If $f\left( x \right) = \mathop \smallint \nolimits_0^x \,\,({\sin ^4}t\,\, + \,\,{\cos ^4}\,\,t)\,dt$, then \[f\,\,\left( {x\,\, + \,\,{\rm{\pi }}} \right)\] will be equal to

$f\left( x \right)\,\, + \,\,f\left( {\rm{\pi }} \right)$ or $f\left( x \right)\,\, + \,\,2f\left( {\frac{{\rm{\pi }}}{2}} \right)$

$f\left( x \right)\,\, + \,\,f\left( {\frac{{\rm{\pi }}}{2}} \right)$

$f\left( x \right)\,\, - \,\,f\left( {\rm{\pi }} \right)$

$f\left( x \right)\,\, - \,\,2\,f\left( {\frac{{\rm{\pi }}}{2}} \right)$

The value of the integral $\mathop \smallint \limits_{1/e}^e \left| {\log \,\,x} \right|dx$, is

$2\,\,\left( {\frac{{1 - e}}{e}} \right)$

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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Hard

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Showing 31-51 of 51 questions

31 Hard

31 of 51

Analyzing +4 | -1

Consider the integrals

{I_1} = \int \limits_0^1 \,{e^{ - x}}{\cos ^2}\,\,x\,dx,\;\,\,{I_2} = \int \limits_0^1 \,{e^{ - {x^2}}}{\cos ^2}\,\,x\,dx,\;\,\,{I_3} = \int \limits_0^1 \,{e^{ - {x^2}}}\;dx

and

{I_4} = \int \limits_0^1 \,{e^{ - \left( {1/2} \right){x^2}}}\,\,dx

The greatest of these integral is

View answer and explanation

Correct option: D

{I_4}

Option D (

{I_4}

) is correct because by comparing the integrands over the interval [0,1],

{e^{ - (1/2){x^2}}}

is the greatest, leading to the largest integral value.

32 Hard

32 of 51

Analyzing +4 | -1

I_1=\int_0^{\pi / 2} \cos (\sin x) d x, I_2=\int_0^{\pi / 2} \sin (\cos x) d x

and

I_3=\int_0^{\pi / 2} \cos x d x

, then

View answer and explanation

Correct option: A

I_1>I_3>I_2

Option C (

I_1>I_2>I_3

) is incorrect. The correct order is

I_1 > I_3 > I_2

because

\cos(\sin x) > \cos x

and

\cos x > \sin(\cos x)

for

x \in (0, \pi/2)

33 Hard

33 of 51

Application +4 | -1

f\left( x \right) = \int \nolimits_0^x \,\,({\sin ^4}t\,\, + \,\,{\cos ^4}\,\,t)\,dt

, then

f\,\,\left( {x\,\, + \,\,\mathrm{\pi }} \right)

will be equal to

View answer and explanation

Correct option: B

f\left( x \right)\,\, + \,\,f\left( \mathrm{\pi } \right)

f\left( x \right)\,\, + \,\,2f\left( {\frac{\mathrm{\pi }}{2}} \right)

Option B (

f(x) + f(\pi)

f(x) + 2f(\frac{{\pi }}{2})

) is correct because the function is periodic and the integral can be split into parts.

34 Hard

34 of 51

Analyzing +4 | -1

Let

I(a)=\int_0^\pi\left(\frac{x}{a}+a \sin x\right)^2 d x

, where '

a

' is positive real. The value of '

a

' for which

I(a)

attains its minimum value, is

View answer and explanation

Correct option: A

\sqrt{\pi \sqrt{\frac{2}{3}}}

Option A (

\sqrt{{\pi \sqrt{{\frac{{2}}{{3}}}}}}

) is correct because the minimum value of I(a) is attained when the derivative with respect to 'a' is zero, leading to this value.

35 Hard

35 of 51

Application +4 | -1

\int \nolimits_{ - 3}^2 \,\,\left\{ {\left| {\,x + 1\,} \right|\,\, + \,\,\left| {\,x + 2\,} \right|\,\, + \,\,\left| {\,x - 1\,} \right|} \right\}dx

is equal to

View answer and explanation

Correct option: C

\frac{{47}}{2}

Option C (

\frac{{47}}{2}

) is correct because it is the result of correctly evaluating the definite integral by breaking it into intervals based on the absolute value function.

