Free Objective Test 02 Practice Test - 11th and 12th

Question 1

Let X be a family of sets and R be a relation on X defined by 'A is disjoint from B'. Then R is

Reflexive

Symmetric

Anti-symmetric

Transitive

SOLUTION

Solution : B

Clearly, the relation is symmetric but it is neither reflexive nor transitive.

Question 2

Let R be a relation on the set N of natural numbers defined by nRm

⇔ n is a factor of m (i.e. n(m). Then R is

Reflexive and symmetric

Transitive and symmetric

Equivalence

Reflexive, transitive but not symmetric

SOLUTION

Solution : D

Since n | n for all n $i n$ N,

therefore R is reflexive.

Since 2 | 6 but 6 | 2, therefore R is not symmetric.

Let n R m and m R p $\Rightarrow$ n|m and m|p $\Rightarrow$ n|p $\Rightarrow$ nRp

So. R is transitive.

Question 3

Inverse exists for a function which is

A. Injective

B. Surjective

C. Bijective

D. Many-one

SOLUTION

Solution : C

We have seen that inverse exists only when function is one-one and onto, i.e. Bijective.

Question 4

If f (x) +f (x+a) + f(x+2a) +....+ f (x+na) = constant; $\forall x$ $ϵ R$ and $a > 0$ and f(x) is periodic,then period of f(x), is

A. (n+1) a

e^{n + 1} a

C. na

e^{n a}

SOLUTION

Solution : A

$f (x) + f (x + a) + f (x + 2 a) + . . . . + f (x + n a) = k$
replacing $x = x + a$
$f (x + a) + f (x + 2 a) + . . . . + f (x + (n + 1) a) = k$
On subtracting second equation from first one -
$f (x) - f (x + (n + 1) a) = 0 f (x) = f (x + (n + 1) a ∴ T = (n + 1) a$

Question 5

The function $f (x) = c o s \sqrt{x}$ is

A. Periodic with period

2 \sqrt{π}

B. Periodic with period

\sqrt{π}

C. Periodic with period

4 π^{2}

D. Not a periodic function

SOLUTION

Solution : D

Try drawing $c o s \sqrt{x}$ graph. It’s not periodic.

Question 6

$T h e i n v e r s e o f f (x) = {(5 - {(x - 8)}^{5})}^{\frac{1}{3}} i s$

$5 - (x - 8)^{5}$

$8 + (5 - x^{3})^{\frac{1}{5}}$

$8 - (5 - x^{3})^{\frac{1}{5}}$

$(5 - (x - 8)^{\frac{1}{5}})^{3}$

SOLUTION

Solution : B

$y = f (x) = {(5 - {(x - 8)}^{5})}^{\frac{1}{3}}$
then $y^{3} = 5 - (x - 8)^{5} \Rightarrow (x - 8)^{5} = 5 - y^{3}$
$\Rightarrow x = 8 + (5 - y^{3})^{\frac{1}{5}}$
Let, $z = g (x) = 8 + (5 - x^{3})^{\frac{1}{5}}$
To check, $f (g (x)) = [5 - {(x - 8)}^{5}]^{\frac{1}{3}}$
$= {(5 - {[{(5 - x^{3})}^{\frac{1}{5}}]}^{5})}^{\frac{1}{3}} = (5 - 5 + x^{3})^{\frac{1}{3}} = x$
similarly , we can show that $g (f (x)) = x$
Hence, $g (x) = 8 + (5 - x^{3})^{\frac{1}{5}}$ is the inverse of $f (x)$

Question 7

Which of the following relations in R is an equivalence relatilon?

x R_{1} y \Leftrightarrow | x | = | y |

x R_{2} y \Leftrightarrow x \geq y

x R_{3} y \Leftrightarrow \frac{x}{y}

x R_{4} y \Leftrightarrow x < y

SOLUTION

Solution : A

It is simple to check that only, $R_{1}$ is an equivalence relation.

Question 8

Let A be the non – empty set of children in a family. The relation ‘x is a brother of y’ in A is

A. reflexive

B. symmetric

C. transitive

D. an equivalence relation

SOLUTION

Solution : C

Let R denotes the relation ‘is brother of ‘. Let X $ϵ$ A. If x is a girl, then we cannot say that x is brother of x.
$∴$ (x,x) $/ ϵ$ R $∴$ R is not reflexive.
Let (x, y) $ϵ$ R $∴$ x is a brother of y.
$∴$ y may or may not be a boy.
$∴$ we cannot say that (y, x) $ϵ$ R.
$∴$ R is not symmetric.
Let (x, y) $ϵ$ and (y, z) $ϵ$ R.
$∴$ x is a brother of y and y is a brother of z
$\Rightarrow$ is brother of z $\Rightarrow$ (x, z) $ϵ$ R
$∴$ R is transitive.
$∴$ The correct answer is (c).

