NCERT Solutions for Miscellaneous Exercise Chapter 1 Class 12

NCERT Solutions for Miscellaneous Exercise Chapter 1 Class 12 - Relations and Functions

Edited By Ramraj Saini | Updated on Dec 04, 2023 11:59 AM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Chapter 1 Miscellaneous Exercise

NCERT Solutions for miscellaneous exercise chapter 1 class 12 Relations and Functions are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for class 12 maths chapter 1 miscellaneous exercise discusses concepts related to inverse, injective, onto functions. Also in Class 12 maths chapter 1 miscellaneous exercise some of the topics of binary operations signum function, inverse functions. All in all it can be said that in miscellaneous exercises, diverse questions are asked unlike in earlier exercises of this chapter. Hence class 12 maths chapter 1 miscellaneous exercise is a good source to cover the syllabus in a comprehensive manner. There are a total of 5 exercises in the NCERT solutions for class 12 maths chapter 1 including miscellaneous exercise which is present in NCERT Class 12th book. Also with NCERT questions, students should practice NCERT exemplar and class 12 CBSE previous year questions for better understanding.

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Miscellaneous exercise class 12 chapter 1 are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise enumerated in NCERT Book together using the link provided below.

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NCERT solutions for Class 12 Maths Chapter 1 Relations and Functions: Miscellaneous Exercise

Question:1 Let $f : R \rightarrow R$ be defined as $f (x) = 10x + 7$ . Find the function $g : R \rightarrow R$ such that $g o f = f o g = I_R$ .

Answer:

$f : R \rightarrow R$

$f (x) = 10x + 7$

$g : R \rightarrow R$ and $g o f = f o g = I_R$

For one-one:

$f(x)=f(y)$

$10x+7=10y+7$

$10x=10y$

$x=y$

Thus, f is one-one.

For onto:

For $y \in R$ , $y=10x+7$

$x= \frac{y-7}{10} \in R$

Thus, for $y \in R$ , there exists $x= \frac{y-7}{10} \in R$ such that

$f(x) = f(\frac{y-7}{10})=10(\frac{y-7}{10})+7=y-7+7=y$

Thus, f is onto.

Hence, f is one-one and onto i.e. it is invertible.

Let $g : R \rightarrow R$ as $f(y)=\frac{y-7}{10}$

$gof(x)=g(f(x))= g(10x+7) =\frac{(10x+7)-7}{10}=\frac{10x}{10}=x$

$fog(x)=f(g(x))= f(\frac{y-7}{10}) =10\frac{y-7}{10}+7=y-7+7=y$

$\therefore$ $gof(x)=I_R$ and $fog(x)=I_R$

Hence, $g : R \rightarrow R$ defined as $g(y)=\frac{y-7}{10}$

Question:2 Let $f : W \rightarrow W$ be defined as $f (n) = n -1$ , if n is odd and $f (n) = n +1$ , if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.

Answer:

$f : W \rightarrow W$

$f (n) = n -1$ if n is odd

$f (n) = n +1$ if n is even.

For one-one:

Taking x as odd number and y as even number.

$f(x)=f(y)$

$x-1=y+1$

$x-y=2$

Now, Taking y as odd number and x as even number.

$f(x)=f(y)$

$x+1=y-1$

$x-y = - 2$

This is also impossible.

If both x and y are odd :

$f(x)=f(y)$

$x-1=y-1$

$x=y$

If both x and y are even :

$f(x)=f(y)$

$x+1=y+1$

$x=y$

$\therefore$ f is one-one.

Onto:

Any odd number 2r+1 in codomain of N is an image of 2r in domain N and any even number 2r in codomain N is the image of 2r+1 in domain N.

Thus, f is onto.

Hence, f is one-one and onto i.e. it is invertible.

Sice, f is invertible.

Let $g : W \rightarrow W$ as $m+1$ if m is even and $m-1$ if m is odd.

When x is odd.

