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NCERT Exemplar Class 12 Maths Solutions Chapter 1 Relations and Functions

NCERT Exemplar Class 12 Maths Solutions Chapter 1 Relations and Functions

Edited By Ravindra Pindel | Updated on Sep 15, 2022 05:48 PM IST | #CBSE Class 12th

NCERT Exemplar Class 12 Maths solutions chapter 1 is crucial to building a base from ground up for the entire academic session. We here will help you to find the solutions of all NCERT questions of the chapter with complete steps and detailed answers. This chapter of NCERT Class 12 Maths Solutions is divided into two parts; relations and functions. We have solutions for questions from the parts explained and solved in detail. NCERT Exemplar Class 12 Maths solutions chapter 1 PDF download is to be available for students for extending learning of the concepts.

NCERT Exemplar Class 12 Maths Solutions Chapter 1: Exercise-1.3

Question:1

Let A = {a, b, c} and the relation R be defined on A as follows:

R = {(a, a), (b, c), (a, b)}.

Then, write the minimum number of ordered pairs to be added in R to make R reflexive and transitive.

Answer:

Here, R = {(a, a), (b, c), (a, b)}

The minimum number of ordered pairs to be added to make R as reflexive is (b, b) and (c, c) to R. Whereas, to make R transitive, the minimum number of ordered pairs to be added is (a, c) to R.

Therefore, we need 3 ordered pairs to add with R to make it reflexive and transitive.

Question:2

Let D be the domain of the real-valued function f defined by f(x) = \sqrt {25 - x\textsuperscript{2}} . Then, write D.

Answer:

f(x) = \sqrt {25 - x\textsuperscript{2}}
Therefore, the function can be defined as:
\\25 - x\textsuperscript{2} \geq 0\\ or, x\textsuperscript{2} \leq 25\\ -5 \leq x \leq 5\\

Therefore, the given function has a domain of [-5, 5].

Question:3

Let f, g:R \rightarrow R be defined by f(x) = 2x + 1 and g (x) = x\textsuperscript{2} - 2, \forall x \in R, respectively. Then, find gof.

Answer:

Here,
f(x) = 2x + 1
g (x) = x\textsuperscript{2} -2, \forall x \in R\\
Therefore, gof= g(f(x)) = g (2x + 1) = (2x + 1)\textsuperscript{2} -2 = 4x\textsuperscript{2} + 4x + 1 -2 = 4x\textsuperscript{2} + 4x - 1\\

Question:4

. Let f: R \rightarrow R be the function defined by f (x) = 2x - 3, \forall x \in R. write f\textsuperscript{-1}.

Answer:

Here, f (x) = 2x - 3, \forall x \in R\\
Let’s say,
y = 2x - 3\\ or, x = (y + 3)/ 2

Therefore,
f\textsuperscript{-1}(x) = (x + 3)/ 2\\

Question:5

If A = \{ a, b, c, d \} and the function f = \{ (a, b), (b, d), (c, a), (d, c) \} , write f\textsuperscript{-1}.

Answer:
Here, A = \{ a, b, c, d \} and the function f = \{ (a, b), (b, d), (c, a), (d, c) \} \\
Therefore, f\textsuperscript{-1} = \{ (b, a), (d, b), (a, c), (c, d) \} \\

Question:6

If f:R \rightarrow R is defined by f (x) = x\textsuperscript{2} - 3x + 2, write f (f (x)).

Answer:

Here, f (x) = x\textsuperscript{2} - 3x + 2
Therefore,
\\ f(f(x)) = f(x\textsuperscript{2} - 3x + 2)\\ = (x\textsuperscript{2} - 3x + 2)\textsuperscript{2} - 3(x\textsuperscript{2} - 3x + 2) + 2,\\ = x\textsuperscript{4} + 9x\textsuperscript{2} + 4 - 6x\textsuperscript{3} + 4x\textsuperscript{2} - 12x - 3x\textsuperscript{2} + 9x - 6 + 2\\ = x\textsuperscript{4} - 6x\textsuperscript{3} + 10x\textsuperscript{2} - 3x\\
Similarly,
\\f(f(x)) = x\textsuperscript{4} - 6x\textsuperscript{3} + 10x\textsuperscript{2} - 3x\\

Question:7

Is g = \{ (1, 1), (2, 3), (3, 5), (4, 7) \} a function? If g is described by g (x) = \alpha x + \beta , then what value should be assigned to \alpha\: \: and\: \: \beta .

Answer:

Here,g = \{ (1, 1), (2, 3), (3, 5), (4, 7) \}
Here, each and every element of a domain has a unique image. Therefore, g is a function.
Also, g (x) = \alpha x + \beta
Therefore,
\\g(1) = \alpha (1) + \beta = 1\\ \alpha + \beta = 1 \ldots \ldots .. (1)\\
Similarly,
g(2) = \alpha (2) + \beta = 2 \alpha + \beta = 3 \ldots \ldots .. (2)\\
By solving (1) and (2), we get
\alpha = 2 and \beta = -1\\
Therefore, g (x) = 2x - 1


Question:8

Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective.
(i) \{ (x, y): \text{x is a person, y is the mother of x} \} .\\
(ii) \{ (a, b): \text{a is a person, b is an ancestor of a} \} .\\

Answer:
(i)Here, \{ (x, y): \text{x is a person, y is the mother of x} \} .\\
Therefore, each person (x) has only one biological mother.
Hence, the given set of ordered pairs make a function.
Therefore, there are more than one person who may have the same mother. Hence, the function is many-one and surjective.
(ii) Here, \{ (a, b): \text{a is a person, b is an ancestor of a} \} .\\
It’s seen that any person ‘a’ has more than one ancestor.
Therefore, it is not a function.

Question:9

If the mappings f and g are given by f = \{ (1, 2), (3, 5), (4, 1) \} and g = \{ (2, 3), (5, 1), (1, 3) \} , write fog.\\

Answer:
Here,
f = \{ (1, 2), (3, 5), (4, 1) \} and g = \{ (2, 3), (5, 1), (1, 3) \}
Therefore,
\\fog (2) = f(g(2)) = f(3) = 5\\ fog (5) = f(g(5)) = f(1) = 2\\ fog (1) = f(g(1)) = f(3) = 5\\
So, we can write that:
\\fog = \{ (2, 5), (5, 2), (1, 5) \} \\


Question:10

Let C be the set of complex numbers. Prove that the mapping f: C \rightarrow R is given by f (z) = \vert z \vert , \forall z \in C, is neither one-one nor onto.

