Q: For x ∈ ℝ , let tan - 1 x ∈ - π 2 , π 2 . Then the minimum value of the function f : ℝ → ℝ defined by f x = ∫ 0 x tan - 1 x e t - cos t 1 + t 2023 d

Given, f x = ∫ 0 x tan - 1 x e t - cos t 1 + t 2023 d t Now differentiating both side we get, f ' x = e x tan - 1 x - cos x tan - 1 x 1 + x tan - 1 x 2023 × x 1 + x 2 + tan - 1 x ⇒ f ' x = g x · h x where g x = e x tan − 1 ⁡ x − cos ⁡ x tan − 1 ⁡ x 1 + x tan − 1 ⁡ x 2023 > 0 ∀ x And h x = x 1 + x 2 + tan − 1 x which is < 0 &#160

Q: The greatest integer less than or equal to ∫ 1 2 log 2 x 3 + 1 d x + ∫ 1 log 2 9 2 x - 1 1 3 d x is

Let I = ∫ 1 2 log 2 x 3 + 1 d x + ∫ 1 log 2 9 2 x - 1 1 3 d x Let ∫ 1 log 2 9 2 x - 1 1 3 d x = I 1 Let 2 x - 1 = t 3 ⇒ 2 x ln 2 d x = 3 t 2 d t ⇒ d x = 3 t 2 t 3 + 1 ln 2 d t i.e. I 1 = ∫ 1 2 3 t 3 ln 2 t 3 + 1 d t So, I = ∫ 1 2 log 2 x 3 + 1 d x + ∫ 1 2 3 t 3 ln 2 t 3 + 1 d t or I = ∫ 1 2 log 2 t 3 + 1 + t · 3 t 2 t 3 + 1 ln 2 d t = t ⋅ log 2 t 3 + 1 1 2 = 2 log 2 9 - 1 log 2 2 = 2 log 2 9 - 1 = 4 log 2 3

Q: Consider the equation ∫ 1 e log e x 1 2 x a - log e x 3 2 2 d x = 1 , a ∈ - ∞ , 0 ∪ 1 , ∞ . Which of the following statements is/are TRUE?

Let I = ∫ 1 e ln x 1 2 d x x a - ln x 3 2 2 Now put a - ln x 3 2 = t ⇒ - 3 2 ln x 1 2 · 1 x d x = d t So, the integral becomes I = ∫ a a - 1 - 2 3 d t t 2 ⇒ I = - 2 3 t - 2 + 1 - 2 + 1 a a - 1 ⇒ I = 2 3 t a a - 1 ⇒ I = 2 3 1 a - 1 - 1 a Now given I = 2 3 1 a a - 1 = 1 ⇒ 2 = 3 a 2 - 3 a ⇒ 3 a 2 - 3 a - 2 = 0 ⇒ a = 3 ± 9 - 4 3 - 2 6 So, a = 3 + 33 6 , 3 - 33 6

Q: Let f : - π 2 , π 2 → ℝ be a continuous function such that f 0 = 1 and ∫ 0 π 3 f t d t = 0 Then which of the following statements is (are

(A) Let g x = f x - 3 cos 3 x Now ∫ 0 π 3 g x d x = ∫ 0 π 3 f x d x - 3 ∫ 0 π 3 cos 3 x d x = 0 - sin 3 x 0 π 3 = 0 Hence g x = 0 has a root in 0 , π 3 (B) Let h x = f x - 3 sin 3 x + 6 π Now ∫ 0 π 3 h x d x = ∫ 0 π 3 f x d x - 3 ∫ 0 π 3 sin 3 x d x + ∫ 0 π 3 6 π d x = 0 - - cos 3 x 0 π 3 + 6 x π 0 π 3 = 0 - 2 + 2 = 0 Hence h x = 0 has a root in 0 , π 3 (C) lim x &#859

Q: Let g_ i : [ 8 , 3 8 ] R , i=1,2, and f: [ 8 , 3 8 ] R be functions such that g_ 1 (x)=1, g_ 2 (x)=|4 x- | and f(x)= ^ 2 x, for all x [ 8 , 3 8 ] Define S_ i = _ 8 ^ 3 8 f(x) g_ i (x) d x, i=1,2 The v

S 1 = ∫ π 8 3 π 8 sin 2 x · 1 d x … … . . . i S 1 = ∫ π 8 3 π 8 sin 2 3 π 8 + π 8 - x d x S 1 = ∫ π 8 3 π 8 cos 2 x d x … … . . ii From (i) and (ii), we get 2 S 1 = ∫ π 8 3 π 8 1 d x 2 S 1 = π 4 16 S 1 π = 2

