 2x} \cr { - \cos x} & {\cos x} & { - \sin x} \cr {\sin x} & {\sin x} & {\cos x} \cr } } \right|$$,

then

f(x) attains its minimum at x = 0

f(x) attains its maximum at x = 0

f'(x) = 0 at more than three points in ($$-$$$$\pi $$, $$\pi $$)

f'(x) = 0 at exactly three points in ($$-$$$$\pi $$, $$\pi $$)

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