Answer:

Here, f: C \rightarrow R is given by f (z) = \vert z \vert , \forall z \in C
If we assume z = 4 + 3i
Then,
f(4 + 3i) = \vert 4 + 3i \vert = \sqrt (4\textsuperscript{2} + 3\textsuperscript{2})= \sqrt 25 = 5\\
Similarly, for z = 4 - 3i
f(4 - 3i) = \vert 4 - 3i \vert = \sqrt (4\textsuperscript{2} + 3\textsuperscript{2}) = \sqrt 25 = 5\\
Therefore, it is clear that f(z) is many-one.
So, \vert z \vert \geq 0, \forall z \in C,\\
However, in the question R is the co-domain given.
Hence, f(z) is not onto. So, f(z) is neither one-one nor onto.

Question:11

Let the function f: R \rightarrow R be defined by f (x) = cos x, \forall x \in R. Show that f is neither one-one nor onto.

Answer:
It is given that, f: R \rightarrow R, f(x) = cos x, \forall x \in R\\
So, we can write:
f(x\textsubscript{1}) = f(x\textsubscript{2})\\ or, cos x\textsubscript{1}= cos x\textsubscript{2}\\
Hence, x\textsubscript{1 }= 2n \pi \pm x\textsubscript{2}, where n \in Z\\
It is understandable that for any value of x\textsubscript{1 }and\: \: x\textsubscript{2}, the above equation has an infinite number of solutions.
Therefore, f(x) is a many one function.
We know the range of cos x is [-1, 1] and it is a subset of the given co-domain R.
Hence, the given function is not onto.

Question:12

Let X = \{ 1, 2, 3 \} and Y = \{ 4, 5 \} . Find whether the following subsets of X \times Y are functions from X to Y or not.
(i) f = \{ (1, 4), (1, 5), (2, 4), (3, 5) \}
(ii) g = \{ (1, 4), (2, 4), (3, 4) \} \\
(iii) h = \{ (1,4), (2, 5), (3, 5) \}
(iv) k = \{ (1,4), (2, 5) \} .\\

Answer:

Here, X = \{ 1, 2, 3 \} and Y = \{ 4, 5 \} \\
Therefore, X \times Y = \{ (1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5) \} \\
(i) f = \{ (1, 4), (1, 5), (2, 4), (3, 5) \} \\
Here, f(1) = 4 and again f(1) = 5.
Therefore, f is not a function here.
As a result, there is no unique of pre- image ‘1’.
(ii) g = \{ (1, 4), (2, 4), (3, 4) \} \\
We can clearly see that g is a function. Here in g, each element of the given domain has a unique image at the given range
(iii) h = \{ (1,4), (2, 5), (3, 5) \} \\
It’s clear that h is a function of each pre-image that has a unique image.
Again, h(2) = h(3) = 5
Therefore, the function h is also many-one.
(iv) k = \{ (1, 4), (2, 5) \} \\
Here, ‘3’ does not have any image under the mapping. Therefore, k is not a function.

Question:13

If functions f: A \rightarrow B \: \: and\: \: g: B \rightarrow A \text{ satisfies } g of = IA, then show that f is one-one and g is onto.

Answer:

Here, it is given:
f: A \rightarrow B \: \: and\: \: g: B \rightarrow A \text{ satisfies } g of = IA
It’s clear here that the function ‘g’ is inverse of ‘f’.
So, ‘f’ has to be both one-one as well as onto.
As a result, ‘g’ is both one-one and onto.

Question:14

Let f: R \rightarrow R be the function defined by f(x) = 1/(2 - cos x) \forall x \in R. Then, find the range off.

Answer:

Here, f(x) = 1/(2 - cos x) \forall x \in R
Let’s say,
\\y = 1/(2 - cos x)\\ \therefore2y - ycos x = 1\\ Or, cos x = (2y - 1)/ y\\ Or, cos x = 2 - 1/y\\
As the range of cos x is, -1 \leq cos x \leq 1\\
Therefore,
\\-1 \leq 2 - 1/y \leq 1\\ -3 \leq - 1/y \leq -1\\ 1 \leq - 1/y \leq 3\\ 1/3 \leq y \leq 1\\
Hence, the range of the given function is [1/3, 1].

Question:15

Let n be a fixed positive integer. Define a relation R in Z as follows: \forall a, b \in Z, aRb if and only if a - b is divisible by n. Show that R is an equivalence relation.

Answer:

Here, we have to a relation R in Z as follows: \forall a, b \in Z, aRb if and only if a - b is divisible by n.
Here, aRa \Rightarrow (a - a) is divisible by n and this is true for all integers.
Therefore, R is reflective.
For aRb, aRb \Rightarrow (a - b) is also divisible by n.
or, - (b - a) is divisible by n.
or, (b - a) is divisible by n
Hence, we can write it as bRa.
Therefore, R is symmetric.
For aRb , (a - b) is divisible by n.
For bRc, (b - c) is divisible by n.
Hence, (a - b) + (b-c) is divisible by n.
Or, (a-c) is divisible by n. This can be expressed as aRc.
Therefore, R is transitive.
So, R is an equivalence relation.

Question:16

If A = {1, 2, 3 }, define relations on A which have properties of being:

(a) reflexive, transitive but not symmetric

(b) symmetric but neither reflexive nor transitive

(c) reflexive, symmetric, and transitive.