Question 1

The value of the definite integral _ 0 ^ 2 1 3^ x+3 2 \, dx is

Accepted Answer

Let I = _ 0 ^ 2 1 3^x + 3 dx Substitute t = 3^x, which gives dt = 3^x 3 dx dx = dt t 3 . When x = 0, t = 1. When x = 2, t = 9. The integral becomes: I = _ 1 ^ 9 1 t + 3 dt t 3 = 1 3 _ 1 ^ 9 1 t(t + 3) dt Using partial fractions: 1 t(t + 3) = 1 3 ( 1 t - 1 t + 3 ) I = 1 3 3 _ 1 ^ 9 ( 1 t - 1 t + 3 ) dt I = 1 3 3 [ t - (t + 3)]_ 1 ^ 9 I = 1 3 3 [ ( t t + 3 ) ]_ 1 ^ 9 I = 1 3 3 ( ( 9 12 ) - ( 1 4 ) ) I = 1 3 3 ( ( 3 4 ) - ( 1 4 ) ) I = 1 3 3 ( 3/4 1/4 ) = 1 3 3 3 I = 1 3

Question 2

If = _ 1 2 ^2 ^ -1 x 2 x^2-3 x+2 d x then the value of 7 ( 2 7 ) is ____________. (Here, the inverse trigonometric function ^ -1 x assumes values in (- 2 , 2 ).)

Accepted Answer

= _ 1 2 ^2 ^ -1 x 2 x^2-3 x+2 d x ....(i) aligned & Let x= 1 t \ & d x=- 1 t^2 d t \ & = _2^ 1 2 ^ -1 ( 1 t ) 2 t^2 - 3 t +2 ( -1 t^2 ) d t aligned = _ 1 2 ^2 ^ -1 t 2 t ^2-3 t +2 dt ....(ii) Now by (i) + (ii) aligned & 2 = _ 1 2 ^2 2 2 x^2-3 x+2 d x \ & = 8 _ 1 2 ^2 d x x^2- 3 x 2 +1 \ & = 8 _ 1 2 ^2 (x- 3 4 )^2+ 7 16 \ & = 8 7 4 [ ^ -1 ( x- 3 4 7 4 ) ]_ 1 2 ^2 aligned aligned & = 2 7 [ ^ -1 4 x-3 7 ]_ 1 2 ^2 \ & = 2 7 [ ^ -1 5 7 - ^ -1 (- 1 7 ) ] \ & = 2 7 ^ -1 ( 5 7 + 1 7 ) 1- 5 7 \ &

Question 3

Paragraph: Let $f: [0, 2 ] [0,1]$ be the function defined by $f(x)= ^2 x$ and let $g: [0, 2 ] [0, )$ be the function defined by $g(x)= x 2 -x^2 $. Question: The value of 2 _0^ 2 f(x) g(x) d x-

Accepted Answer

I=2 _0^ 2 ^2 x x 2 -x^2 _ I_1 - _0^ 2 g(x) d x Let I_1= _0^ 2 ^2 x ( 4 )^2- (x- 4 )^2 (making perfect square) apply kings I_1= _0^ 2 ^2 x ( 4 )^2- ( 2 -x )^2 add both 2 I_1= _0^ 2 ( 4 )^2- (x- 4 )^2 i.e. 2 I_1= _0^ 2 g(x) Now I=2 I_1- _0^ 2 g(x)=0

Question 4

Let $f: [0, 2 ] [0,1]$ be the function defined by $f(x)= ^2 x$ and let $g: [0, 2 ] [0, )$ be the function defined by $g(x)= x 2 -x^2 $. The value of 16 ^3 _0^ 2 f(x) g(x) d x is

Accepted Answer

Now I_1= _0^ 2 f(x) g(x) d x= 1 2 _0^ 2 g(x) d x (it is explained in previous question solution) i.e. 1 2 _0^ 2 ( 4 )^2- (x- 4 )^2 dx Using a^2-x^2 = 1 2 (x a^2-x^2 +a^2 ^ -1 ( x a ) )+C 1 2 [ (x- 4 ) 2 x 2 -x^2 + ^2 2 2 ^ -1 ( x- 4 4 ) ]_0^ / 2 aligned & 1 2 [ (0+ ^3 64 )- (0+ ( - ^3 64 ) ) ] \ & 1 2 ^3 32 aligned Now 16 ^3 ^3 64 = 1 4 =0.25