36 Hard

36 of 51

Application +4 | -1

\int \nolimits_{ - 1}^1 \,\,\left[ {x + \,} \right[x + \,[x]]]\,\,dx

where

\left[ {\, \cdot \,} \right]

denotes greatest integer function, is equal to

View answer and explanation

Correct option: C

- 3

Option D (-4) is correct because the integral of the greatest integer function is evaluated by splitting the integral into intervals where the greatest integer function is constant.

37 Hard

37 of 51

Application +4 | -1

\int_{-1}^3\left[x+\frac{1}{2}\right] \mathrm{dx}=

(where [.] represents G.I.F.)

View answer and explanation

Correct option: A

Option A (4) is correct because the integral of the greatest integer function [x + 1/2] from -1 to 3 evaluates to 4 after breaking the integral into intervals where the function is constant.

38 Hard

38 of 51

Analyzing +4 | -1

Let

I_{1}=\int_{0}^{\pi / 2} e^{-x^{2}} \sin (x) d x ; I_{2}=\int_{0}^{\pi / 2} e^{-x^{2}} d x ; I_{3}=\int_{0}^{\pi / 2} e^{-x^{2}}(1+x) d x

and consider the statements

I.

{I}_{1}<{I}_{2}

\quad

II.

{I}_{2}<{I}_{3}

\quad

III.

{I}_{1}={I}_{3}

Which of the following is(are) true?

View answer and explanation

Correct option: D

Both I and II

Option D (Both I and II) is correct because in the interval (0, π/2), 0 < sin(x) < 1 and 1+x > 1, leading to the inequalities I1 < I2 < I3.

39 Hard

39 of 51

Understanding +4 | -1

Let

f: R \rightarrow R

be defined as

f(x)=\int\limits_{0}^1 \frac{x^2+t^2}{2-t} d t

. Then the curve

y=f(x)

is -

View answer and explanation

Correct option: D

A parabola

Option D (A parabola) is correct because after evaluating the integral, the resulting equation is a quadratic in

x

, which represents a parabola.

40 Hard

40 of 51

Application +4 | -1

The value of

\frac{\pi^2}{\ln 3} \int_{7 / 6}^{5 / 6} \sec (\pi x) d x

, is

View answer and explanation

Correct option: C

\pi

The correct answer is C (

\pi

) because after evaluating the definite integral of the secant function and simplifying, the result is

\pi

41 Hard

41 of 51

Application +4 | -1

f(x)=e^{g(x)}

and

g(x)=\int_2^x \frac{t d t}{1+t^4}

, then

f^{\prime}(2)

is equal to

View answer and explanation

Correct option: A

2 / 17

Option A is correct because using the chain rule and the fundamental theorem of calculus, we find

f'(2) = 2/17

42 Hard

42 of 51

Analyzing +4 | -1

For

x \in R, x \neq 0

, if

y(x)

is differentiable function such that

x \int_1^x y(t) d t=(x+1) \int_1^x t y(t) d t

, then

y(x)

equals (where

C

is a constant).

View answer and explanation

Correct option: D

\frac{C}{x^3} e^{-1 / x}

Option D (

\frac{C}{x^3} e^{-1 / x}

) is correct because it is the solution to the differential equation obtained after differentiating the given integral equation twice.

43 Hard

43 of 51

Analyzing +4 | -1

Let

f

be a differentiable function from

R

R

such that

\left| {f(x) - f(y)} \right| \le 2{\left| {x - y} \right|^{\frac{3}{2}}},

for all

x,y \in R.

If f

(0) = 1

, then

\mathrm{ }\int\limits_0^1 {{f^2}(x)dx}

is equal to

View answer and explanation

Correct option: C

Option C (1) is correct because the given condition implies that f'(x) = 0, making f(x) a constant function. Since f(0) = 1, f(x) = 1 for all x, and the integral of f²(x) from 0 to 1 is 1.

