Q: Let R denote the set of all real numbers. For a real number x, let [x] denote the greatest integer less than or equal to x. Let n denote a natural number. Match each entry in List-I to the correct ent

aligned & P(x)=10 x^3-45 x^2+60 x+35 \\ & P^ (x)=30(x-1)(x-2) aligned P ( x ) decreases in [1,2] Range of P ( x )=[55,60] f( x )= [ P ( x ) n ] value of n =9 (Q) For g(x) to be increasing 2 n^2-13 n-15 0 (R) aligned & h(x)= (x^2-9 )^n (x^2+2 x+3 ) \\ & h^ (x)= (x^2-9 )^n(2 x+2)+ (x^2+2 x+3 ) n (x^2-9 )^ n-1 2 x \\ & = (x^2-9 )^ n-1 [2 (x^2-9 )(x+1)+2 n x (x^2+2 x+3 ) ] \\ & =(x+3)^ n-1 (x-3)^ n-1 q(x) aligned Derivative must change sign at x=3 aligned & n -1= odd \\ & n = even \\ & n =6 aligned

Q: Let R denote the set of all real numbers. Let f: R R be defined by f(x)= cases 6 x+ x 2 x+ x & if x 0 \\ 7 3 & if x=0 cases Then which of the following statements is (are) TRUE?

since 3> 7 3 3>f(0) x=0 is local minima ....option (B) is correct Now, aligned & f(x)= 6 x+ x 2 x+ x =1+ 4 x 2 x+ x \\ & f^ (x)= 4[(2 x+ x) 1-x(2+ x)] (2 x+ x)^2 = 4( x-x x) (2 x+ x)^2 \\ & = 4 x( x-x) (2 x+ x)^2 aligned Options (C) \& (D) are also correct.

Q: Let S be the set of all twice differentiable functions f from ℝ to ℝ such that d 2 f d x 2 x > 0 for all x ∈ - 1 , 1 . For f ∈ S , let X f be the number of points

Given, f " ( x ) > 0 & f ( x ) - x = 0 Now let, g x = f x − x Now differentiating above equation we get, ⇒ g ' x = f ' x − 1 Again differentiating we get, g " ( x ) = f " ( x ) > 0 ⇒ concave up So, possible graphs will be, Here in above graph X f = 0 as there is no point for which f x = x Now here X f ≤ 2 as there will be minimum two points when y = x &nb

Q: Let f_ 1 :(0, ) R and f_ 2 :(0, ) R be defined by f_ 1 (x)= _ 0 ^ x _ j=1 ^ 21 (t-j)^ j d t, x>0 and f_ 2 (x)=98(x-1)^ 50 -600(x-1)^ 49 +2450, x>0 where, for any positive integer n and real numbers a_

Given f 1 x = ∫ 0 x t − 1 t − 2 2 ...... t − 2 t 21 d t ⇒ f ' 1 x = x − 1 x − 2 2 ..... x − 21 21 Checking critical points At all odd integers from 1 to 21 f x will have an extrema with 1 , 5 , 9 , 13 , 17 , 21 being points of minima & 3 , 7 , 11 , 15 , 19 being points of maxima. So m 1 = 6 & n 1 = 5 Hence 2 m 1 + 3 n 1 + m 1 n 1 = 57

Q: Let f_ 1 :(0, ) R and f_ 2 :(0, ) R be defined by f_ 1 (x)= _ 0 ^ x _ j=1 ^ 21 (t-j)^ j d t, x > 0 and f_ 2 (x)=98(x-1)^ 50 -600(x-1)^ 49 +2450, x > 0 where, for any positive integer n and real number

f 2 x = 98 x - 1 50 - 600 x - 1 49 + 2450 , x > 0 f 2 ' x = 98 × 50 x − 1 49 − 600 × 49 x − 1 48 = 4900 x − 1 48 x − 1 − 6 = 4900 x − 1 48 x − 7 So extrema is at x = 7 only, which is minima m 2 = 1 , n 2 = 0 Hence 6 m 2 + 4 n 2 + 8 m 2 n 2 = 6

