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Let a ∈ R and let f : R → R be given by f (x) = x5 – 5x + a. Then- a)f(x) has three real roots if a > 4
- b)f(x) has only real root if a > 4
- c)f x) has three real roots if a < – 4
- d)f(x) has three real roots if – 4 < a < 4
Correct answer is option 'B,D'. Can you explain this answer?
Let a ∈ R and let f : R → R be given by f (x) = x5 – 5x + a. Then
a)
f(x) has three real roots if a > 4
b)
f(x) has only real root if a > 4
c)
f x) has three real roots if a < – 4
d)
f(x) has three real roots if – 4 < a < 4
|
Anushka Roy answered |
f (x) = x5 - 5 x+a
f (x) = 0 ⇒ x5 - 5 x +a = 0
⇒ a = 5x – x5 = g(x)
⇒ g(x) = 0 when x = 0, 51/4 - 51/4 and g' (x ) = 0 ⇒ x = 1, – 1
Also g (– 1) = – 4 and g(1) = 4
∴ graph of g(x) will be as shown below.
From graph
f (x) = 0 ⇒ x5 - 5 x +a = 0
⇒ a = 5x – x5 = g(x)
⇒ g(x) = 0 when x = 0, 51/4 - 51/4 and g' (x ) = 0 ⇒ x = 1, – 1
Also g (– 1) = – 4 and g(1) = 4
∴ graph of g(x) will be as shown below.
From graph
if a ∈ ( -4,-4)
then g(x) = a or f (x) = 0 has 3 real roots If a > 4 or a < – 4
then f(x) = 0 has only one real root.
∴ (b) and (d) are the correct options.
then g(x) = a or f (x) = 0 has 3 real roots If a > 4 or a < – 4
then f(x) = 0 has only one real root.
∴ (b) and (d) are the correct options.
The entire graphs of the equation y = x2 + kx – x + 9 is strictly above the x-axis if and only if- a)k < 7
- b)– 5 < k < 7
- c)k > – 5
- d)None of these.
Correct answer is option 'B'. Can you explain this answer?
The entire graphs of the equation y = x2 + kx – x + 9 is strictly above the x-axis if and only if
a)
k < 7
b)
– 5 < k < 7
c)
k > – 5
d)
None of these.
|
Pallavi Das answered |
For entire graph to be above x-axis, we should have
The domain of sin-1 [log3 (x/3)] is- a)[1, 9]
- b)[–1, 9]
- c)[–9, 1]
- d)[–9, –1]
Correct answer is option 'A'. Can you explain this answer?
The domain of sin-1 [log3 (x/3)] is
a)
[1, 9]
b)
[–1, 9]
c)
[–9, 1]
d)
[–9, –1]
|
Asha Nair answered |
Domain of sin-1x is [-1,1]
Therefore, -1 ≤ log3(x/3) ≤ 1
3-1 ≤ x/3 ≤ 3
1 ≤ x ≤ 9
Therefore, the domain of sin-1[log3(x/3)] is [1,9]
Let f(x) = ( x + 1)2 – 1,x > –1Statement -1 : The set {x : f(x) = f –1(x) = {0, –1}Statement-2 : f is a bijection.- a)Statement-1 is true, Statement-2 is true.Statement-2 is not a correct explanation for Statement-1.
- b)Statement-1 is true, Statement-2 is false.
- c)Statement-1 is false, Statement-2 is False.
- d)Statement-1 is False, Statement-2 is true.Statement-2 is not a correct explanation for Statement-1.
Correct answer is option 'B'. Can you explain this answer?
Let f(x) = ( x + 1)2 – 1,x > –1
Statement -1 : The set {x : f(x) = f –1(x) = {0, –1}
Statement-2 : f is a bijection.
a)
Statement-1 is true, Statement-2 is true.Statement-2 is not a correct explanation for Statement-1.
b)
Statement-1 is true, Statement-2 is false.
c)
Statement-1 is false, Statement-2 is False.
d)
Statement-1 is False, Statement-2 is true.Statement-2 is not a correct explanation for Statement-1.
|
Sai Chakraborty answered |
Given that f (x) = (x + 1)2 –1, x > –1 Clearly Df = [–, ∞) but co-demain is not given.