44 Hard

44 of 51

Application +4 | -1

\int \limits_{ - \pi }^\pi \,\,{\left[ {\cos \,\,px\,\, - \,\,\sin \,\,qx} \right]^2}dx

, where

p,\;\,q

are integers is equal to

View answer and explanation

Correct option: D

2\,\pi

Option D (

2\pi

) is correct because when integrating the square of

(\cos px - \sin qx)

from

-\pi

\pi

, the cross term vanishes due to symmetry, and the integrals of

\cos^2 px

and

\sin^2 qx

each contribute

\pi

45 Hard

45 of 51

Application +4 | -1

f(x)

is a continuous function for all real values of

x

and satisfies

\int\limits_{n}^{n+1} f(x) d x=\frac{n^2}{2} \,\forall\, n \in I

, then

\int\limits_{{-3}}^5 f(|x|) d x

is equal to:

View answer and explanation

Correct option: B

\frac{35}{2}

Option B (

\frac{{35}}{{2}}

) is correct because it is obtained by summing the integrals over the given intervals using the provided condition.

Explore Other Chapters in MATHEMATICS

View All →

SEQUENCE OF SERIES 1764 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs DIFFERENTIAL EQUATIONS 977 MCQs

46 Hard

46 of 51

Application +4 | -1

Let

{A_1}

be the area of the parabola

{y^2} = 4ax

lying between vertex and latusrectum and

{A_2}

be the area between latusrectum and double ordinate

x = 2a

. Then,

{A_1}/{A_2} =

View answer and explanation

Correct option: B

\left( {2\sqrt 2 + 1} \right)/7

Option B is correct because the ratio of the areas A1 and A2 is calculated using definite integrals and simplification, resulting in (2

\sqrt{2}

+ 1)/7.

47 Hard

47 of 51

Application +4 | -1

The value of the integnal

\int\limits_{0}^3\left(x^2+1\right) \, d[x]

, is where

[\cdot]

is the greatest integer function

View answer and explanation

Correct option: C

Option C (17) is correct because the integral is evaluated using integration by parts and the properties of the greatest integer function.

48 Hard

48 of 51

Application +4 | -1

The value of the integral

\int \limits_{1/e}^e \left| {\log \,\,x} \right|dx

, is

View answer and explanation

Correct option: A

2\,\,\left( {\frac{{e - 1}}{e}} \right)

Option A (

2\,\,\left( {\frac{{e - 1}}{e}} \right)

) is correct because the integral of the absolute value of log x from 1/e to e evaluates to this expression.

49 Hard

49 of 51

Application +4 | -1

Find the value of

\int\limits_{0}^{\pi / 4}(\sin x)^4 d x

View answer and explanation

Correct option: B

\frac{3 \pi}{32}-\frac{1}{4}

Option B (

\frac{{3 \pi}}{{32}} - \frac{1}{4}

) is correct because the integral of

(\sin x)^4

from 0 to

\frac{\pi}{4}

is evaluated using trigonometric identities to simplify the integrand.

50 Hard

50 of 51

Analyzing +4 | -1

f\,\,\left( x \right)

is a function satisfying

f\left( {\frac{1}{x}} \right)\,\, + \,\,{x^2}\,f\left( x \right) = 0

for all non-zero

x

, then

\int \nolimits_{\sin \,\,\mathrm{\theta }}^{\mathrm{cosec \theta }} \,\,f\left( x \right)\,\,dx

is equal to

View answer and explanation

Correct option: A

\mathrm{0}

Option A (0) is correct because using the substitution x = 1/t and the given functional equation, the integral becomes its negative, implying it equals zero.

51 Hard

51 of 51

Application +4 | -1

The value of the integral

\int_1^{{e^6}} {\left[ {\frac{{\log x}}{3}} \right]} \,dx,

where [.] denotes the greatest integer function, is

View answer and explanation

Correct option: B

e^6-e^3

Option B (

e^6-e^3

) is correct because the integral is solved by breaking it into intervals based on the greatest integer function.

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