Q: Paragraph: Let _ 1 :[0, ) R , _ 2 :[0, ) R , f:[0, ) R and g:[0, ) R be functions such that f(0)=g(0)=0, _ 1 (x)=e^ -x +x, x 0, _ 2 (x)=x^ 2 -2 x-2 e^ -x +2, x 0, f(x)= _ -x ^ x (|t|-t^ 2 ) e^ -t^ 2 d

(A) Given f x = 2 ∫ 0 x t - t 2 e - t 2 d t ; x > 0 g x = ∫ 0 x 2 t e - 1 d t ; x > 0 put t = u 2 ∴ g x = 2 ∫ 0 x u 2 e - u 2 d u = 2 ∫ 0 x t 2 e - t 2 d t now, f x + g x = ∫ 0 x 2 t e - t 2 d t = 1 - e - x 2 ⇒ f ( ln 3 ) + g ( ln 3 ) = 1 - e - ln 3 = 2 3 (B) for α ∈ ( 1 , x ) , ψ 1 ( x ) = 1 + α x ⇒ e - x + x - 1 - α x = 0 ⇒ e - x - 1 = x ( α - 1 ) Which is not possible because LHS < 0

Question 1

Consider the function f: (0, ) (- , ) given by f(x) = x \, _e(x) - x + 1. Then which one of the following statements is TRUE?

Accepted Answer

Given f(x) = x _e(x) - x + 1 Differentiating with respect to x: f'(x) = 1 2 x _e(x) + x ( 1 x ) - 1 f'(x) = _e(x) 2 x + 1 x - 1 Differentiating again with respect to x: f''(x) = d dx ( 1 2 x^ -1/2 _e(x) + x^ -1/2 - 1 ) f''(x) = 1 2 ( - 1 2 x^ -3/2 _e(x) + x^ -1/2 1 x ) - 1 2 x^ -3/2 f''(x) = - _e(x) 4x^ 3/2 + 1 2x^ 3/2 - 1 2x^ 3/2 f''(x) = - _e(x) 4x^ 3/2 For x (0, 1), _e(x) 0. Thus, f'(x) is strictly increasing in (0, 1). For x (1, ), _e(x) > 0, which implies f''(x) Therefore, f'(x) attains its

Question 2

Let P be the point on the parabola y = x^2 such that the slope of the tangent to the parabola at the point P is 4. Let Q be the point in the first quadrant lying on the circle x^2 + y^2 = 2 such that

Accepted Answer

For the point P on the parabola y = x^2, the slope of the tangent is dy dx = 2x. Given 2x = 4 x = 2. Substituting x = 2 in y = x^2, we get y = 4. Thus, the coordinates of P are (2, 4). For the point Q on the circle x^2 + y^2 = 2, differentiating with respect to x gives 2x + 2y dy dx = 0 dy dx = - x y . Given - x y = -1 x = y. Since Q lies in the first quadrant, substituting x = y in x^2 + y^2 = 2 gives 2x^2 = 2 x = 1, y = 1. Thus, the coordinates of Q are (1, 1). For the point R on the ellipse x

Question 3

Let R denote the set of all real numbers. Define the function f: R R by f( x )= \ array cc 2-2 x^2-x^2 1 x & if x 0 \ 2 & if x=0 array . Then which one of the following statements is TRUE ?

Accepted Answer

(A) RHD at x=0: _ h 0 (2-2 h^2-h^2 1 h )-2 h =0 Similarly LHD at x =0 is also equal to 0 . Differentiable at x =0 (B) f^ (x)=-4 x-2 x 1 x -x^2 ( 1 x ) ( -1 x^2 ) aligned & f^ (x)=- (4 x+2 x 1 x )+ 1 x \ & f^ (x)=- (2 x (4- 1 x ) )+ 1 x aligned for x (0, ) We can't say f(x) is decreasing on (0, ) as 1 x oscillates. (C) for x (- , 0), for any >0 f(x) is not increasing on (- , 0) as 1 x oscillates from -1 to 1. (D) f(0)=2 aligned & f(0+h) <2 \ & f(0-h) <2 aligned x =0 is local maxima