Answer:

Here, A = \{ 1, 2, 3 \} .\\
(i) Assume R\textsubscript{1}= \{ (1, 1), (1, 2), (1, 3), (2, 3), (2, 2), (1, 3), (3, 3) \} \\
Here, (1, 1), (2, 2) and (3, 3) \in \ R\textsubscript{1}. R\textsubscript{1} is reflexive.
(1, 2) \in R\textsubscript{1}, (2, 3) \in R\textsubscript{1} \Rightarrow (1, 3) \in R\textsubscript{1}
. Hence,R\textsubscript{1} is transitive.
Now, (1, 2) \in R\textsubscript{1 } \Rightarrow (2, 1) \notin R\textsubscript{1}.
Therefore, R\textsubscript{1} is not symmetric.
(ii) Let say, R\textsubscript{2}= \{ (1, 2), (2, 1) \}
So, (1, 2) \in R\textsubscript{2}, (2, 1) \in R\textsubscript{2}\\
Therefore, R\textsubscript{2 } is symmetric,
(1, 1) \notin R\textsubscript{2}. Therefore, R\textsubscript{2 } is not reflexive.
(1, 2) \in R\textsubscript{2}, (2, 1) \in R\textsubscript{2 } but (1, 1) \notin R\textsubscript{2}. Hence, R\textsubscript{2 } is not transitive.
(iii) Let R\textsubscript{3 } = \{ (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3) \} \\
R\textsubscript{3 } is reflexive as (1, 1) (2, 2) and (3, 3) \in R\textsubscript{3 }\\
R\textsubscript{3 } is symmetric as (1, 2), (1, 3), (2, 3) \in R\textsubscript{3 } \Rightarrow (2, 1), (3, 1), (3, 2) \in R\textsubscript{3 }\\
Therefore, R\textsubscript{3 } is reflexive, symmetric and transitive.

Question:17

Let R be relation defined on the set of natural number N as follows:
R = \{ (x, y): x \in N, y \in N, 2x + y = 41 \} . Find the domain and range of the relation R. Also verify whether R is reflexive, symmetric, and transitive.

Answer:

Here, R = \{ (x, y): x \in N, y \in N, 2x + y = 41 \}
So, the domain D= \{ 1, 2, 3, \ldots .., 20 \} \\
And the Range = \{ 1, 3, 5, \ldots .., 39 \} \\
Here, (2, 2) \notin \: \: R\: \: as\: \: 2 \times 2 + 2 \neq 41. Therefore, R is not reflexive.
Again, (1, 39) \in R \: \: but\: \: (39, 1) \: \: \notin R. So, R is not symmetric.
Again, (11, 19) \notin R, (19, 3) \notin R; but (11, 3) \notin R. So, Further R is not transitive.
Therefore, R is neither reflexive nor symmetric and nor transitive.

Question:18

Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:

(a) an injective mapping from A to B

(b) a mapping from A to B which is not injective

(c) a mapping from B to A.

Answer:

Here, A = \{ 2, 3, 4 \} , B = \{ 2, 5, 6, 7 \} \\
(i) Assume, f: A \rightarrow B \text{ denote a mapping}
Then, f = \{ (x, y): y = x + 3 \} or\\
f = \{ (2, 5), (3, 6), (4, 7) \} , this is an injective mapping.
(ii) Again assume, g: A \rightarrow B denotes a mapping: g = \{ (2, 2), (3, 2), (4, 5) \} . Hence, it is not an injective mapping.
(iii) Again assume, h: B \rightarrow A denotes a mapping: h = \{ (2, 2), (5, 3), (6, 4), (7, 4) \} . Hence, it is a mapping from B to A.

Question:19

Give an example of a map

(i) which is one-one but not onto

(ii) which is not one-one but onto

(iii) which is neither one-one nor onto.
Answer:
(i) Say, f: N \rightarrow N, be a mapping defined by f (x) = x\textsuperscript{2}\\
If, f(x\textsubscript{1}) = f (x\textsubscript{2})\\
Then, x\textsubscript{1}\textsuperscript{2}= x\textsubscript{2}\textsuperscript{2}\\
So,x\textsubscript{1 }= x\textsubscript{2 }(as, x\textsubscript{1 }+ x\textsubscript{1 }\text{cannot be } 0)
f(x\textsubscript{1 }) = f(x\textsubscript{2 }), hence, f(x) is one-one.
However, ‘f’ is not onto, as for 1 \in N, therefore, there is no existence of x in N : f(x) = 2x + 1.
(ii) Let f: R \rightarrow [0, \infty ), is a mapping that is defined by f(x) = \vert x \vert \\
Then, we can conclude that f(x) is not a one-one as f(2) and f(-2) are the same here.
But \vert x \vert \geq 0, so the range is [0, \infty ].\\
Therefore, f(x) is onto.
(iii) Assume, f: R \rightarrow R , be a mapping which is defined by f(x) = x\textsuperscript{2}\\
Then we can say that f(x) is not one-one as f(1) and f (-1) are the same.
The range of f(x) is [0, \infty ).
Therefore, f (x) is neither one-one nor onto.

Question:20

Let A = R - \{ 3 \} , B = R - \{ 1 \} . Let f : A \rightarrow B be defined by f(x) = x - 2/ x - 3 \forall x \in A . Then show that f is bijective.

Answer:

Here,
A = R - \{ 3 \} , B = R - \{ 1 \} \\
f : A \rightarrow B be defined by f (x) = x - 2/ x - 3 \forall x \in A\\
Hence, f (x) = (x - 3 + 1)/ (x - 3) = 1 + 1/ (x - 3)\\
Say f(x1) = f (x2)\\
1+\frac{1}{x_{1}-3}=1+\frac{1}{x_{2}-3}
\\\frac{1}{x_{1}-3}=\frac{1}{x_{2}-3} \\\\ x_{1}=x_{2}
So, f (x) is an injective function.
\\If, y = (x - 2)/ (x -3)\\ Or, x - 2 = xy - 3y\\ Or, x(1 - y) = 2 - 3y\\ Or, x = (3y - 2)/ (y - 1)\\ Or, y \in R - \{ 1 \} = B\\
So, f (x) is onto or subjective.
Therefore, f(x) is a bijective function.