Question 5

For x &#8712; &#8477; , let tan - 1 x &#8712; - &#960; 2 , &#960; 2 . Then the minimum value of the function f : &#8477; &#8594; &#8477; defined by f x = &#8747; 0 x tan - 1 x e t - cos t 1 + t 2023 d

Accepted Answer

Given, f x = &#8747; 0 x tan - 1 x e t - cos t 1 + t 2023 d t Now differentiating both side we get, f ' x = e x tan - 1 x - cos x tan - 1 x 1 + x tan - 1 x 2023 &#215; x 1 + x 2 + tan - 1 x &#8658; f ' x = g x &#183; h x &#160;where&#160; g x = e x tan &#8722; 1 &#8289; x &#8722; cos &#8289; x tan &#8722; 1 &#8289; x 1 + x tan &#8722; 1 &#8289; x 2023 &#62; 0 &#160; &#8704; &#160; x And  h x = x 1 + x 2 + tan &#8722; 1 &#8197;&#8197;&#8197;&#8197; x &#160; which&#160;is &#160; &#60; 0 &#160

Question 6

The greatest integer less than or equal to &#8747; 1 2 log 2 x 3 + 1 d x + &#8747; 1 log 2 9 2 x - 1 1 3 d x is

Accepted Answer

Let  I = &#8747; 1 2 log 2 x 3 + 1 d x + &#8747; 1 log 2 9 2 x - 1 1 3 d x Let &#8747; 1 log 2 9 2 x - 1 1 3 d x = I 1 Let 2 x - 1 = t 3 &#8658; 2 x ln 2 d x = 3 t 2 d t &#8658; d x = 3 t 2 t 3 + 1 ln 2 d t i.e.  I 1 = &#8747; 1 2 3 t 3 ln 2 t 3 + 1 d t   So,  I = &#8747; 1 2 log 2 x 3 + 1 d x + &#8747; 1 2 3 t 3 ln 2 t 3 + 1 d t or I = &#8747; 1 2 log 2 t 3 + 1 + t &#183; 3 t 2 t 3 + 1 ln 2 d t = t &#8901; log 2 t 3 + 1 1 2 = 2 log 2 9 - 1 log 2 2 = 2 log 2 9 - 1 = 4 log 2 3

Question 7

For positive integer n , define f n = n + 16 + 5 n - 3 n 2 4 n + 3 n 2 + 32 + n - 3 n 2 8 n + 3 n 2 + 48 - 3 n - 3 n 2 12 n + 3 n 2 + &#8230; + 25 n - 7 n 2 7 n 2 . Then, the value of lim n &#8594; &#

Accepted Answer

Given  f n = n + 16 + 5 n - 3 n 2 4 n + 3 n 2 + 32 + n - 3 n 2 8 n + 3 n 2 + &#8230; . + 25 n - 7 n 2 7 n 2 = 16 + 5 n - 3 n 2 4 n + 3 n 2 + 1 + 32 + n - 3 n 2 8 n + 3 n 2 + 1 + &#8230; &#8230; + 25 n - 7 n 2 7 n 2 + 1 f n = 9 n + 16 4 n + 3 n 2 + 9 n + 32 8 n + 3 n 2 + &#8230; &#8230; + 25 n 7 n 2 i.e.  f n = &#8721; r = 1 n 9 n + 16 r 4 r n + 3 n 2 = 1 n &#8721; r = 1 n 9 + 16 r n 4 r n + 3 Now,  lim n &#8594; &#8734; f n = &#8747; 0 1 9 + 16 x 4 x + 3 d x = &#8747; 0 1 16 x + 1

Question 8

Consider the equation &#8747; 1 e log e x 1 2 x a - log e x 3 2 2 d x = 1 , a &#8712; - &#8734; , 0 &#8746; 1 , &#8734; . Which of the following statements is/are TRUE?