$gof(x)=g(f(x))= g(n-1) =n-1+1=n$

When x is even

$gof(x)=g(f(x))= g(n+1) =n+1-1=n$

Similarly, m is odd

$fog(x)=f(g(x))= f(m-1) =m-1+1=m$

m is even ,

$fog(x)=f(g(x))= f(m+1) =m-1+1=m$

$\therefore$ $gof(x)=I_W$ and $fog(x)=I_W$

Hence, f is invertible and the inverse of f is g i.e. $f^{-1}=g$ , which is the same as f.

Hence, inverse of f is f itself.

Question:3 If f : R → R is defined by f(x) = x ² – 3x + 2, find f (f (x)).

Answer:

This can be solved as following

f : R → R

$f(x) = x^{2}-3x+2$

$f(f(x)) = f(x^{2}-3x+2) = (x^{2}-3x+2)^{2} - 3(x^{2}-3x+2)+2$

$= (x^{4}+9x^{2}+4-6x^{3}-12x+4x^{2})-3x^{2}+9x-6+2$

$= x^{4} - 6x^{3}+10x^{2} - 3x$

Question:4 Show that the function $f : R \rightarrow \{x \in R : - 1 < x < 1\}$ defined by $f(x) = \frac{x}{1 + |x|}$ $x \in R$ is one one and onto function.

Answer:

The function $f : R \rightarrow \{x \in R : - 1 < x < 1\}$ defined by

$f(x) = \frac{x}{1 + |x|}$ , $x \in R$

One- one:

Let $f(x)=f(y)$ , $x,y \in R$

$\frac{x}{1+\left | x \right |}=\frac{y}{1+\left | y \right |}$

It is observed that if x is positive and y is negative.

$\frac{x}{1+x}= \frac{y}{1+y}$

Since x is positive and y is negative.

$x> y\Rightarrow x-y> 0$ but 2xy is negative.

$x-y\neq 2xy$

Thus, the case of x is positive and y is negative is removed.

Same happens in the case of y is positive and x is negative so this case is also removed.

When x and y both are positive:

$f(x)=f(y)$

$\frac{x}{1+x}= \frac{y}{1+y}$

$x(1+y)=y(1+x)$

$x+xy=y+xy$

$x=y$

When x and y both are negative : $f(x)=f(y)$

$\frac{x}{1-x}= \frac{y}{1-y}$

$x(1-y)=y(1-x)$

$x-xy=y-xy$

$x=y$

$\therefore$ f is one-one.

Onto:

Let $y \in R$ such that $-1< y< 1$

If y is negative, then $x= \frac{y}{y+1} \in R$

$f(x)=f(\frac{y}{y+1} )= \frac{\frac{y}{1+y}}{1+ | \frac{y}{1+y}|}$ $= \frac{\frac{y}{1+y}}{1+ \frac{-y}{1+y}}=\frac{y}{1+y-y}=y$

If y is positive, then $x= \frac{y}{1-y} \in R$

$f(x)=f(\frac{y}{1-y} )= \frac{\frac{y}{1-y}}{1+ | \frac{y}{1-y}|}$ $= \frac{\frac{y}{1-y}}{1+ \frac{-y}{1-y}}=\frac{y}{1-y+y}=y$

Thus, f is onto.

Hence, f is one-one and onto.

Question:5 Show that the function $f : R \rightarrow R$ given by $f (x) = x ^3$ is injective.

Answer:

$f : R \rightarrow R$

$f (x) = x ^3$

One-one:

Let $f(x)=f(y)\, \, \, \, \, \, x,y \in R$

$x^{3}=y^{3}$

We need to prove $x=y$ .So,

Let $x\neq y$ then there cubes will not be equal i.e. $x^{3}\neq y^{3}$ .
It will contradict given condition of cubes being equal.
Hence, $x=y$ and it is one -one which means it is injective

Question:6 Give examples of two functions $f : N \rightarrow Z$ and $g : Z \rightarrow Z$ such that $gof$ is injective but g is not injective. (Hint : Consider $f (x) = x$ and $g (x) = | x |$ ).