Therefore f (x) need not be necessarily onto.
But if f (x) is onto then as f (x) is one one also, (x + 1) being something +ve, f–1(x) will exist where (x + 1)2 –1 = y
Therefore f (x) need not be necessarily onto.
But if f (x) is onto then as f (x) is one one also, (x + 1) being something +ve, f–1(x) will exist where (x + 1)2 –1 = y
∴ The statement-1 is correct but statement-2 is false.
Let f : R → R be any function. Define g : R → R by g(x) = |f(x)| for all x. Then g is- a)onto if f is onto
- b)one-one if f is one-one
- c)continuous if f is continuous
- d)differentiable if f is differentiable.
Correct answer is option 'C'. Can you explain this answer?
Let f : R → R be any function. Define g : R → R by g(x) = |f(x)| for all x. Then g is
a)
onto if f is onto
b)
one-one if f is one-one
c)
continuous if f is continuous
d)
differentiable if f is differentiable.
|
|
Ameya Menon answered |
Let h (x) = |x| then g (x) = |f (x)| = h (f (x))
Since composition of two continuous functions is continuous, therefore g is continuous if f is continuous.
Since composition of two continuous functions is continuous, therefore g is continuous if f is continuous.
The function f(x) = 2|x| + |x + 2| – | |x + 2| – 2 |x| has a local minimum or a local maximum at x = - a)-2
- b)-2/3
- c)2
- d)2/3
Correct answer is option 'A,B'. Can you explain this answer?
The function f(x) = 2|x| + |x + 2| – | |x + 2| – 2 |x| has a local minimum or a local maximum at x =
a)
-2
b)
-2/3
c)
2
d)
2/3
|
Anirudh Yadav answered |
the critical points can be obtained by solving |x| = 0
The graph of y = f(x) is as follows
From graph f(x) has local minimum at –2 and 0 and
Th e fun ction f(x) = |px – q| + r | x |, x ∈ (-∞,∞) where p > 0, q > 0, r > 0 assumes its minimum value only on one point if- a)p ≠ q
- b)r ≠ q
- c)r ≠ p
- d)p = q = r
Correct answer is option 'C'. Can you explain this answer?
Th e fun ction f(x) = |px – q| + r | x |, x ∈ (-∞,∞) where p > 0, q > 0, r > 0 assumes its minimum value only on one point if
a)
p ≠ q
b)
r ≠ q
c)
r ≠ p
d)
p = q = r
|
|
Akanksha Sarkar answered |
From graph (i) infinite many points for min value of f (x)
From graph (ii) only pt. of min of f (x) at x = q/p
From graph (iii) only one pt. of min of f (x) at x = 0
If x satisfies |x - 1| + |x - 2| + |x - 3| > 6 , then- a)0 < x < 4
- b)x < -2 or x > 4
- c)x < 0 or x > 4
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
If x satisfies |x - 1| + |x - 2| + |x - 3| > 6 , then
a)
0 < x < 4
b)
x < -2 or x > 4
c)
x < 0 or x > 4
d)
None of these
|
Mehul Chavan answered |
Graph of f (x) shows f (x) > 6 for x < 0 or x > 4
Let R be the set of real numbers. If f : R → R is a function defined by f (x) = x2, then f is :- a)Injective but not surjective
- b)Surjective but not injective
- c)Bijective
- d)None of these.
Correct answer is option 'D'. Can you explain this answer?
Let R be the set of real numbers. If f : R → R is a function defined by f (x) = x2, then f is :
a)
Injective but not surjective
b)
Surjective but not injective
c)
Bijective
d)
None of these.
|
Rhea Joshi answered |
f (x) = x2 is many one as f (1) = f (–1) = 1 Also f is into as – ve real number have no pre-image.
∴ F is neither injective nor surjective.
∴ F is neither injective nor surjective.
X an d Y are two sets and f : X → Y. If {f(c) = y; c ⊂ X, y ⊂ Y} and {f–1(d) = x; d ⊂ Y, x ⊂ X}, then the true statement is- a)f(f–1(b)) = b
- b)f–1(f(a)) = a
- c)f(f–1(b)) = b, b ⊂ y
- d)f–1(f(a)) = a, a ⊂ x
Correct answer is option 'D'. Can you explain this answer?