Question:21

Let A = [-1, 1]. Then, discuss whether the following functions defined on A are one-one, onto, or bijective:
(i) f(x) = x/2 (ii) g(x) = \vert x \vert \\
(iii) h(x) = x \vert x \vert (iv) k(x) = x\textsuperscript{2}\\

Answer:

Here, A = [-1, 1]
(i) f: [-1, 1] \rightarrow [-1, 1], f (x) = x/2\\
If, f(x\textsubscript{1}) = f(x\textsubscript{2})\\
x\textsubscript{1}/2 = x\textsubscript{2}/2\\
So, f(x) is one-one.
Also, x \in [-1, 1]\\
x/2 = f(x) = [-1/2, 1/2]\\
Hence, the range is a subset of co-domain A
So, f(x) is not onto.
Therefore, f (x) is not bijective.
(ii)
\\g(x) = \vert x \vert \\ \text{Let} g(x\textsubscript{1}) = g (x\textsubscript{2})\\ \vert x\textsubscript{1} \vert = \vert x\textsubscript{2} \vert \\ x\textsubscript{1}= \pm x\textsubscript{2}\\
So, g(x) is not one-one.
Also, g(x) = \vert x \vert \geq 0, \text{for all real x}\\
Hence, the range is [0, 1], which is subset of co-domain ‘A’
So, f(x) is not onto.
Therefore, f(x) is not bijective.
(iii)
\\\\h(x) = x \vert x \vert \\ \text{Let } h(x\textsubscript{1}) = h(x\textsubscript{2})\\ x\textsubscript{1} \vert x\textsubscript{1} \vert = x\textsubscript{2} \vert x\textsubscript{2} \vert \\ \text{If }x\textsubscript{1}, x\textsubscript{2}> 0\\ x\textsubscript{1}\textsuperscript{2}= x\textsubscript{2}\textsuperscript{2}\\ x\textsubscript{1}\textsuperscript{2}- x\textsubscript{2}\textsuperscript{2}= 0\\ (x\textsubscript{1} - x\textsubscript{2})(x\textsubscript{1} + x\textsubscript{2}) = 0\\ x\textsubscript{1 }= x\textsubscript{2 }(as x\textsubscript{1}+ x\textsubscript{2} \neq 0)\\ \text{Again }, x\textsubscript{1}, x\textsubscript{2 }< 0, and x\textsubscript{1}= x\textsubscript{2}\\ \text{Therefore }, x\textsubscript{1 }and x\textsubscript{2 }\text{of opposite sign}, x\textsubscript{1 } \neq x\textsubscript{2 }.\\
Hence, f(x) is one-one.
\\\text{For x} \in [0, 1], f(x) = x\textsubscript{2} \in [0, 1]\\ \text{For x} < 0, f(x) = -x\textsubscript{2} \in [-1, 0)\\
Hence, the range is [-1, 1].
So, h(x) is onto.
Therefore, h(x) is bijective.
(iv)
\\k(x) = x^{2}\\ \text{Let }k (x\textsubscript{1}) = k (x\textsubscript{2})\\ x\textsubscript{1}\textsuperscript{2}= x\textsubscript{2}\textsuperscript{2}\\ x\textsubscript{1 }= \pm x\textsubscript{2}\\
Therefore, k(x) is not one-one.

Question:22

Each of the following defines a relation on N:
(i) x is greater than y, x, y \in N\\
(ii) x + y = 10, x, y \in N\\
(iii) x y is square of an integer x, y \in N\\
(iv) x + 4y = 10 x, y \in N\\
Determine which of the above relations are reflexive, symmetric, and transitive.

Answer:

(i)Here, x is greater than y; x, y \in N\\
If (x, x) \in R, then x > x, that does not satisfy for any x \in N.\\
Therefore, R is not reflexive.
Say,(x, y) \in R
\\Or, xRy\\ Or, x > y
\Rightarrow y > x \\ For any x, y \in N, the above condition is not true.
Hence, R is not symmetric.
Again, xRy and yRz
\\Or, x > y and y > z\\ Or, x > z\\ Or, xRz\\
Hence, R is transitive.
(ii) x + y = 10;x, y \in N
Thus,
\\R = \{ (x, y); x + y = 10, x, y \in N \} \\ R = \{ (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1) \} \\
Therefore, (1, 1) \notin R\\
So, R is not reflexive.
Again, (x, y) \in R \Rightarrow (y, x) \in R\\
Therefore, R is symmetric.
And, (1, 9) \in R, (9, 1) \in R, but (1, 1) \notin R\\
Therefore, R is not transitive.
(iii)Here, xy is square of an integer x, y \in N
R = \{ (x, y) : \text{xy is a square of an integer} x, y \in N \} \\
So, (x, x) \in R, \forall x \in N\\
For any x \in N, x\textsuperscript{2 } is an integer.
Thus, R is reflexive.
If (x, y) \in R \Rightarrow (y, x) \in R\\
So, R is symmetric.
Again, if xy and yz both are square of an integer.
Then, xy = m\textsuperscript{2} \: \: and \: \: yz = n\textsuperscript{2} \text{for some m, n} \in Z\\
\\x = m\textsuperscript{2} /y and z = n\textsuperscript{2} /y\\ xz = m\textsuperscript{2} n\textsuperscript{2} / y\textsuperscript{2}, this must be the square of an integer.
Therefore, R is transitive.
(iv) x + 4y = 10; x, y \in N\\
\\R = \{ (x, y): x + 4y = 10; x, y \in N \} \\ R = \{ (2, 2), (6, 1) \} \\ Here, (1, 1) \notin R\\
Hence, R is not symmetric.
\\(x, y) \in R \Rightarrow x + 4y = 10\\ And, (y, z) \in R \Rightarrow y + 4z = 10\\ Or,\ x - 16z = -30\\ Or,\ (x, z) \notin R\\
Therefore, R is not transitive.

Question:23

Let A = {1, 2, 3, … 9} and R be the relation in A \times A defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in A \times A. Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)].

Answer:

\\Here, A = \{ 1, 2, 3, \ldots 9 \} \: \: and \: \: (a, b) R (c, d)\: \: if a + d = b + c for (a, b), (c, d) \in A \times A.\\ Say (a, b) R (a, b)\\ Therefore, a + b = b + a, \forall a, b \in A .
This must be true for any a, b \in A.\\
Hence, R is reflexive.
Say, (a, b) R (c, d)
Then,
\\a + d = b + c\\ c + b = d + a\\ (c, d) R (a, b)\\
Therefore, R is symmetric.
Let \\(a, b) R(c, d) ,and, (c, d) R(e, f)
\\ a + d = b + c ,and, c + f = d + e\\ a + d = b + c ,and, d + e = c + f\\ Or, (a + d) - (d + e = (b + c) - (c + f)\\ Or, a - e = b - f\\ Or, a + f = b + e\\ Or, (a, b) R(e, f)\\
So, R is transitive.
Therefore, R is an equivalence relation.