Question 9

Given, f(x) = $l o g \frac{1 + x}{1 - x}$ and g(x) $= \frac{3 x + x^{3}}{1 + 3 x^{2}}$ then (fog) (x) equals

A. -f(x)

B. 3f(x)

[f (x)]^{3}

D. None of these

SOLUTION

Solution : B

$(f o g) (x) = f (g (x)) = f (\frac{3 x + x^{3}}{1 + 3 x^{2}}) = l o g \frac{1 + 3 x^{2} + 3 x + x^{3}}{1 + 3 x^{2} - 3 x - x^{3}} = l o g \frac{(1 + x)^{3}}{(1 - x)^{3}} = 3 f (x)$

Question 10

Let $f : [0, \infty) \to$ [0,2] be defined by $f (x) = \frac{2 x}{1 + x}$ , then f is

A. one – one but not onto

B. onto but not one – one

C. both one – one and onto

D. neither one – one nor onto

SOLUTION

Solution : A

We can draw the graph and see that the given function is one - one.
Since, Range $⊏$ Codomain $\Rightarrow$ f is into
$∴$ f is only injective.

Question 11

Which one of the following function is not invertible?

f : R \to R, f (x) = 3 x + 1

f : R \to [0, \infty), f (x) = x^{2}

f : R^{+} \to R^{+}, f (x) = \frac{1}{x^{3}}

N o n e o f t h e a b o v e

SOLUTION

Solution : B

The function f(x) = $x^{2}$ , x $ϵ$ R is not one – one because
f(-4)=f (4) = 16
$∴$ It is not invertible

Question 12

The range of the function $f (x) = \frac{2 + x}{2 - x}, x \neq 2$ is

R

R - {- 1}

R - {1}

R - {- 2}

SOLUTION

Solution : B

$y = \frac{2 + x}{2 - x} \Rightarrow 2 y - y x = 2 + x \Rightarrow x (y + 1) = 2 y - 2 \Rightarrow x = \frac{2 y - 2}{y + 1} \Rightarrow f^{- 1} (x) = \frac{2 x - 2}{x + 1}$
$∴$ Range of $f =$ Domain of $f^{- 1} = R - {- 1}$

Question 13

If $2 f (s i n x) + f (c o s x) = x \forall x ϵ$ R then range of f(x) is

[\frac{- π}{3}, \frac{π}{3}]

[\frac{- 2 π}{3}, \frac{π}{3}]

[\frac{- 2 π}{3}, \frac{π}{6}]

[\frac{- π}{6}, \frac{π}{6}]

SOLUTION

Solution : B

Put $x = s i n^{- 1} x$
$2 f (x) + f (\sqrt{1 - x^{2}}) = s i n^{- 1} x \to (1)$
On Putting $x = c o s^{- 1} x$
$\Rightarrow 2 f (\sqrt{1 - x^{2}}) + f (x) = c o s^{- 1} x \to (2)$
Eq. $(1) \times 2 \Rightarrow 4 f (x) + 2 f (\sqrt{1 - x^{2}}) = 2 s i n^{- 1} x \to (3)$
On subtracting Eq. 2 from Eq. 3 we get -
$3 f (x) = 2 s i n^{- 1} x - c o s^{- 1} x$
$f (x) = \frac{2}{3} s i n^{- 1} x - \frac{1}{3} (\frac{π}{2} - s i n^{- 1} x)$
$= s i n^{- 1} x - \frac{π}{6}$
$f_{m a x} = \frac{π}{2} - \frac{π}{6} = \frac{π}{3}, f_{m i n} = - \frac{π}{2} - \frac{π}{6} = \frac{- 4 π}{6} = \frac{- 2 π}{3}$
$= [\frac{- 2 π}{3}, \frac{π}{3}]$

Question 14

The range of $f (x) = t a n^{- 1} (x^{2} + x + a)$ $\forall x ϵ$ R is a subset of $[0, \frac{π}{2})$ then the range of a is -

[- \sqrt{3}, \frac{1}{4}]

(\frac{- π}{2}, \frac{π}{2})

[- \sqrt{3}, - 1]

[\frac{1}{4}, \infty)

SOLUTION

Solution : D

$t a n^{- 1} (x^{2} + x + a) \geq 0 \Rightarrow x^{2} + x + a \geq 0$
$\Rightarrow D \leq 0 \Rightarrow 1 - 4 a \leq 0 \Rightarrow a \geq \frac{1}{4}$ ; D is discriminant of quadratic equation.
$\Rightarrow a ϵ [\frac{1}{4}, \infty)$

Question 15

The range of the function $f (x) = \frac{x + 3}{| x + 3 |}, x \neq - 3$ is

{3, - 3}

R - {- 3}

C. All positive integers

{- 1, 1}

SOLUTION

Solution : D

$f (x) = 1$ when $x + 3 > 0$
$f (x) = - 1$ when $x + 3 < 0$
Range $= {- 1, 1}$

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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