Answer:

$f : N \rightarrow Z$

$g : Z \rightarrow Z$

$f (x) = x$

$g (x) = | x |$

One - one:

Since $g (x) = | x |$

$f(1)=\left | 1 \right | = 1$

$f(-1)=\left |- 1 \right | = 1$

As we can see $f(1)=f(-1)$ but $1\neq -1$ so $g(x)$ is not one-one.

Thus , g(x) is not injective.

$gof : N \rightarrow Z$

$gof(x) = g(f(x)) = g(x) =$ $\left | x \right |$

Let $gof(x)=gof(y)\, \, \, \, \, x,y \in N$

$\left | x \right |=\left | y \right |$

Since, $x,y \in N$ so x and y are both positive.

$\therefore x=y$

Hence, gof is injective.

Question:7 Give examples of two functions $f : N\rightarrow N$ and $g : N\rightarrow N$ such that $gof$ is onto but $f$ is not onto.

(Hint : Consider $f(x) = x + 1$ and $g(x) = \left\{\begin{matrix} x -1 & if x > 1\\ 1 & if x = 1 \end{matrix}\right.$

Answer:

$f : N\rightarrow N$ and $g : N\rightarrow N$

$f(x) = x + 1$ and $g(x) = \left\{\begin{matrix} x -1 & if x > 1\\ 1 & if x = 1 \end{matrix}\right.$

Onto :

$f(x) = x + 1$

Consider element in codomain N . It is clear that this element is not an image of any of element in domain N .

$\therefore$ f is not onto.

$gof : N\rightarrow N$

$gof(x)= g(f(x))= g(x+1)= x+1-1 =x \, \, \, \, \, \, \, \, \, since\, x \in N\Rightarrow x+1> 1$

Now, it is clear that $y \in N$ , there exists $x=y \in N$ such that $gof(x)=y$ .

Hence, $gof$ is onto.

Question:8 Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if $A \subset B$ . Is R an equivalence relation on P(X)? Justify your answer.

Answer:

Given a non empty set X, consider P(X) which is the set of all subsets of X.

Since, every set is subset of itself , ARA for all $A \in P(x)$

$\therefore$ R is reflexive.

Let $ARB \Rightarrow A\subset B$

This is not same as $B\subset A$

If $A =\left \{ 0,1 \right \}$ and $B =\left \{ 0,1,2 \right \}$

then we cannot say that B is related to A.

$\therefore$ R is not symmetric.

If $ARB \, \, \, and \, \, \, BRC, \, \, then \, \, A\subset B \, \, \, and \, \, B\subset C$

this implies $A\subset C$ $= ARC$

$\therefore$ R is transitive.

Thus, R is not an equivalence relation because it is not symmetric.

Question:9 Given a non-empty set X, consider the binary operation $* : P(X) \times P(X) \rightarrow P(X)$ given by $A * B = A \cap B\;\; \forall A ,B\in P(X)$ , where P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation ∗.

Answer:

Given $* : P(X) \times P(X) \rightarrow P(X)$ is defined as $A * B = A \cap B\;\; \forall A ,B\in P(X)$ .

As we know that $A \cap X=A=X\cap A\forall A \in P(X)$

$\Rightarrow$ $A*X =A=X*A \forall A in P(X)$

Hence, X is the identity element of binary operation *.

Now, an element $A \in P(X)$ is invertible if there exists a $B \in P(X)$ ,

such that $A*B=X=B*A$ (X is identity element)

i.e. $A\cap B=X=B\cap A$

This is possible only if $A=X=B$ .

Hence, X is only invertible element in $P(X)$ with respect to operation *

Question:10 Find the number of all onto functions from the set $\{1, 2, 3, ... , n\}$ to itself.

Answer:

The number of all onto functions from the set $\{1, 2, 3, ... , n\}$ to itself is permutations on n symbols 1,2,3,4,5...............n.

Hence, permutations on n symbols 1,2,3,4,5...............n = n

Thus, total number of all onto maps from the set $\{1, 2, 3, ... , n\}$ to itself is same as permutations on n symbols 1,2,3,4,5...............n which is n.

Question:11(i) Let $S = \{a, b, c\}$ and $T = \{1, 2, 3\}$ . Find $F^{-1}$ of the following functions F from S to T, if it exists.