Question:24

Using the definition, prove that the function f: A \rightarrow B is invertible if and only if f is both one-one and onto.

Answer:

Say, f: A \rightarrow B be many-one functions.
If, f(a) = p \text{ and } f(b) = p\\
Then f\textsuperscript{-1}(p) = a \: \: and \: \: f\textsuperscript{-1}(p) = b\\
In this case, we have two images ‘a and b’ for one pre-image ‘p’. This is because the inverse function is not defined here.
However, to be one-one, f must be invertible.
Say, f: A \rightarrow B is not onto function.
B = \{ p, q, r \} and \{ p, q \} is the range of f.
There is no pre-image for the image r, which will have no image in set A.
And, f must be onto to be invertible.
Thus, to be both one-one and onto f must be invertible
If f is a bijective function, then f = X \rightarrow Y is invertible.

Question:26

Let \ast be the binary operation defined on Q. Find which of the following binary operations are commutative
\\(i) a \ast b = a - b\: \: \forall a, b \in Q\\ (ii) a \ast b = a^2 + b^2 \: \: \forall a, b \in Q\\ (iii) a \ast b = a + ab \: \: \forall a, b \in Q\\ (iv) a \ast b = (a - b)\textsuperscript{2}\: \: \forall a, b \in Q\\

Answer:

Here, \ast is a binary operation defined on Q.
(i)
\\a \ast b = a - b, \forall a, b \in Q \: \: and \: \: b \ast a = b - a\\ So, a \ast b \neq b \ast a\\
Hence, \ast is not commutative.
(ii)
\\a \ast b = a\textsuperscript{2}+ b\textsuperscript{2}\\ b \ast a = b\textsuperscript{2}+ a\textsuperscript{2}\\
Therefore, \ast is commutative.
(iii)
\\a \ast b = a + ab\\ b \ast a = b + ab\\ Hence, a + ab \neq b + ab\\
Therefore, \ast is not commutative.
(iv)
\\a \ast b = (a - b)\textsuperscript{2}, \forall a, b \in Q\\ b \ast a = (b -a)\textsuperscript{2}\\ As, (a - b)\textsuperscript{2} = (b - a)\textsuperscript{2}\\
Therefore, \ast is commutative.

Question:27

Let \ast be binary operation defined on R by a \ast b = 1 + ab, \forall a, b \in R. Then the operation \ast is
(i) commutative but not associative
(ii) associative but not commutative
(iii) neither commutative nor associative
(iv) both commutative and associative

Answer:

(i) Here,
\ast is a binary operation defined on R by a \ast b = 1 + ab, \forall a, b \in R
So, a \ast b = ab + 1 = b \ast a\\
Therefore, \ast is a commutative binary operation.
\\Now, a \ast (b \ast c) = a \ast (1 + bc) = 1 + a (1 + bc) = 1 + a + abc\\ Again, (a \ast b) \ast c = (1 + ab) \ast c = 1 + (1 + ab) c = 1 + c + abc\\ Therefore, a \ast (b \ast c) \neq (a \ast b) \ast c\\
Hence, \ast is not associative.
Therefore, \ast is commutative but not associative.

Question:28

Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to b \forall a, b \in T. Then R is
(A) reflexive but not transitive (B) transitive but not symmetric
(C) equivalence (D) none of these

Answer:

(C) equivalence
Here aRb, if a is congruent to b, \forall a, b \in T.\\
So, in aRa, a is congruent to a. This must always be true.
Therefore, R is reflexive.
Say, aRb \Rightarrow a \sim b\\
b \sim a \Rightarrow bRa\\
Therefore, R is symmetric.
Say, aRb and bRc
\\a \sim b \: \: and\: \: b \sim c\\ a \sim c \Rightarrow aRc\\
Therefore, R is transitive.
Therefore, R is an equivalence relation.

Question:29

Consider the non-empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then R is

(A) symmetric but not transitive (B) transitive but not symmetric

(C) neither symmetric nor transitive (D) both symmetric and transitive

Answer:

(B) transitive but not symmetric

If, aRb means a is brother of b.

Then, it does not mean b is also a brother of a. Because, b can be a sister of a too.

Therefore, R is not symmetric.

If, aRb implies that a is the brother of b.

and bRc implies that b is the brother of c.

Therefore, a must be the brother of c.

Hence, R is transitive.

Question:30

The maximum number of equivalence relations on the set A = {1, 2, 3} are

(A) 1 (B) 2

(C) 3 (D) 5

Answer:

(D) 5
Given, set A = \{ 1, 2, 3 \} \\
Now, the number of equivalence relations as follows
\\R1 = \{ (1, 1), (2, 2), (3, 3) \} \\ R2 = \{ (1, 1), (2, 2), (3, 3), (1, 2), (2, 1) \} \\ R3 = \{ (1, 1), (2, 2), (3, 3), (1, 3), (3, 1) \} \\ R4 = \{ (1, 1), (2, 2), (3, 3), (2, 3), (3, 2) \} \\ R5 = \{ (1, 2, 3) \Leftrightarrow A \times A = A^2 \} \\
Thus, the maximum number of equivalence relations is ‘5’.

Question:31

If a relation R on the set {1, 2, 3} be defined by R = {(1, 2)}, then R is

(A) reflexive (B) transitive

(C) symmetric (D) none of these

Answer:

(D) none of these

If, R be defined on the set {1, 2, 3} by R = {(1, 2)}

Then, we can say that R is not reflexive, transitive, and symmetric.

Question:32

Let us define a relation R in R as aRb if a \geq b. Then R is

(a) an equivalence relation (b) reflexive, transitive but not symmetric (c) symmetric, transitive but not reflexive (d) neither transitive nor reflexive but symmetric.

Answer:

(b).
The defined relation R in R as aRb if a \geq b.\\
Similarly, aRa implies a \geq a which is true.
Therefore, it is reflexive.
Let aRb \rightarrow a \geq b, but, b \ngtr a.\\
Therefore, we can’t write it as Rba
Hence, R is not symmetric.
Now, a \geq b, b \geq c. So, a \geq c, and this is true.
Therefore, R is transitive.