Question 4

Let R denote the set of all real numbers. For a real number x, let [x] denote the greatest integer less than or equal to x. Let n denote a natural number. Match each entry in List-I to the correct ent

Accepted Answer

aligned & P(x)=10 x^3-45 x^2+60 x+35 \ & P^ (x)=30(x-1)(x-2) aligned P ( x ) decreases in [1,2] Range of P ( x )=[55,60] f( x )= [ P ( x ) n ] value of n =9 (Q) For g(x) to be increasing 2 n^2-13 n-15 0 (R) aligned & h(x)= (x^2-9 )^n (x^2+2 x+3 ) \ & h^ (x)= (x^2-9 )^n(2 x+2)+ (x^2+2 x+3 ) n (x^2-9 )^ n-1 2 x \ & = (x^2-9 )^ n-1 [2 (x^2-9 )(x+1)+2 n x (x^2+2 x+3 ) ] \ & =(x+3)^ n-1 (x-3)^ n-1 q(x) aligned Derivative must change sign at x=3 aligned & n -1= odd \ & n = even \ & n =6 aligned

Question 5

Let R denote the set of all real numbers. Let f: R R be defined by f(x)= cases 6 x+ x 2 x+ x & if x 0 \ 7 3 & if x=0 cases Then which of the following statements is (are) TRUE?

Accepted Answer

since 3> 7 3 3>f(0) x=0 is local minima ....option (B) is correct Now, aligned & f(x)= 6 x+ x 2 x+ x =1+ 4 x 2 x+ x \ & f^ (x)= 4[(2 x+ x) 1-x(2+ x)] (2 x+ x)^2 = 4( x-x x) (2 x+ x)^2 \ & = 4 x( x-x) (2 x+ x)^2 aligned Options (C) \& (D) are also correct.

Question 6

Let the function f: R R be defined by f(x)= x e^ x (x^ 2023 +2024 x+2025 ) (x^2-x+3 ) + 2 e^ x (x^ 2023 +2024 x+2025 ) (x^2-x+3 ) . Then the number of solutions of f(x)=0 in R is

Accepted Answer

f(x)= (x^ 2023 +2024 x+2025 ) e^ x (x^2-x+3 ) ( x+2) ( x+2) is never zero for x^ 2023 +2024 x+2025=0 aligned & let ( x )= x ^ 2023 +2024 x +2025 \ & ^ ( x )=2023 x ^ 2022 +2024>0 x R aligned ( x ) is an Strictly Increasing function ( x )=0 for exactly one value of x f ( x )=0 has one solution

Question 7

Let S be the set of all twice differentiable functions f from &#8477; to &#8477; such that d 2 f d x 2 x &#62; 0 for all x &#8712; - 1 , 1 . For f &#8712; S , let X f be the number of points

Accepted Answer

Given, f &#34; ( x ) &#62; 0 &#160; &#160; &#38; &#160; &#160; f ( x ) - x = 0 Now let, g x = f x &#8722; x Now differentiating above equation we get, &#8658; g ' x = f ' x &#8722; 1 Again differentiating we get, g &#34; ( x ) = f &#34; ( x ) &#62; 0 &#8658; &#160;concave&#160;up&#160; So, possible graphs will be, Here in above graph  X f = 0  as there is no point for which  f x = x Now here  X f &#8804; 2  as there will be minimum two points when  y = x &nb

Question 8

Let f_ 1 :(0, ) R and f_ 2 :(0, ) R be defined by f_ 1 (x)= _ 0 ^ x _ j=1 ^ 21 (t-j)^ j d t, x>0 and f_ 2 (x)=98(x-1)^ 50 -600(x-1)^ 49 +2450, x>0 where, for any positive integer n and real numbers a_

