Free Objective Test 01 Practice Test - 11th and 12th
Question 1
Let P = {(x,y) x2+y2=1,x,y∈R}. Then P is.
Reflexive
Symmetric
Transitive
Anti-symmetric
SOLUTION
Solution : B
Here we can see that the relation is neither reflexive nor transitive but it is symmetric,
because x2+y2=1⇒y2+x2=1
Question 2
R is relation over the set of integers and it is given by (x, y) ϵ R ⇔ R |x - y| ≤ 1. Then, R is
SOLUTION
Solution : B
As (x,x) ϵ R ⇒ |x−x|≤ 1
⇒ 0 ≤ 1 (True),
Thus, reflexive.
As (x,y) ϵ R ⇒ |x−y|≤ 1
⇒ |y−x||≤1 ⇒ (y,x) ϵ R,
Thus, symmetric.
Again, (x, y) ϵ R and (y, z) ϵ R
⇒|x−y|≤ 1 and |y−z|1/⇒|x−z|≤ 1
∴ Not transitive
Question 3
Let R be a relation over the set N×n and it is defined by (a, b) R (c, d) ⇒ a+ d = b + c. Then, R is
SOLUTION
Solution : D
(a, b) R (a, b) because a + b = b + a. So, r is reflexive.
(a, b)R (c, d) ⇒ a+d = b+c ⇒ c+b = d+a
⇒ (c,d) R (a,b)
So, R is symmetric.
(a, b) R (c, d) and (c, d) R (e, f)
⇒ a + d = b + c, c + f = d + e
Adding, a + d + c + f = b + c + d +e
⇒ a + f = b + e
⇒ (a, b) R (e, f).
∴ R is transitive.
Question 4
The period of the function |sinx|+|cosx| is
SOLUTION
Solution : A
The smallest of π2,2π,4π,π is π2
Let f(x)=|sin x|+|cos x|.
∴ f(x+π2)=∣∣sin(x+π2)∣∣+∣∣cos(x+π2)∣∣
=|cos x|+|−sin x|
=|cos x|+|sin x|
=f (x)
∴ The period of given function is π2
Question 5
Which of the following functions are periodic?
SOLUTION
Solution : C
f(x) = log x, is not periodic.
f(x) = ex, is not periodic.
f(x) = x - [x] = {x}, has period 1
f(x) = x + [x], is not periodic
Question 6
Let f:R→R,g:R→R, be two functions, such that f(x) =2x – 3, g (x) = x3 + 5.
The function (fog)−1 (x) is equal to
SOLUTION
Solution : D
We have, f : R → R, g: R → R defined by f(x) = 2x - 3 and g (x) = x3 + 5
It can be checked that f(x) and g(x) are bijective functions
∴ fo g is also bijective and (fog) = f(g(x)) = f (x3+5)=2(x3+5)−3=2x3+7
(fog)(x)=y⇒2x3+7=y⇒x=(y−72)13
∴ (fog)−1(x)=(x−72)13,x ϵ R
∴ The correct answer is (d).
Question 7
If f: R → R ,g : R → R and h: R → R are such that f(x)=x2 , g(x)=tanx and h(x)=logx, then the value of (h(g(f(x))))
if x=√π4 will be
SOLUTION
Solution : A
(ho(gof))(x)=h{g{f(x)}}
=h{tan x2}=log{tan x2}
∴ At x=√π4⇒(ho(gof))(x)=log tan π4
= log 1 = 0
Question 8
If the function f:[1,∞)→[1,∞) is defined by f(x)=2x(x−1), then f−1(x) is
SOLUTION
Solution : B
f(x)=y⇒2x(x−1)=y⇒x(x−1) log2 2=log2y
⇒x(x−1)=log2 y⇒x2−x−log2 y=0
⇒x=1±√1+4 log2 y2
∴x=1+√1+4 log2 y2
∴f−1(x)=12(1+√1+4 log2 x)
∴ The correct answer is (b).
Question 9
f:R×R→R such that f(x + iy) = √x2+y2. Then, f is
SOLUTION
Solution : A
Since, f (x + iy) = f (x - iy) ∴ f is many – one
Also, Range =R+ and codomain = R.
∴ Range ⊏ codomain ⇒ f is into.
Hence, f is many – one into.
Question 10
With reference to a universal set, the inclusion of a subset in another, is relation, which is
SOLUTION
Solution : D
Since A⊆A.
∴ Relation ′⊆′ is reflexive.
Since A⊆B,B⊆C⇒A⊆C
∴ Relation ′⊆′ is transitive.
But If A⊆B, Doesn't imply B⊆A,
∴ Relation is not symmetric
Question 11
If f:R→R and g:R→R are given by f(x) = |x| and g(x) = [x], then g(f(x))≤f(g(x) is true for -
SOLUTION
Solution : D
g(f(x))≤f(g(x))⇒g(|x|)≤f[x]⇒[|x|]≤|[x]|
This is true for xϵR..
Question 12
f(x)=x2−3x+4x2+3x+4 the range of f(x) is
SOLUTION
Solution : D
y=x2−3x+4x2+3x+4
yx2+3xy+4y=x2−3x+4
x2(y−1)+3x(y+1)+4(y−1)=0
D ≥0⇒9(y+1)2−4.4(y−1)2≥0
(3(y+1)−4(y−1)) (3(y+1)+4(y−1))≥0
(−y+7)(7y−1)≥0
(y−7)(y−17)≤0
17≤y≤7
Question 13
The range of the function f(x)=cos2x4+sinx4,x ϵ R is
SOLUTION
Solution : D
f(x)=1−sin2x4+sinx4=−{sin2x4−sinx4}+1=−{(sinx4−12)2−14}+1
=54−(sinx4−12)2
Maximun f(x)=54
Minimum f(x)=54−(−1−12)2=54−94=−1
Range of f(x)=[−1,54]
Question 14
Let f(x)=x−[x]1+x−[x],x ϵ R, where [ x] denotes the greatest integer function. Then, the range of f is
SOLUTION
Solution : B
The graph of y=x−[x] is as shown below
When x is an integer, x−[x]=0
Hence, f(x) = 0 when x is an integer
x→[x] as x tends to an integer.
Let X = x−[x]
So, f(x)=X1+X,X ϵ [0,1)
As X→1,X1+X→12
Hence, the range of f(x) is [0,12) .
Question 15
Range of f(x)=tan(π[x2−x])1+sin(cos x) is (where [x] denotes the greatest integer function)
SOLUTION
Solution : D
f(x)=tan(π[x2−x])1+sin(cos x)={0} because of [x2−x] is integer