X an d Y are two sets and f : X → Y. If {f(c) = y; c ⊂ X, y ⊂ Y} and {f–1(d) = x; d ⊂ Y, x ⊂ X}, then the true statement is
a)
f(f–1(b)) = b
b)
f–1(f(a)) = a
c)
f(f–1(b)) = b, b ⊂ y
d)
f–1(f(a)) = a, a ⊂ x
|
Aaditya Datta answered |
Given that X and Y are two sets and f : X → Y.
{f (c) = y; c ⊂ X, y ⊂ Y} and
{f –1(d) = x : d ⊂ Y, x ⊂ X }
The pictorial representation of given information is as shown:
{f (c) = y; c ⊂ X, y ⊂ Y} and
{f –1(d) = x : d ⊂ Y, x ⊂ X }
The pictorial representation of given information is as shown:
Since f –1 (d) = x ⇒ f (x) = d Now if a ⊂ x
⇒ f (a) ⊂ f (x) = d ⇒ f –1 [f (a)] = a
∴ f –1 (f (a)) = a, a ⊂ x is the correct option.
⇒ f (a) ⊂ f (x) = d ⇒ f –1 [f (a)] = a
∴ f –1 (f (a)) = a, a ⊂ x is the correct option.
L et f : (– 1, 1) → B , be a function defined by 
then f is both one - one and onto when B is the interval- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
L et f : (– 1, 1) → B , be a function defined by then f is both one - one and onto when B is the interval
then f is both one - one and onto when B is the intervala)
b)
c)
d)
|
|
Asha Nair answered |
Clearly, range of f (x)
For f to be onto, codomain = range
∴ Co-domain of function
∴ Co-domain of function
The graph of the function y = f(x) is symmetrical about the line x = 2, then- a)f (x) = - f (-x)
- b)f (2 +x) = f (2-x)
- c)f (x) = f (-x)
- d)f (x +2) = f (x-2)
Correct answer is option 'B'. Can you explain this answer?
The graph of the function y = f(x) is symmetrical about the line x = 2, then
a)
f (x) = - f (-x)
b)
f (2 +x) = f (2-x)
c)
f (x) = f (-x)
d)
f (x +2) = f (x-2)
|
Ashwini Chakraborty answered |
Let us consider a graph symm. with respect to line x = 2 as shown in the figure.
From the figure
f (x1) = f (x2), where x1 = 2-x and x2 = 2+x
∴ f (2 - x) = f (2+x)
∴ f (2 - x) = f (2+x)
A function f from the set of natural numbers to integers defined by
- a) neither one -one nor onto
- b) one-one but not onto
- c) onto but not one-one
- d) one-one and onto both.
Correct answer is option 'D'. Can you explain this answer?
A function f from the set of natural numbers to integers defined by
a)
neither one -one nor onto
b)
one-one but not onto
c)
onto but not one-one
d)
one-one and onto both.
|
Ishita Deshpande answered |
We have f: N →I
If x and y are two even natural numbers,
If x and y are two even natural numbers,
Again if x and y are two odd natural numbers then
Also each negative integer is an image of even natural number and each positive integer is an image of odd natural number.
∴ f is onto.
Hence f is one one and onto both.
Hence f is one one and onto both.
If the functions f(x) and g(x) are defined on R → R such that
- a)one-one & on to
- b)neither one-one nor onto
- c)one-one but not on to
- d)on to but not one-one
Correct answer is option 'A'. Can you explain this answer?
If the functions f(x) and g(x) are defined on R → R such that
a)
one-one & on to
b)
neither one-one nor onto
c)
one-one but not on to
d)
on to but not one-one
|
Parth Yadav answered |
Since f – g : R → R for any x there is only one value of (f (x) – g(x)) whether x is rational or irrational.
Moreover as x∈R, f (x) – g (x) also belongs to R.
Therefore, (f – g) is one-one onto.
Moreover as x∈R, f (x) – g (x) also belongs to R.
Therefore, (f – g) is one-one onto.
R be given by f (x) = (log(sec x + tan x))3.Then- a)f (x) is an odd function
- b)f (x) is one-one function
- c)f (x) is an onto function
- d)f (x) is an even function
Correct answer is option 'A,B,C'. Can you explain this answer?
