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

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.

NCERT solutions for class 12 maths chapter 1 exercise 1.1

NCERT solutions for class 12 maths chapter 1 exercise 1.2

NCERT solutions for class 12 maths chapter 1 exercise 1.3

NCERT solutions for class 12 maths chapter 1 exercise 1.4

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 $g o f = f o g = I_R$ .

$f : R \rightarrow R$

$f (x) = 10x + 7$

$g : R \rightarrow R$ 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.

$f : W \rightarrow W$

$f (n) = n -1$ 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)).

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.

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.

$f : R \rightarrow R$

$f (x) = x ^3$

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 $g (x) = | x |$ ).

$f : N \rightarrow Z$

$g : Z \rightarrow Z$

$f (x) = x$

$g (x) = | x |$

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

$f : N\rightarrow N$ 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.

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 ∗.

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.

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

$F:S\rightarrow T$

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

$F:S\rightarrow T$

$F : \{a, b, c\}\rightarrow \left \{ 1,2,3 \right \}$ 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.

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 :

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

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

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).

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

$%u200B%u200B%u200B%u200B%u200B%u200B%u200BA * A = (A -A) \cup (A - A) = \phi \cup \phi = \phi , A \ \epsilon \ P(X).$ $A * A = \left ( A - A \right ) \cup \left ( A-A \right ) = \phi \cup \phi = \phi ,$ 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 $a$ .

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).

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

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

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

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

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 $(0, 1]$ ?

$f : R \rightarrow R$ 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 $[x]=0\, \, \, \, \, if\, 0< x< 1$

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

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

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.

### Also see-

• NCERT exemplar solutions class 12 maths chapter 1

• NCERT solutions for class 12 maths chapter 1

NCERT Solutions Subject Wise

Can You Predict Whether A Cricket Ball Will Hit The Stumps With Class 11 Maths? Learn How 3 min read Mar 05, 2022 Read More How To Ace Class 8 NSTSE: 5-Year Analysis, Pattern, Syllabus, Topics And Weightage 8 min read Apr 05, 2022 Read More

• NCERT solutions class 12 chemistry

• NCERT solutions for class 12 physics

• NCERT solutions for class 12 biology

• NCERT solutions for class 12 mathematics

## Subject wise NCERT Exemplar solutions

• NCERT Exemplar Class 12th Maths

• NCERT Exemplar Class 12th Physics

• NCERT Exemplar Class 12th Chemistry

• NCERT Exemplar Class 12th Biology

Happy learning!!!