Question:33

Let A = {1, 2, 3} and consider the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}, then R is

(a) reflexive but not symmetric (b) reflexive but not transitive (c) symmetric and transitive (d) neither symmetric nor transitive.

Answer:

(a)

Given, R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}

Therefore, it can be written as: 1R1, 2R2 and 3R3.

Therefore, R is reflexive.

Here, 1R2 is not the same as 2R1 and 2R3 is not the same as 3R2.

Therefore, R is not symmetric.

Again, 1R1 and 1R2 \rightarrow 1R3\\

Therefore, R is transitive.

Question:34

The identity element for the binary operation * defined on Q ~ {0} as a * b = 2 ab a, b Q ~ {0} is

(a) 1 (b) 0 (c) 2 (d) None of these

Answer:

(c)
Here: a \ast b = ab/2 a, b Q \sim \{ 0 \} \\
Assume ‘e’ as the identity element.
Therefore, a \ast e = ae/2 = a \rightarrow e = 2 \\

Question:35

If the set A contains 5 elements and set B contains 6 elements, then the number of one-one and onto mapping from A to B is

(a) 720 (b) 120 (c) 0 (d) None of these

Answer:

Let, the number of elements in A and B set are m and n respectively. Therefore, one-one and onto mapping

from A to B is n! when m = n

and, 0 if m ≠ n

It is given that, m = 5 and n = 6.

As,5 ≠ 6 So, from A to B mapping = 0

Question:36

Let A = \{ 1, 2, 3, ..., n \} and B = \{ a, b \} . Then the number of surjections from A to B is
(a) \textsuperscript{n}P\textsubscript{2}
(b) 2\textsuperscript{n} - 2
(c)2\textsuperscript{n} - 1
(d) None of these

Answer:

(d)
It is given that, A = \{ 1, 2, 3, ..., n \} and B = \{ a, b \} \\
Say, the number of elements in set A and B are m and n respectively.
Therefore, \textsuperscript{n}C\textsubscript{m} \times m! is the number of surjections from A to B, for n \ngeq m \\
Given, m=2.
Therefore, the number of surjections from A to B
= nC_2 \times 2! = n!/[2! \times (n-2)!] = n(n - 1) = n\textsuperscript{2} - n \\

Question:37

Let f : R \rightarrow R be defined by f(x) = 1/x \forall x \in R then f is
(a) one-one (b) onto (c) bijective (d) f is not defined

Answer:

(d)
Here, f(x) = 1/x
Say, x = 0. Then f(x) = 1/0 = undefined
Hence, f(x) is not defined.

Question:38

Let f : R \rightarrow R be defined by f(x) = 3x\textsuperscript{2} - 5 and g : R \rightarrow R by g(x) = x/(x\textsuperscript{2}+1), then gof is
\\(a) (3x\textsuperscript{2}-5)/(9x\textsuperscript{4}-30x\textsuperscript{2}+ 26)\\ (b) (3x\textsuperscript{2}-5)/(9x\textsuperscript{4}-6x\textsuperscript{2}+ 26) \\(c) 3x\textsuperscript{2}/(x\textsuperscript{4}+2x\textsuperscript{2}-4) \\(d) (3x\textsuperscript{2})/(9x\textsuperscript{4}-30\textsuperscript{2}+2)\\

Answer:

(a)
Given, f(x) = 3x\textsuperscript{2} - 5 \ and \ g(x) = x/(x\textsuperscript{2}+1)\\
Therefore,
\\gof = gof (x) = g(3x\textsuperscript{2}-5) = (3x\textsuperscript{2}-5)/[(3x\textsuperscript{2}-5)\textsuperscript{2}+1]\\ = (3x\textsuperscript{2}-5)/(9x\textsuperscript{4}-30x\textsuperscript{2}+ 26)\\

Question:39

Which of the following functions from Z to Z are bijections?
\\(a) f(x) = x\textsuperscript{3 }\\(b) f(x) = x + 2 \\(c) f(x) = 2x + 1 \\(d) f(x) = x\textsuperscript{2} + 1 \\

Answer:

(b)
Given that f : Z \rightarrow Z \\
Say, x\textsubscript{1}, x\textsubscript{2} \in f(x) \rightarrow f(x\textsubscript{1}) = x\textsubscript{1}+ 2; f(x\textsubscript{2}) = x\textsubscript{2}+ 2;\\
For, f(x\textsubscript{1}) = f(x\textsubscript{2})\ \rightarrow x\textsubscript{1}+ 2 = x\textsubscript{2}+\ 2 \\
or, x\textsubscript{1 }= x\textsubscript{2}\\
Therefore, the function f(x) is one-one.
Say, y = x + 2
or, x = y - 2 \forall y \in Z \\
Hence, f(x) is onto.
Therefore, the function f(x) is bijective.

Question:40

Let f : R \rightarrow R be the functions defined by f(x) = x\textsuperscript{3} + 5. Then f\textsuperscript{-1}(x) is
\\(a) (x + 5)\textsuperscript{1/3}\\ (b) (x - 5)\textsuperscript{1/3}\\ (c) (5 - x)\textsuperscript{1/3}\\ (d) 5 - x \\

Answer:

(b)Here,
\\f(x) = x\textsuperscript{3} + 5\\ Say, y = x\textsuperscript{3} + 5\\ or, x\textsuperscript{3} = y - 5 \\ or, x = (y - 5)\textsuperscript{1/3}\ f\textsuperscript{-1}1(x) = (x - 5)\textsuperscript{1/3} \\

Question:41

Let f : A \rightarrow B and g : B \rightarrow C be the bijective functions. Then(gof)\textsuperscript{-1} is
\\(a) f\textsuperscript{-1 }og\textsuperscript{-1} \\(b) fog \\(c) g\textsuperscript{-1}of\textsuperscript{-1}\\ (d) gof \\

Answer:

(a)
Given that, f : A \rightarrow B \text{ and } g : B \rightarrow C \\
Then, (gof)\textsuperscript{-1} = f\textsuperscript{-1}og\textsuperscript{-1}\\