Accepted Answer

Let  I = &#8747; 1 e ln x 1 2 d x x a - ln x 3 2 2 Now put  a - ln x 3 2 = t &#8658; - 3 2 ln x 1 2 &#183; 1 x d x = d t So, the integral becomes  I = &#8747; a a - 1 - 2 3 d t t 2 &#8658; I = - 2 3 t - 2 + 1 - 2 + 1 a a - 1 &#8658; I = 2 3 t a a - 1 &#8658; I = 2 3 1 a - 1 - 1 a Now given  I = 2 3 1 a a - 1 = 1 &#8658; 2 = 3 a 2 - 3 a &#8658; 3 a 2 - 3 a - 2 = 0 &#8658; &#160; a = 3 &#177; 9 - 4 3 - 2 6 So,  a = 3 + 33 6 , 3 - 33 6

Question 9

Let  f : - &#960; 2 , &#960; 2 &#8594; &#8477;  be a continuous function such that  f 0 = 1  and  &#8747; 0 &#960; 3 f t d t = 0 Then which of the following statements is (are

Accepted Answer

(A) Let g x = f x - 3 cos 3 x Now &#8747; 0 &#960; 3 g x d x = &#8747; 0 &#960; 3 f x d x - 3 &#8747; 0 &#960; 3 cos 3 x d x = 0 - sin 3 x 0 &#960; 3 = 0 Hence g x = 0  has a root in 0 , &#960; 3 (B) Let  h x = f x - 3 sin 3 x + 6 &#960; Now &#8747; 0 &#960; 3 h x d x = &#8747; 0 &#960; 3 f x d x - 3 &#8747; 0 &#960; 3 sin 3 x d x + &#8747; 0 &#960; 3 6 &#960; d x = 0 - - cos 3 x 0 &#960; 3 + 6 x &#960; 0 &#960; 3 = 0 - 2 + 2 = 0 Hence h x = 0 has a root in 0 , &#960; 3 (C) lim x &#859

Question 10

Let g_ i : [ 8 , 3 8 ] R , i=1,2, and f: [ 8 , 3 8 ] R be functions such that g_ 1 (x)=1, g_ 2 (x)=|4 x- | and f(x)= ^ 2 x, for all x [ 8 , 3 8 ] Define S_ i = _ 8 ^ 3 8 f(x) g_ i (x) d x, i=1,2 The v

Accepted Answer

S 1 = &#8747; &#960; 8 3 &#960; 8 sin 2 x &#183; 1 d x &#160; &#160; &#160; &#160; &#8230; &#8230; . . . i S 1 = &#8747; &#960; 8 3 &#960; 8 sin 2 3 &#960; 8 + &#960; 8 - x d x S 1 = &#8747; &#960; 8 3 &#960; 8 cos 2 x d x &#160; &#160; &#160; &#8230; &#8230; . . ii From (i) and (ii), we get 2 S 1 = &#8747; &#960; 8 3 &#960; 8 1 d x 2 S 1 = &#960; 4 16 &#160; S 1 &#960; = 2

Question 11

Let g_ i : [ 8 , 3 8 ] R , i=1,2, and f: [ 8 , 3 8 ] R be functions such that g_ 1 (x)=1, g_ 2 (x)=|4 x- | and f(x)= ^ 2 x, for all x [ 8 , 3 8 ] Define S_ i = _ 8 ^ 3 8 f(x) g_ i (x) d x, i=1,2 The v

Accepted Answer

S 2 = &#8747; &#960; 8 3 &#960; 8 sin 2 x 4 x - &#960; d x &#160; &#160; &#160; &#8230; &#8230; &#8230; i S 2 = &#8747; &#960; 8 3 &#960; 8 sin 2 3 &#960; 8 + &#960; 8 - x 4 3 &#960; 8 + &#960; 8 - x - &#960; d x S 2 = &#8747; &#960; 8 3 &#960; 8 cos 2 x &#960; - 4 x d x &#160; &#160; &#160; &#160; &#8230; &#8230; &#8230; ii From (i) and (ii), we get 2 S 2 = &#8747; &#960; 8 3 &#960; 8 4 x - &#960; d x 2 &#160; S 2 = &#960; 2 16 S 2 = &#960; 2 32 &#8658; 48 S 2 &#960; 2 = 48 32 &#8658; 48 S 2 &#

Question 12

For any real number  x , let  x  denotes the largest integer less than or equal to  x . If  I = &#8747; 0 10 10 x x + 1 d x , then the value of  9 I  is ____.