(i) $F = \{(a, 3), (b, 2), (c, 1)\}$

Answer:

$F:S\rightarrow T$

$F : \{a, b, c\}\rightarrow \left \{ 1,2,3 \right \}$ is defined as $F = \{(a, 3), (b, 2), (c, 1)\}$

$F(a)=3,F(b)=2,F(c)=1$

$\therefore \, \, F^{-1}:T\rightarrow S$ is given by

$F^{-1} = \{(3, a), (2, b), (1, c)\}$

Question:11(ii) Let $S = \{a, b, c\}$ and $T = \{1, 2, 3\}$ . Find $F^{-1}$ of the following functions F from S to T, if it exists.

(ii) $F = \{(a, 2), (b, 1), (c, 1)\}$

Answer:

$F:S\rightarrow T$

$F : \{a, b, c\}\rightarrow \left \{ 1,2,3 \right \}$ is defined as $F = \{(a, 2), (b, 1), (c, 1)\}$

$F(a)=2,F(b)=1,F(c)=1$ , F is not one-one.

So inverse of F does not exists.

Hence, F is not invertible i.e. $F^{-1}$ does not exists.

Question:12 Consider the binary operations $* : R \times R \rightarrow R$ and $\circ : R \times R \rightarrow R$ defined as $a *b = |a - b|$ and $a \circ b = a \;\forall a \in R$ . Show that ∗ is commutative but not associative, $\circ$ is associative but not commutative. Further, show that $\forall a,b,c \in R$ , $a*(b\circ c) = (a*b)\circ (a*c)$ . [If it is so, we say that the operation ∗ distributes over the operation $\circ$ ]. Does $\circ$ distribute over ∗? Justify your answer.

Answer:

Given $* : R \times R \rightarrow R$ and $\circ : R \times R \rightarrow R$ is defined as
$a *b = |a - b|$ and $a \circ b = a \;\forall a,b \in R$

For $a,b \in R$ , we have

$a *b = |a - b|$

$b *a = |b - a| = \left | -(a-b) \right |=\left | a-b \right |$

$\therefore a*b = b *a$

$\therefore$ the operation is commutative.

$(1*2)*3 = (\left | 1-2 \right |)*3=1*3=\left | 1-3 \right |=2$

$1*(2*3) = 1*(\left | 2-3 \right |)=1*1=\left | 1-1 \right |=0$

$\therefore (1*2)*3\neq 1*(2*3)$ where $1,2,3 \in R$

$\therefore$ the operation is not associative

Let $a,b,c \in R$ . Then we have :

$a*(b \circ c) = a *b =\left | a-b \right |$

Hence, $a*(b \circ c)=(a*b )\circ(a* c)$

Now,

$1\circ (2*3) = 1\circ(\left | 2-3 \right |)=1\circ1=1$

$(1\circ 2)*(1 \circ 3) =1*1=\left | 1-1 \right |=0$

$\therefore 1 \circ(2*3)\neq (1\circ 2)*(1 \circ 3)$ for $1,2,3 \in R$

Hence, operation o does not distribute over operation *.

Question:13 Given a non-empty set X, let ∗ : P(X) × P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ∀ A, B ∈ P(X). Show that the empty set φ is the identity for the operation ∗ and all the elements A of P(X) are invertible with A–1 = A. (Hint : (A – φ) ∪ (φ – A) = A and (A – A) ∪ (A – A) = A ∗ A = φ).

Answer:

It is given that

$*: P(X) \times P(X) \rightarrow P(X)$ be defined as

$A * B = (A - B) \cup (B - A), A, B \ \epsilon \ P(X).$

Now, let $A \ \epsilon \ P(X)$ .
Then,

$A * \phi = (A - \phi) \cup (\phi - A) = A \cup \phi = A$
And

$\phi *A = (\phi - A) \cup (A - \phi) = \phi \cup A = A$

Therefore,

$A * \phi = \phi *A = A , A \ \epsilon \ P(X)$

Therefore, we can say that $\phi$ is the identity element for the given operation *.