Question:42

Let f: \{ R-3/5 \} \rightarrow R be defined by f(x) = (3x + 2)/(5x - 3) then
\\(a) f(x) = f(x); \\(b) f\textsuperscript{-1}(x) = - f(x) \\(c) (fof)x = - x \\(d) f\textsuperscript{-1}(x) = f(x)/19\\

Answer:

(a)
Here,f(x) = (3x + 2)/(5x - 3) \forall x \neq 3/5\\
\\So, y = (3x + 2)/(5x - 3)\\ or, y (5x-3) = (3x + 2)\\ or, 5xy-3y = 3x + 2\\ or, x = (3y+2)/(5y-3)\\ Therefore, f\textsuperscript{-1}(x) = (3x+2)/(5x-3)\\ Hence, f\textsuperscript{-1}(x) = f(x)\\

Question:43

Let f : [0, 1] \rightarrow [0, 1] be defined by

f(x) = { x, if is rational

{ 1-x , if is irrational
Then (fof)x is

(a) constant

(b) 1 + x

(c) x

(d) None of these

Answer:

(c)
Here, f : [0, 1] \rightarrow [0, 1]\\
\\f = f\textsuperscript{-1} \\ Therefore, (fof)x = x\\

Question:46

Let f : R \rightarrow R be defined by

\\f(x) = \{2x, when x> 3 \\ \ \ \ \: \: \: \: \: \: \: \: \ \ \ \ \ \{ x\textsuperscript{2}, when 1< x \leq 3\\ \ \ \ \ \ \ \ \ \ \ \ \{ 3x, when x \leq 1\\ then f(- 1) + f(2) + f(4) is\\

\\ (a) 9 \\(b) 14 \\(c) 5 \\(d) None of these \\

Answer:

(a)
Here, f(- 1) + f(2) + f(4) = 3(- 1) + (2)\textsuperscript{2} + 2(4) = - 3 + 4 + 8 = 9 \\

Question:47

If f : R \rightarrow R be given by f(x) = tan x, then f-1 is
\\(a) \frac{\pi}{4} \\ (b) (n \pi + \pi /4) where\ n \in Z \\(c) \ does \ not \ exist\\ (d) None\ of \ these \\

Answer:

(a)
\\Here, f(x) = tan x \\ Assume, f(x) = y = tan x\\ Then,\ \ x = tan\textsuperscript{-1} (y)\\ or, f\textsuperscript{-1}(x) = tan\textsuperscript{-1}\ (x) \\ or, f\textsuperscript{-1}(1) = tan\textsuperscript{-1} (1)\\ or,\ f\textsuperscript{-1}(1) = tan\textsuperscript{-1} tan ( \pi /4)=\frac{\pi}{4}\\

Question:48

Let the relation R be defined in N by aRb if 2a + 3b = 30. Then R = ............

Answer:

Here, aRb : 2a + 3b = 30

Or, 3b = 30 – 2a

Or, b = (30 - 2a)/ 3

for a = 3, b = 8

a = 6, b = 6

a = 9, b = 4

a = 12, b = 2

Therefore, R = {(3, 8), (6, 6), (9, 4), (12, 2)}

Question:51

Let f : R \rightarrow R be defined by f(x) = x/ \sqrt (1+x\textsuperscript{2}) then (fofof)(x) = ................. \\

Answer:

\begin{aligned} (\text { fofof })(x) &=f[f(x))] \\ &=f\left[\begin{array}{l} f\left(\frac{x}{\sqrt{1+x^{2}}}\right) \end{array}\right] \\ &=f\left(\frac{\frac{x}{\sqrt{1+x^{2}}}}{\sqrt{1+\frac{x^{2}}{1+x^{2}}}}\right) \\ &=f\left(\frac{x}{\sqrt{1+2 x^{2}}}\right) \end{aligned}
\\\text { fofof }(x) =\frac{\frac{x}{\sqrt{1+2 x^{2}}}}{\sqrt{1+\frac{x^{2}}{1+2 x^{2}}}} \\ =\frac{x}{\sqrt{1+3 x^{2}}}

Question:53

Let R = {(3,1), (1,3), (3,3)} be a relation defined on the set A = {1, 2, 3}. Then R is symmetric, transitive and not reflexive.

Answer:

False Here in the question, R = {(3,1), (1,3), (3,3)} which is defined on the set A = {1, 2, 3}

As (1 ,1) R, R is not a reflexive one.

As (3, 1) R and (1, 3) belongs to R, then R is symmetric.

Again, (1, 3) R, (3, 1) R. However (1, 1) does not belong to R. Then R is not transitive.

Question:54

Let f: R→R be the function defined by f(x) = sin(3x+2), x R. Then f is invertible.

Answer:

False
Here, f(x) = sin(3x+2), x R. This function is not a one-one onto function for all x \in R.\\
Again, f(x) = sin(3x+2) = 0\\
or, 3x + 2 = n\pi, n \in \ Z.
Hence, the function is not invertible.

Question:55

Every relation which is symmetric and transitive is also reflexive.

Answer:

False

Assume, a relation R : R= {(1,2), (2,1), (2,2)} on the set A = {1,2}

Therefore, it is clear that (1,1) ∉ R. Hence, it is not reflexive.

Question:56

An integer m is said to be related to another integer n, if m is an integral multiple of n. This relation Z is reflexive, symmetric, and transitive.

Answer:

False.

Here, the given relation in the question is reflexive and transitive. However, it is not symmetric

Question:57

Let A = \{ 0,1 \} and N be the set of natural numbers. Then the mapping f: N \rightarrow A defined by f(2n - 1) = 0, f(2n) = 1, n \in N, \\is onto.

Answer:

True
Here, A = \{ 0,1 \} , and\: \: f(2n - 1) = 0, f(2n) = 1, \forall A n \in N\\
Therefore, the range of f is \{ 0, 1 \} \\
Hence, f: N \rightarrow A mapping is onto.

Question:58

The relation R on the set A = {1, 2, 3} defined as R = {(1,1), (1,2), (2,1), (3,3)}

Answer:

False

As, R = \{ (1,1), (1,2), (2,1), (3,3) \} \\

Therefore, (2,2) \notin R\\

Hence, R is not reflexive.

Question:60

The composition of function is associative.