Accepted Answer

Given f 1 x = &#8747; 0 x t &#8722; 1 t &#8722; 2 2 ...... t &#8722; 2 t 21 d t &#8658; &#160; &#160; &#160; f ' 1 x = x &#8722; 1 x &#8722; 2 2 ..... x &#8722; 21 21 Checking critical points At all odd integers from  1  to  21   f x  will have an extrema with  1 , 5 , 9 , 13 , 17 , 21  being points of minima &  3 , 7 , 11 , 15 , 19  being points of maxima. So  m 1 = 6  &  n 1 = 5 Hence  2 m 1 + 3 n 1 + m 1 n 1 = 57

Question 9

Let f_ 1 :(0, ) R and f_ 2 :(0, ) R be defined by f_ 1 (x)= _ 0 ^ x _ j=1 ^ 21 (t-j)^ j d t, x > 0 and f_ 2 (x)=98(x-1)^ 50 -600(x-1)^ 49 +2450, x > 0 where, for any positive integer n and real number

Accepted Answer

f 2 x = 98 x - 1 50 - 600 x - 1 49 + 2450 , &#160;&#160; x &#62; 0 f 2 ' x = 98 &#215; 50 x &#8722; 1 49 &#8722; 600 &#215; 49 x &#8722; 1 48 = 4900 x &#8722; 1 48 x &#8722; 1 &#8722; 6 = 4900 x &#8722; 1 48 x &#8722; 7 So extrema is at  x = 7  only, which is minima m 2 = 1 , n 2 = 0 Hence  6 m 2 + 4 n 2 + 8 m 2 n 2 = 6

Question 10

Paragraph: Let _ 1 :[0, ) R , _ 2 :[0, ) R , f:[0, ) R and g:[0, ) R be functions such that f(0)=g(0)=0, _ 1 (x)=e^ -x +x, x 0, _ 2 (x)=x^ 2 -2 x-2 e^ -x +2, x 0, f(x)= _ -x ^ x (|t|-t^ 2 ) e^ -t^ 2 d

Accepted Answer

(A) Given   f x = 2 &#8747; 0 x t - t 2 e - t 2 d t ; x &#62; 0 g x = &#8747; 0 x 2 t e - 1 d t ; x &#62; 0 put t = u 2 &#8756; g x = 2 &#8747; 0 x u 2 e - u 2 d u = 2 &#8747; 0 x t 2 e - t 2 d t now, f x + g x = &#8747; 0 x 2 t e - t 2 d t = 1 - e - x 2 &#8658; f ( ln 3 ) + g ( ln 3 ) = 1 - e - ln 3 = 2 3 (B) for &#945; &#8712; ( 1 , x ) ,   &#968; 1 ( x ) = 1 + &#945; x &#8658; e - x + x - 1 - &#945; x = 0 &#8658; e - x - 1 = x ( &#945; - 1 ) Which is not possible because LHS &#60; 0

Question 11

Paragraph: Let _ 1 :[0, ) R , _ 2 :[0, ) R , f:[0, ) R and g:[0, ) R be functions such that f(0)=g(0)=0, _ 1 (x)=e^ -x +x, x 0, _ 2 (x)=x^ 2 -2 x-2 e^ -x +2, x 0, f(x)= _ -x ^ x (|t|-t^ 2 ) e^ -t^ 2 d

Accepted Answer

(A) Given &#968; 1 x = e - x + x , &#160;&#160; x &#8805; 0 &#968; ' 1 x = - e - x + 1 , &#160;&#160; x &#8805; 0 &#968; ' 1 x  is increasing function. &#968; 1 x &#62; &#968; 1 0 &#160; &#8704; &#160; x &#8805; 0 &#968; 1 x &#8805; 1 So, e - x + x &#60; 1 for x &#8712; ( 0 , &#8734; ) is incorrect. LHS is increasing and unbounded function. (B) Given &#968; 2 x = x 2 - 2 x - 2 e - x + 2 , &#160;&#160; x &#8805; 0 &#968; ' 2 x = 2 x - 2 + 2 e - x , &#160;&#160; x &#8805; 0 &#968;

Question 12

Let f : R &#8594; R be defined by  f x = x 2 - 3 x - 6 x 2 + 2 x + 4 Then which of the following statements is(are) TRUE?