Now, an element A $\epsilon$ P(X) will be invertible if there exists B $\epsilon$ P(X) such that

$A*B = \phi = B*A \ \ \ \ \ \ \ \ \ \ \ \ (\because \phi \ is \ an \ identity \ element)$

Now, We can see that

$???????A * A = (A -A) \cup (A - A) = \phi \cup \phi = \phi , A \ \epsilon \ P(X).$ $A * A = \left ( A - A \right ) \cup \left ( A-A \right ) = \phi \cup \phi = \phi ,$ such that $A \ \epsilon \ P(X)$

Therefore, by this we can say that all the element A of P(X) are invertible with $A^{-1}= A$

Question:14 Define a binary operation ∗ on the set $\{0, 1, 2, 3, 4, 5\}$ as $a * b = \left\{\begin{matrix} a + b &if\;a+b < 6 \\ a+ b -6 & if\;a+b\geq6 \end{matrix}\right.$ Show that zero is the identity for this operation and each element $a\neq 0$ of the set is invertible with $6 - a$ being the inverse of $a$ .

Answer:

X = $\{0, 1, 2, 3, 4, 5\}$ as

$a * b = \left\{\begin{matrix} a + b &if\;a+b < 6 \\ a+ b -6 & if\;a+b\geq6 \end{matrix}\right.$

An element $c \in X$ is identity element for operation *, if $a*c=a=c*a \, \, \forall \, \, a \in X$

For $a \in X$ ,

$a *0 = a+0=a\, \, \, \, \, \, \, \, \, \, \, \, \left [ a \in X \Rightarrow a+0< 6 \right ]$

$0 *a = 0+a=a\, \, \, \, \, \, \, \, \, \, \, \, \left [ a \in X \Rightarrow a+0< 6 \right ]$

$\therefore \, \, \, \, \, \, \, \, \, a*0=a=0 *a$ $\forall a \in X$

Hence, 0 is identity element of operation *.

An element $a \in X$ is invertible if there exists $b \in X$ ,

such that $a*b=0=b*a$ i.e. $\left\{\begin{matrix} a + b =0=b+a &if\;a+b < 6 \\ a+ b -6=0=b+a-6 & if\;a+b\geq6\end{matrix}\right.$

means $a=-b$ or $b=6-a$

But since we have X = $\{0, 1, 2, 3, 4, 5\}$ and $a,b \in X$ . Then $a\neq -b$ .

$\therefore b=a-x$ is inverse of a for $a \in X$ .

Hence, inverse of element $a \in X$ , $a\neq 0$ is 6-a i.e. , $a^{-1} = 6-a$

Question:15 Let $A = \{- 1, 0, 1, 2\}$ , $B = \{- 4, - 2, 0, 2\}$ and $f, g : A \rightarrow B$ be functions defined by $f(x) = x^2 -x, x\in A$ and $g(x) = 2\left |x - \frac{1}{2} \right | - 1, x\in A$ . Are $f$ and $g$ equal? Justify your answer. (Hint: One may note that two functions $f : A \rightarrow B$ and $g : A \rightarrow B$ such that $f (a) = g (a) \;\forall a \in A$ , are called equal functions).

Answer:

Given :

$A = \{- 1, 0, 1, 2\}$ , $B = \{- 4, - 2, 0, 2\}$

$f, g : A \rightarrow B$ are defined by $f(x) = x^2 -x, x\in A$ and $g(x) = 2\left |x - \frac{1}{2} \right | - 1, x\in A$ .

