Q: Let f : 0 , 1 → ℝ be the function defined as f x = 4 x x - 1 4 2 x - 1 2 , where [ x ] denotes the greatest integer less than or equal to x . Then which of the following is(are) true

Given, f : ( 0 , 1 ) → ℝ f x = 4 x x - 1 4 2 x - 1 2 We know that, 4 x will be discontinuous function at integer points, So, critical point = 1 4 , 1 2 , 3 4 as x ∈ 0 , 1 And function will be discontinuous at x = 3 4 because for 1 2 & 1 4 , x - 1 2 & x - 1 4 2 will make the function continuos Hence, the function is discontinuous exactly at one point in 0 , 1 , so option&nb

Question 1

Let R denote the set of all real numbers. Let f : R R be an arbitrary function and let g : R R be the function defined by g(x) = x f(x), for all x R . Then which of the following statements is (are) T

Accepted Answer

Given g(x) = x f(x) for all x R . For option A: Let f(x) = 1 x for x 0 and f(0) = 0. Then g(x) = 1 for x 0 and g(0) = 0. _ x 0 g(x) = 1 g(0). Thus, g is not necessarily continuous at x = 0. Option A is false. For option B: The derivative of g at x = 0 is given by: g'(0) = _ x 0 g(x) - g(0) x - 0 = _ x 0 x f(x) - 0 x = _ x 0 f(x). If f is continuous at x = 0, then _ x 0 f(x) = f(0). Since f(0) is a finite real number, g'(0) exists and g is differentiable at x = 0. Option B is true. For option C:

Question 2

Consider the function f : (- 2 , 2 ) (- , ) defined by f(x) = (|x| + |x - 1|) x + [x x], where [x x] is the greatest integer less than or equal to x x. Let be the total number of points in the interva

Accepted Answer

Let f(x) = g(x) + h(x), where g(x) = (|x| + |x - 1|) x and h(x) = [x x]. First, we analyze h(x) = [x x] on the interval (- 2 , 2 ). The function y = x x is an even function. For x [0, 2 ), x x is strictly increasing. At x = 0, x x = 0. As x 2 , x x 2 1.57. Thus, x x [0, 1.57) for x (- 2 , 2 ). The value of [x x] will be 0 when 0 x x Let x_1 (0, 2 ) be the unique point where x_1 x_1 = 1. By symmetry, -x_1 is the unique point in (- 2 , 0 ) where (-x_1) (-x_1) = 1. Therefore, h(x) has jump disconti

Question 3

For a real number , let [ ] denote the greatest integer less than or equal to . For a finite set S, let |S| denote the number of elements in the set S. Consider the functions f : (-3, 3) (- , ) and g

Accepted Answer

For the function f(x) = [x^3] _e(1 + ^2( (x - [x]))), we can simplify the argument of the sine function. Since [x] is an integer, ( (x - [x])) = ( x - [x]) = ( x). Thus, ^2( (x - [x])) = ^2( x). The function f(x) can be rewritten as f(x) = [x^3] _e(1 + ^2( x)). The term _e(1 + ^2( x)) is continuous everywhere. The term [x^3] has jump discontinuities where x^3 is an integer. For x (-3, 3), x^3 (-27, 27). The integer values of x^3 in this interval are k \ -26, -25, , 26\ , which gives 53 points of

Question 4

Let f : 0 , 1 &#8594; &#8477; be the function defined as f x = 4 x x - 1 4 2 x - 1 2 , where [ x ]  denotes the greatest integer less than or equal to x . Then which of the following is(are) true

Accepted Answer

Given, f : ( 0 , 1 ) &#8594; &#8477; f x = 4 x x - 1 4 2 x - 1 2 We know that,  4 x  will be discontinuous function at integer points, So,  critical point&#160; = 1 4 , 1 2 , 3 4  as  x &#8712; 0 , 1 And function will be discontinuous at x = 3 4  because for  1 2 &#160; &#38; &#160; 1 4 ,  x - 1 2 &#160; &#38; &#160; x - 1 4 2  will make the function continuos Hence, the function is discontinuous exactly at one point in  0 , 1 , so option&nb

Question 5

Let the functions f : - 1 , 1 &#8594; &#8477;  and  g : - 1 , 1 &#8594; - 1 , 1 be defined by  f x = &#160; 2 x - 1 + 2 x + 1 &#160; &#38; &#160; g x = x - x , where x denotes the great

Accepted Answer

f x = 2 x - 1 + 2 x + 1 = - 4 x , x - 1 2 2 , - 1 2 ≤ x ≤ 1 2 4 x , x > 1 2 g x = x f g x = - 4 g x , g x - 1 2 2 , - 1 2 ≤ g x ≤ 1 2 4 g x g x > 1 2 Graph of g x is f g x = 2 , x ∈ - 1 , - 1 2 ∪ 0 , 1 2 4 x + 1 x ∈ - 1 2 , 0 4 x x ∈ 1 2 , 1 c = 1 at x = 0 d = 3 at x = - 1 2 , 0 , 1 2 , ∴ c + d = 4 .

Question 6

Let f : &#8477; &#8594; &#8477;  and  g : &#8477; &#8594; &#8477; be functions satisfying  f x + y = f x + f y + f x f y &#160;and&#160; f x = x g x  for all x , y &#8712; &#8477;

Accepted Answer

Given  f x + y = f x + f y + f x f y Put  x = y = 0 in given relation. &#8658; &#160;&#160; f 0 = f 0 + f 0 + f 2 0 &#8658; &#160;&#160; f 0 = 0 or - 1 &#8757; &#160;&#160; f x + y = f x + f y + f x &#183; f y &#8658; &#160;&#160; f x + y - f x y = f y 1 + f x y &#8658; &#160;&#160; lim y &#8594; 0 f ( x + y ) - f ( x ) y = lim y &#8594; 0 1 + f x &#183; f y y &#8757; &#160;&#160; lim x &#8594; 0 &#160; g x = lim x &#8594; 0 f x x = 1 &#8658; &#160;&#160; f ' x = 1 + f x &#8658; &#

Question 7

Let the function f : R &#8594; R be defined by f x = x 3 &#8722; x 2 + x &#8722; 1 sin x and let g : R &#8594; R be an arbitrary function. Let f g : R &#8594; R be the product function defined by f g

Accepted Answer

Differentiability of f g at x = 1 Left-hand derivative : f g ' 1 - = lim h &#8594; 0 f g 1 - h &#8722; f g 1 - h = lim h &#8594; 0 1 - h 3 - 1 - h 2 - h sin 1 - h g 1 - h &#8722; 0 - h = lim h &#8594; 0 1 - h 2 + sin 1 - h g 1 - h &#160; . . . . . . . i Right-hand derivative : f g ' 1 + = lim h &#8594; 0 f g 1 + h &#8722; f g 1 h = lim h &#8594; 0 1 + h 3 - 1 + h 2 + h sin 1 + h g 1 + h &#8722; 0 h = lim h &#8594; 0 1 + h 2 + sin 1 + h g 1 + h &#160; . . . . . . . i i If g is continuous

Continuity and Differentiability — Previous Year Questions & Solutions