R be given by f (x) = (log(sec x + tan x))3.Then
a)
f (x) is an odd function
b)
f (x) is one-one function
c)
f (x) is an onto function
d)
f (x) is an even function
|
|
Abhishek Majumdar answered |
∴ f is an odd function.
(a) is correct and (d) is not correct.
Also
We know that strictly increasing function is one one.
∴ f is one one
∴ (b) is correct
∴ Range of f = (–∞, ∞) = R
∴ f is an onto function.
∴ (c) is correct.
∴ f is an onto function.
∴ (c) is correct.
For real x, let f (x) = x3 + 5x + 1, then- a)f is onto R but not one-one
- b)f is one-one and onto R
- c)f is neither one-one nor onto R
- d)f is one-one but not onto R
Correct answer is option 'B'. Can you explain this answer?
For real x, let f (x) = x3 + 5x + 1, then
a)
f is onto R but not one-one
b)
f is one-one and onto R
c)
f is neither one-one nor onto R
d)
f is one-one but not onto R
|
Ameya Tiwari answered |
Given that f (x) = x3 + 5x + 1
⇒ f (x) is strictly increasing on R
⇒ f (x) is one one
∴ Being a polynomial f (x) is cont. and inc.
Hence f is onto also. So, f is one one and onto R.
A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?
- a)a
- b)b
- c)c
- d)d
Correct answer is option 'C'. Can you explain this answer?
A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?
a)
a
b)
b
c)
c
d)
d
|
Baishali Chakraborty answered |
Clearly function f (x) = 3x2 - 2x+1 is increasing when
f ' (x) = 6x – 2 > 0 ⇒ x ∈[1 / 3,∞)
f ' (x) = 6x – 2 > 0 ⇒ x ∈[1 / 3,∞)
∴ f (x) is incorrectly matched with 
Which of the following functions is periodic?- a)f(x) = x – [x] where [x] denotes the largest integer less than or equal to the real number x
- b)
- c)f(x) = x cosx
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
Which of the following functions is periodic?
a)
f(x) = x – [x] where [x] denotes the largest integer less than or equal to the real number x
b)
c)
f(x) = x cosx
d)
none of these
|
|
Siddharth Ghosh answered |
Clearly it is a periodic function with period 1.
∴ (a) is the correct alternative.
∴ (a) is the correct alternative.
Then, for what value of a is f (f(x)) = x ?
the value
for which the function, 
, is defined, is
Let g (x) be a function defined on [– 1, 1]. If the area of the equilateral triangle with two of its vertices at (0,0) and 
then the function g(x) is- a)
- b)
- c)
- d)
Correct answer is option 'B,C'. Can you explain this answer?
Let g (x) be a function defined on [– 1, 1]. If the area of the equilateral triangle with two of its vertices at (0,0) and then the function g(x) is
then the function g(x) isa)
b)
c)
d)
|
Palak Joshi answered |
As (0, 0) and (x, g (x)) are two vertices of an equilateral triangle; therefore, length of the side of D is
∴ (b), (c) are the correct answers as (a) is not a function
(∴ image of x is not unique)
Let f(x) = sin x and g(x) = ln | x |. If th e ran ges of th e composition functions fog and gof are R1 and R2 respectively, th en- a)R1 = {u : -1 < u < 1}, R2 ={v : - ∞ < v < 0}
- b)R1 = {u : - ∞ < u < 0}, R2 = {v : -1 < v < 0}
- c)R1 = {u : -1 < u < 1}, R2 ={v : - ∞ < v < 0}
- d)R1 = {u : -1 < u < 1}, R2 ={v : - ∞ < v < 0}
Correct answer is option 'D'. Can you explain this answer?