Accepted Answer

Given I = &#8747; 0 10 10 x x + 1 d x Check the values of  x  at which the greatest integer function becomes integer. 10 x x + 1 = 1 &#160;&#160; &#8658; &#160;&#160; x = 1 9 10 x x + 1 = 4 &#160;&#160; &#8658; &#160;&#160; x = 2 3 10 x x + 1 = 9 &#160;&#160; &#8658; &#160;&#160; x = 9 I = &#8747; 0 1 9 0 d x + &#8747; 1 9 2 3 1 d x + &#8747; 2 3 9 2 d x + &#8747; 9 10 3 d x I = 0 + [ x ] 1 9 2 3 + [ 2 x ] 2 3 9 + [ 3 x ] 9 10 I = 2 3 - 1 9 + 18 - 4 3 + 30 - 27 I = 21 - 2 3 - 1 9 = 182

Question 13

Let  f : &#8477; &#8594; &#8477; be a differentiable function such that its derivative f ' is continuous and  f &#960; = - 6 . If  F : 0 , &#960; &#8594; &#8477; is defined by

Accepted Answer

I = &#8747; 0 &#960; f ' x &#183; cos &#160; x + F x &#183; cos &#160; x d x = 2 = &#8747; 0 &#960; f ' x &#183; cos &#160; x &#183; d x + &#8747; 0 &#960; F x cos &#160; x &#183; d x = 2 = cos x &#183; f ( x ) 0 &#960; - &#8747; 0 &#960; - sin x &#183; f x d x + &#8747; 0 &#960; F x &#183; cos &#160; x &#183; d x = 2 &#8658; &#160;&#160; cos &#960; . f &#960; - cos 0 . f 0 + &#8747; 0 &#960; sin x &#183; f x d x + &#8747; 0 &#960; F x &#183; cos &#160; x d x = 2 &#8658; &#160;&#160; - 1

Question 14

Which of the following inequalities is/are TRUE?

Accepted Answer

We know that  cos x = 1 &#8722; x 2 2 ! + x 4 4 ! - x 6 6 ! + &#8230; &#8230; &#8230; &#8230; &#8658; cos x &#8805; 1 &#8722; x 2 2 &#8756; &#8747; 0 1 x cos x d x &#8805; &#8747; 0 1 x 1 &#8722; x 2 2 d x = x 2 2 &#8722; x 4 8 0 1 = 1 2 &#8722; 1 8 = 3 8 So, option  A  is correct. Now, we also know that  sin x = x &#8722; x 3 3 ! + x 5 5 ! - x 7 7 ! + &#8230; &#8230; &#8230; &#8230; &#8230; &#8658; sin x &#8805; x &#8722; x 3 3 ! , &#160; x &#8712; 0 , &#160; 1 &#8756;

Question 15

The value of the integral &#8747; 0 &#960; / 2 3 cos &#8289; &#952; cos &#8289; &#952; + sin &#8289; &#952; 5 d &#952; equals

Accepted Answer

I = &#8747; 0 &#960; 2 3 cos &#8289; &#952; sin &#8289; &#952; + cos &#8289; &#952; 5 . d &#952; &hellip;(i) I = &#8747; 0 &#960; 2 3 sin &#8289; &#952; cos &#8289; &#952; + sin &#8289; &#952; 5 . d &#952; &hellip;(ii) using &#160; &#8747; a b f x . d x = &#8747; a b f a + b - x . d x (i) + (ii) &#8658; 2 I = &#8747; 0 &#960; 2 3 sin &#8289; &#952; + cos &#8289; &#952; sin &#8289; &#952; + cos &#8289; &#952; 5 . d &#952; &#8658; 2 I = 3 &#8747; 0 &#960; 2 d &#952; sin &#8289; &#952; + cos &#8289

Question 16

If I = 2 &#960; &#8747; - &#960; 4 &#960; 4 d x 1 + e s i n x 2 - c o s 2 x then find 27 I 2 equals____

Accepted Answer

I = 2 &#960; &#8747; - &#960; 4 &#960; 4 1 1 + e s i n x 2 - c o s 2 x &#8901; d x Using &#8747; - a a f x &#8901; d x = &#8747; 0 a f x + f - x &#8901; d x I = 2 &#960; &#8747; 0 &#960; 4 1 1 + e s i n x 2 - c o s 2 x + 1 1 + e - s i n x 2 - c o s 2 x &#8901; d x I = 2 &#960; &#8747; 0 &#960; 4 1 1 + e s i n x 2 - c o s 2 x + e s i n x e s i n x + 1 2 - c o s 2 x &#8901; d x I = 2 &#960; &#8747; 0 &#960; 4 d x 2 - c o s 2 x I = 2 &#960; &#8747; 0 &#960; 4 d x 2 - 1 - t a n 2 x 1 + t a n 2 x = 2

Definite Integration — Previous Year Questions & Solutions