It can be observed that

$f(-1) = (-1)^2 -(-1)=1+1=2$

$g(-1) = 2\left |-1 - \frac{1}{2} \right | - 1= 2\left | \frac{-3}{2} \right |-1=3-1=2$

$f(-1)=g(-1)$

$f(0) = (0)^2 -(0)=0+0=0$

$g(0) = 2\left |0 - \frac{1}{2} \right | - 1= 2\left | \frac{-1}{2} \right |-1=1-1=0$

$f(0)=g(0)$

$f(1) = (1)^2 -(1)=1-1=0$

$g(1) = 2\left |1 - \frac{1}{2} \right | - 1= 2\left | \frac{1}{2} \right |-1=1-1=0$

$f(1)=g(1)$

$f(2) = (2)^2 -(2)=4-2=2$

$g(2) = 2\left |2 - \frac{1}{2} \right | - 1= 2\left | \frac{3}{2} \right |-1=3-1=2$

$f(2)=g(2)$

$\therefore \, \, \, f(a)=g(a) \forall a\in A$

Hence, f and g are equal functions.

Question:16 Let $A = \{1, 2, 3\}$ . Then number of relations containing $(1, 2)$ and $(1, 3)$ which are reflexive and symmetric but not transitive is

(A) 1
(B) 2
(C) 3
(D) 4

Answer:

$A = \{1, 2, 3\}$

The smallest relations containing $(1, 2)$ and $(1, 3)$ which are
reflexive and symmetric but not transitive is given by

$R = \left \{ (1,1),(2,2),(3,3),(1,2),(1,3),(2,1),(3,1) \right \}$

$(1,1),(2,2),(3,3) \in R$ , so relation R is reflexive.

$(1,2),(2,1) \in R$ and $(1,3),(3,1) \in R$ , so relation R is symmetric.

$(2,1),(1,3) \in R$ but $(2,3) \notin R$ , so realation R is not transitive.

Now, if we add any two pairs $(2,3)$ and $(3,2)$ to relation R, then relation R will become transitive.

Hence, the total number of the desired relation is one.

Thus, option A is correct.

Question:17 Let $A = \{1, 2, 3\}$ . Then number of equivalence relations containing $(1, 2)$ is
(A) 1
(B) 2
(C) 3
(D) 4

Answer:

$A = \{1, 2, 3\}$

The number of equivalence relations containing $(1, 2)$ is given by

$R = \left \{ (1,1),(2,2),(3,3),(1,2),(2,1) \right \}$

We are left with four pairs $(2,3)$ , $(3,2)$ , $(1,3),(3,1)$ .

$(1,1),(2,2),(3,3) \in R$ , so relation R is reflexive.

$(1,2),(2,1) \in R$ and $(2,3),(3,2) \notin R$ , so relation R is not symmetric.

$(1,3),(3,1) \notin R$ , so realation R is not transitive.

Hence , equivalence relation is bigger than R is universal relation.

Thus the total number of equivalence relations cotaining $(1,2)$ is two.

Thus, option B is correct.

Question:18 Let $f : R \rightarrow R$ be the Signum Function defined as $f(x) = \left\{\begin{matrix} 1 & x> 0 \\ 0 &x = 0 \\ -1& x < 0 \end{matrix}\right.$ and $g : R \rightarrow R$ be the Greatest Integer Function given by $g (x) = [x]$ , where $[x]$ is greatest integer less than or equal to $x$ . Then, does $fog$ and $gof$ coincide in $(0, 1]$ ?

Answer:

$f : R \rightarrow R$ is defined as $f(x) = \left\{\begin{matrix} 1 & x> 0 \\ 0 &x = 0 \\ -1& x < 0 \end{matrix}\right.$

$g : R \rightarrow R$ is defined as $g (x) = [x]$

Let $x \in (0, 1]$

Then we have , $[1]=1$ if x=1 and $[x]=0\, \, \, \, \, if\, 0< x< 1$

1627294208851

$\therefore \, \, gof(x)=g(f(x))=g(1)=\left [ 1 \right ]=1\, \, \, \,(since \, \, x> 0)$

Hence,for $x \in (0, 1]$ , $fog(x)=0$ and $gof(x)=1$ .

Hence , gof and fog do not coincide with $(0, 1]$ .