Let f(x) = sin x and g(x) = ln | x |. If th e ran ges of th e composition functions fog and gof are R1 and R2 respectively, th en
a)
R1 = {u : -1 < u < 1}, R2 ={v : - ∞ < v < 0}
b)
R1 = {u : - ∞ < u < 0}, R2 = {v : -1 < v < 0}
c)
R1 = {u : -1 < u < 1}, R2 ={v : - ∞ < v < 0}
d)
R1 = {u : -1 < u < 1}, R2 ={v : - ∞ < v < 0}
|
Om Verma answered |
Let f, g and h be real-valued functions defined on the interval 
and 
. If a, b and c denote, respectively, the absolute maximum of f, g and h on [0, 1], then- a)a = b and c ≠ b
- b)a = c and a ≠ b
- c)a ≠ b and c ≠ b
- d)a = b = c
Correct answer is option 'D'. Can you explain this answer?
Let f, g and h be real-valued functions defined on the interval and . If a, b and c denote, respectively, the absolute maximum of f, g and h on [0, 1], then
and . If a, b and c denote, respectively, the absolute maximum of f, g and h on [0, 1], thena)
a = b and c ≠ b
b)
a = c and a ≠ b
c)
a ≠ b and c ≠ b
d)
a = b = c
|
Mira Yadav answered |
∴ f (x) is an increasing function on [ 0,1]
∴ g (x) is an increasing function on [ 0,1]
∴ h (x) is an increasing function on [ 0,1]
Hence a = b = c.
Let E = {1, 2, 3, 4} and F = {1, 2}. Then the number of onto functions from E to F is- a)14
- b)16
- c)12
- d)8
Correct answer is option 'A'. Can you explain this answer?
Let E = {1, 2, 3, 4} and F = {1, 2}. Then the number of onto functions from E to F is
a)
14
b)
16
c)
12
d)
8
|
|
Nikhil Sengupta answered |
From E to F we can define, in all, 2 × 2 × 2 × 2 = 16 functions (2 options for each element of E) out of which 2 are into, when all the elements of E either map to 1 or to 2.
∴ No. of onto functions = 16 – 2 = 14
∴ No. of onto functions = 16 – 2 = 14
Let f(x) = | x – 1 |. Then- a)f(x2) = (f(x))2
- b)f(x + y) = f(x) + f(y)
- c)f(| x |) = | f(x) |
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
Let f(x) = | x – 1 |. Then
a)
f(x2) = (f(x))2
b)
f(x + y) = f(x) + f(y)
c)
f(| x |) = | f(x) |
d)
None of these
|
Swara Ahuja answered |
∴ Put x = 2
LHS = f (22) = |4 – 1| = 3 and RHS = ( f (2))2 = 1
∴ (a) is not correct
Consider f (x + y) = f (x) + f (y)
Put x = 2, y = 5 we get
f (7) = 6; f (2) + f (5) = 1 + 4 = 5
∴ (b) is not correct
Consider f (| x |) = | f (x) |
Put x = – 5 then f (| –5 |) = f (5) = 4
| f (– 5) | = | – 5 – 1| = 6
∴ (c) is not correct.
Hence (d) is the correct alternative.
Let f : (0, 1) → R be defined by 
where b is a constant such that 0 < b < 1. Then- a)f is not invertible on (0, 1)
- b)
- c)
- d)f –1 is differentiable (0, 1)
Correct answer is option 'A,B'. Can you explain this answer?
Let f : (0, 1) → R be defined by where b is a constant such that 0 < b < 1. Then
where b is a constant such that 0 < b < 1. Thena)
f is not invertible on (0, 1)
b)
c)
d)
f –1 is differentiable (0, 1)
|
|
Keerthana Bose answered |
∴ f is neither onto nor invertible
Hence a and b are the correct options.
g(x) = f '(x) and given that F(5) = 5, then F(10) is equal to
Let f(x) be defined for all x > 0 and be continuous. Let f(x) 
for all x, y and f(e) = 1. Then- a)f(x) is bounded
- b)
- c)x f(x) → 1 as x → 0
- d)f(x) = ln x
Correct answer is option 'D'. Can you explain this answer?
Let f(x) be defined for all x > 0 and be continuous. Let f(x) for all x, y and f(e) = 1. Then
for all x, y and f(e) = 1. Thena)
f(x) is bounded
b)
c)
x f(x) → 1 as x → 0
d)
f(x) = ln x
|
Siddharth Basak answered |
f (x) is continuous and defined for all x > 0 and
⇒ Clearly f (x) = ln x which satisfies all these properties
∴ f (x) = ℓnx
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