Accepted Answer

f x = x 2 - 3 x - 6 x 2 + 2 x + 4 f ' x = x 2 + 2 x + 4 2 x - 3 - x 2 - 3 x - 6 2 x + 2 x 2 + 2 x + 4 2 = 5 x 2 + 20 x x 2 + 2 x + 4 2 = 5 x ( x + 4 ) x 2 + 2 x + 4 2 Hence, it is increasing in ( - &#8734; , - 4 ] &#8746; [ 0 , &#8734; )   Now as  x &#8594; - &#8734; then f x = 1 , &#160; also,  x &#8594; &#8734; then f x = 1 f - 4 = 16 + 12 - 6 16 - 8 + 4 = 22 12 = 11 6 f 0 = - 3 2 Hence, range is  - 3 2 , &#160; 11 6 , and the function is into.

Question 13

Let the function  f : 0 , &#960; &#8594; &#8477; be defined by  f &#952; = sin &#952; + cos &#952; 2 + sin &#952; - cos &#952; 4 . Suppose the function f has a local minimum at &#952; precis

Accepted Answer

f &#952; = 1 + sin 2 &#952; + 1 - sin 2 &#952; 2 = sin 2 2 &#952; - sin 2 &#952; + 2 = sin 2 &#952; - 1 2 2 + 7 4 It has minima when sin 2 &#952; = 1 2 = sin &#960; 6 &#952; &#8712; 0 , &#960; 2 &#952; = &#960; 6 , 5 &#960; 6 &#952; = &#960; 12 , 5 &#960; 12 &#955; 1 = 1 12 , &#160; &#955; 2 = 5 12 sum = 1 2 .

Question 14

Let a , &#160; b and &#955; be positive real numbers. Suppose P is an end point of the latus rectum of the parabola y 2 = 4 &#955; x , and suppose the ellipse x 2 a 2 + y 2 b 2 = 1 passes through the

Accepted Answer

y 2 = 4 λ x ⇒ d y d x P = 1 = m 1 Also, x 2 a 2 + y 2 b 2 = 1 ⇒ d y d x P = − b 2 2 a 2 = m 2 Given both tangents are perpendicular. ⇒ m 1 ⋅ m 2 = − 1 ⇒ b 2 = 2 a 2 And a 2 = b 2 1 − e 2 ⇒ 1 = 2 1 − e 2 ⇒ e = 1 2

Question 15

Consider all rectangles lying in the region x , &#160; y &#8712; R &#215; R : 0 &#8804; x &#8804; &#960; 2 &#8201; a n d &#8201; 0 &#8804; y &#8804; 2 sin 2 x and having one side on the x -axis. The a

Accepted Answer

2 sin 2 θ 1 = 2 sin 2 θ 2 ⇒ 2 θ 1 = π − 2 θ 2 ⇒ θ 2 = π 2 − θ 1 … . ( 1 ) Now perimeter p θ 1 , θ 2 = 2 θ 2 − θ 1 + 2 sin 2 θ 1 ⇒ p θ 1 = 2 π 2 − 2 θ 1 + 2 sin 2 θ 1 ⇒ p ' θ 1 = 2 − 2 + 4 cos 2 θ 1 Also, p " θ 1 = 2 − 8 sin 2 θ 1 0 For maximum perimeter, p ' θ 1 = 0 & p " θ 1 0 ⇒ cos 2 θ 1 = 1 2 ⇒ 2 θ 1 = π 3 ⇒ θ 1 = π 6 Now area at θ 1 = π 6 will be θ 2 − θ 1 × 2 sin 2 θ 1 = π 2 − 2 θ 1 ⋅ 2 sin 2 θ 1 = π 2 − π 3 × 2 sin π 3 = π 6 ⋅ 3 = π 2 3

Question 16

For a polynomial g x with real coefficients, let m g denote the number of distinct real roots of g x . Suppose S is the set of polynomials with real coefficients defined by  S = x 2 &#8722; 1 2 a