Question:19 Number of binary operations on the set are(A) 10(B) 16(C) 20(D ) 8

Answer:

Binary operations on the set $\{a, b\}$ are is a function from $\{a, b\}\times \{a, b\}\rightarrow \{a, b\}$

i.e. * is a function from $\left \{ (a,b),(b,a),(a,a),(b,b) \right \} \rightarrow \{a, b\}$

Hence, the total number of binary operations on set $\{a, b\}$ is $2^{4}=16.$

Hence, option B is correct

More About NCERT Solutions for Class 12 Maths Chapter 1 Miscellaneous Exercise

The NCERT Class 12 Maths chapter Relations and Functions has a total of 5 exercises including miscellaneous exercise. NCERT solutions for Class 12 Maths chapter 1 miscellaneous exercise covers solutions to 19 main questions and their sub-questions. This exercise covers more NCERT syllabus topics than preceding exercises and holds more weightage for exams also. Hence NCERT solutions for Class 12 Maths chapter 1 miscellaneous exercise is recommended for students to strengthen their conceptual understanding.

Also Read| NCERT Notes For Class 12 Mathematics Chapter 1

Benefits of NCERT Solutions for Class 12 Maths Chapter 1 Miscellaneous Exercises

By using NCERT book miscellaneous exercise chapter 1 Class 12 students will be able to get an holistic understanding of the topics mentioned in this chapter.
Many questions mentioned in Class 12 Maths chapter 1 miscellaneous solutions are useful for the CBSE class 12 board exam and also for exams like JEE main, VITEEE, etc.
Revision can also be facilitated by using NCERT solutions for Class 12 Maths chapter 10 miscellaneous exercises
This exercise can also be used in similar physics topics also.

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Key Features Of NCERT Solutions For Class 12 Chapter 1 Miscellaneous Exercise

Comprehensive Coverage: The solutions encompass all the topics covered in miscellaneous exercise class 12 chapter 1, ensuring a thorough understanding of the concepts.
Step-by-Step Solutions: In this class 12 chapter 1 maths miscellaneous solutions, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
Accuracy and Clarity: Solutions for class 12 maths miscellaneous exercise chapter 1 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
Conceptual Clarity: In this class 12 maths ch 1 miscellaneous exercise solutions, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
Inclusive Approach: Solutions for class 12 chapter 1 miscellaneous exercise cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
Relevance to Curriculum: The solutions for miscellaneous exercise class 12 chapter 1 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

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NCERT Solutions Subject Wise

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Frequently Asked Questions (FAQs)

1. Which concepts are covered in class 12 maths chapter 1 miscellaneous exercises

This exercise discusses concepts related to inverse, injective, onto functions.

2. What are the important topics in chapter relations and functions ?

Definitions of relations and functions, types of relations, types of functions, composition of functions, invertible function and binary operations are the important topics in this chapter.

3. What is the weightage held by the chapter relations and functions for CBSE board exam ?

The weightage of chapter relation and function is more than 5% in the CBSE board examination.

4. How are the NCERT solutions helpful in the board exam ?

As students can assess from previous year questions that many questions are asked directly from NCERT exercise. Hence to score well in the examination, it is required to practice NCERT exercise for class 12 maths before the exam.

5. What are relations in Class 12 Maths?

In maths, relation defines the relationship between sets of values of ordered pairs

6. Mention the number of questions in class 12 maths chapter 1 miscellaneous exercise?

Total 19 questions are the in miscellaneous exercise Class 12 Maths chapter 1

7. How many exercises are there in the NCERT class 12 maths chapter 1 ?

There are a total of 5 exercises including a miscellaneous exercise in the NCERT class 12 maths chapter 1.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer

Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

Read Complete Answer

Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer

Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer

Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer

kumardurgesh1802 10 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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

NCERT Solutions for Miscellaneous Exercise Chapter 1 Class 12 - Relations and Functions

NCERT Solutions For Class 12 Chapter 1 Miscellaneous Exercise

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Access NCERT Solutions for Miscellaneous Exercise Chapter 1 Class 12

Question:1 Let be defined as . Find the function such that .

Key Features Of NCERT Solutions For Class 12 Chapter 1 Miscellaneous Exercise

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Question:1 Let $f : R \rightarrow R$ be defined as $f (x) = 10x + 7$ . Find the function $g : R \rightarrow R$ such that $g o f = f o g = I_R$ .