Answer:

True.

fo(goh)(x) = (fog)oh

Question:61

Every function is invertible.

Answer:

False.

Only bijective functions are invertible.

Question:62

A binary operation on a set always has the identity element.

Answer:

False

‘+’ is a binary operation on the set N but it has no identity element.

Subtopics Covered in NCERT Exemplar Solutions for Class 12 Maths Chapter 1

This chapter is divided into various sub-topics that are related to relations and functions and its operations. They are mentioned below:

  • Introduction
  • Types of relations
  • Types of functions
  • Composition of functions & invertible functions
  • Binary operations

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What student will learn in NCERT Exemplar Class 12 Maths Solutions Chapter 1?

  • NCERT Exemplar solutions for Class 12 Maths chapter 1, students will learn about the basic concept of functions and relations. They will also get a better view of domains, co-domains, range etc. They will solve questions that will help in understanding these relations in a much more detailed way with help on lemmas and theorems.
  • NCERT Exemplar Class 12 Maths solutions chapter 1 will throw light on how to operate these functions by adding, multiplying, subtracting etc. Students will also get a more detailed view of various function types, injectivity and surjectivity. One can understand what all elements and factors comprise a function. Also, it throws light on a topic called invertible functions.
  • Topics about binary operations of functions using two operands will be covered in Class 12 Maths NCERT Exemplar solutions chapter 1 as well. It will explain in detail how commutativity, symmetry, associativity is used, in binary operations.
  • We have a team of expert teachers and guides who have solved these questions and have provided the same for the students to understand.
  • Solving NCERT questions is important, to understand the topic from the very beginning. We have solved the questions in a very simple way so that it is easier for the student to understand and grasp the topic and idea with much ease.
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NCERT Exemplar Class 12 Maths Solutions

Important Topics To Cover From NCERT Exemplar Class 12 Maths Solutions Chapter 1

· This is the very start of high school algebra, which holds the very root of higher education mathematics. In NCERT Exemplar Class 12 Maths chapter 1 solutions, students will learn about relations and functions, which is a type of relation. The relation is nothing but the relationship between two groups or the relationship between input and output.

· The function is a relation, but is used for obtaining the output from a set input. Learning the relations and functions at the very start of 12 Class will help in setting one's roots in calculus mathematics for cracking entrance exams and higher education.

· Several things are covered in Class 12 Maths NCERT Exemplar solutions chapter 1 which goes into the types of functions and relations. One will learn about various relations that will be used, like universal relations, reflective relations, empty relation, symmetric relation, etc.

NCERT Exemplar Class 12 Solutions

Also, check NCERT Solutions for questions given in the book:

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Frequently Asked Question (FAQs)

1. Are these solutions helpful for board examinations?

Yes, NCERT exemplar solutions for Class 12 Maths chapter 1  are designed well to prepare you for board examinations.

2. What are the important topics of this chapter?

The important topics of this NCERT Exemplar Class 12 Maths solutions chapter 1 include Types of Relations, Types of Functions, Binary Operations and Composition of Functions and Invertible Function.

3. How to remember the formulae well?

The best ways to remember formulae is to keep revising them and solving related problems. Other methods include reading them on a daily basis. You can also prepare charts to stick near your study table or bed to go through in your leisure time.

4. How many questions are there in this chapter?

There are a total 70 questions in NCERT Exemplar Class 12 Maths solutions chapter 1 based on different concepts.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

  • Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

  • Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

  • Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.

  • Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.


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In India, the design and coding fields offer exciting career options that can leverage your interest in both. Here's how you can navigate this path:

Choosing Your Stream:

  • Graphic Design Focus: Consider a Bachelor's degree in Graphic Design or a design diploma. Build a strong portfolio showcasing your creative skills. Learn the basics of HTML, CSS, and JavaScript to understand web development better. Many online resources and bootcamps offer these introductory courses.

  • Coding Focus: Pursue a Computer Science degree or a coding bootcamp in India. These programs are intensive but can equip you with strong coding skills quickly. While building your coding prowess, take online courses in graphic design principles and UI/UX design.

Engineering Subjects (for a Degree):

  • Information Technology (IT): This offers a good mix of web development, networking, and database management, all valuable for web design/development roles.

  • Human-Computer Interaction (HCI): This is a specialized field that bridges the gap between design and computer science, focusing on how users interact with technology. It's a perfect choice if you're interested in both aspects.

  • Passing NIOS in October 2024 will make you eligible for NIT admissions in 2025 . NIT admissions are based on your performance in entrance exams like JEE Main, which typically happen in January and April. These exams consider the previous year's Class 12th board results (or equivalent exams like NIOS).

Here's why 2025 is more likely:

  • JEE Main 2024 Admissions: The application process for NITs through JEE Main 2024 is likely complete by now (May 2024). They consider your 2023 Class 12th marks (CBSE in this case).
  • NIOS Results: Since NIOS results typically come out after the NIT admission process, your October 2024 NIOS marks wouldn't be available for JEE Main 2024.

Looking Ahead (2025 Admissions):

  • Focus on JEE Main: Since you have a computer science background, focus on preparing for JEE Main 2025. This exam tests your knowledge in Physics, Chemistry, and Mathematics, crucial for engineering programs at NITs.
  • NIOS Preparation: Utilize the time between now and October 2024 to prepare for your NIOS exams.
  • Eligibility Criteria: Remember, NITs typically require a minimum of 75% marks in Class 12th (or equivalent) for general category students (65% for SC/ST). Ensure you meet this criteria in your NIOS exams.

Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore,  with your academic background. Here's why:

  • High CAT Score: A score exceeding  99.9 percentile is exceptional and puts you amongst the top candidates vying for admission. IIM Bangalore prioritizes  CAT scores heavily in the shortlisting process.

  • Strong Academics: Your 96% in CBSE 12th and a B.Tech degree demonstrate a solid academic foundation, which IIM Bangalore also considers during shortlisting.

However, the shortlisting process is multifaceted:

  • Other Factors: IIM Bangalore considers other factors beyond CAT scores, such as your work experience (if any), XAT score (if you appear for it), academic diversity, gender diversity, and performance in the interview and Written Ability Test (WAT) stages (if shortlisted).

Here's what you can do to strengthen your application:

  • Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.

  • Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.

Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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