Accepted Answer

f x is a polynomial in x such that f x = x 2 &#8722; 1 2 &#160; a 0 + a 1 x + a 2 x 2 + a 3 x 3 &#160; : &#160; a 0 , &#160; a 1 , &#160; a 2 , &#160; a 3 &#8712; R So,  f ' x = x 2 &#8722; 1 &#160; q x where q x is polynomial roots of f &#8242; x = &#8722; 1 , &#160; 1 So, by Rolle&rsquo;s theorem,  f 1 = 0 = f &#8722; 1 &#8658; f ' &#945; = 0 , where  &#945; &#8712; &#8722; 1 , &#160; 1 &#8658; minimum m f &#8242; = 3 Again by Rolle&rsquo;s theorem, at least one real roo

Question 17

Let f x = sin &#8289; &#960; x x 2 , &#160; x &#62; 0 . Let x 1 &#60; x 2 &#60; x 3 &#60; &#8230; &#60; x n &#60; &#8230; be all the points of local maximum of f and y 1 &#60; y 2 &#60; y 3 &#60; &#82

Accepted Answer

f ' x = 2 x cos ⁡ π x π x 2 - tan ⁡ π x x 4 …(i) ⇒ for maxima/minima f ' x = 0 ⇒ cos ⁡ π x = 0 o r π x 2 = tan ⁡ π x ∵ cos ⁡ π x ≠ 0 ∵ tan ⁡ π x will not be defined ∴ maxima/minima will occur Where tan ⁡ π x = π x 2 Case I : i For example In x ∈ 2 n , 2 n + 1 2 at point P 2 , x ∈ 2 , 5 2 cos ⁡ π x > 0 and at P 2 + , tan ⁡ π x > π x 2 at P 2 - , tan ⁡ π x π x 2 hence from equation (i) in x ∈ 2 n , 2 n + 1 2 , f ' x goes from positive to negative hence P 2 is maxima Similarly P 2 , P 4 , P 6 … are

Question 18

Let f : R &#8594; R be given by f x = x - 1 x - 2 x - 5 . Define F x = &#8747; 0 x f t d t , &#160; x &#62; 0 . Then which of the following options is/are correct?

Accepted Answer

F x = ∫ 0 x f t . d t F ' x = f x = x - 1 x - 2 x - 5 ⇒ F x has maxima at x = 2 and minima at x = 1 and x = 5 now F 2 = ∫ 0 2 x 3 - 8 x 2 + 17 x - 10 . d x = x 4 4 - 8 x 3 3 + 17 x 2 2 - 10 x 0 2 F 2 = - 10 3 which is a maxima in x ∈ 0 , 5 F 0 = 0 Hence F x 0 ∀ x ∈ 0 , 5

Question 19

Let f : R &#8594; R be given by f x = x 5 + 5 x 4 + 10 x 3 + 10 x 2 + 3 x + 1 , x &#60; 0 ; x 2 - x + 1 , 0 &#8804; x &#60; 1 ; 2 3 x 3 - 4 x 2 + 7 x - 8 3 , 1 &#8804; x &#60; 3 ; x - 2 l o g e x - 2

Accepted Answer

As given f : R → R f x = x 5 + 5 x 4 + 10 x 3 + 10 x 2 + 3 x + 1 x 0 x 2 - x + 1 0 ≤ x 1 2 3 x 3 - 4 x 2 + 7 x - 8 3 1 ≤ x 3 ( x - 2 ) l n ( x - 2 ) - x + 10 3 x ≥ 3 f ( 0 ) = f ( 0 + ) = f ( 0 ) = 1 f 1 = f 1 + = f 1 = 1 f 3 = f 3 + = f 3 = 1 3 Hence f x is continuous everywhere. Now, l i m x → ∞ f x = - ∞ and l i m x → ∞ f x = + ∞ Hence, range of f ( x ) is - ∞ , ∞ Consider f ' x = 5 x 4 + 20 x 3 + 30 x 2 + 20 x + 3 = 5 ( x + 1 ) 4 - 2 x 0 2 x - 1 0 x 1 2 x 2 - 8 x + 7 1 x 3 l n ( x - 2 ) x >

Application of Derivatives — Previous Year Questions & Solutions