NCERT Solutions for Exercise 1.3 Class 12 Maths Chapter 1

NCERT Solutions for Exercise 1.3 Class 12 Maths Chapter 1 - Relations and Functions

Edited By Ramraj Saini | Updated on Dec 03, 2023 01:28 PM IST | #CBSE Class 12th

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NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.3

NCERT Solutions for Exercise 1.3 Class 12 Maths Chapter 1 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 exercise 1.3 introduces a few more concepts of relations and functions which includes inverse function, imposition of one function on other eg. fog(x) or gof(x) etc. Exercise 1.3 Class 12 Maths will help students to grasp the basic concepts related to finding the inverse of a function. Practicing this exercise is of utmost importance because most of the questions related to finding the inverse of a function are asked hence students can go through NCERT Solutions for Class 12 Maths chapter 1 exercise 1.3 to score well in CBSE class 12 board exam. In competitive exams also like JEE main ,some questions can be asked from class 12 ex 1.3. Concepts related to inverse functions discussed in Class 12th Maths chapter 1 exercise 1.3 can be useful for NEET physics syllabus, as the problems in physics use the concepts of functions.

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NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions: Exercise 1.3

Question:1 Let $f : \{1, 3, 4\}\rightarrow \{1, 2, 5\}$ and $g : \{1, 2, 5\} \rightarrow \{1, 3\}$ be given by $f = \{(1, 2), (3, 5), (4, 1)\}$ and $g = \{(1, 3), (2, 3), (5, 1)\}$ . Write down $gof$ .

Answer:

Given : $f : \{1, 3, 4\}\rightarrow \{1, 2, 5\}$ and $g : \{1, 2, 5\} \rightarrow \{1, 3\}$

$f = \{(1, 2), (3, 5), (4, 1)\}$ and $g = \{(1, 3), (2, 3), (5, 1)\}$

$gof(1) = g(f(1))=g(2) = 3$ $\left [ f(1)=2 \, and\, g(2)=3 \right ]$

$gof(3) = g(f(3))=g(5) = 1$ $\left [ f(3)=5 \, and\, g(5)=1 \right ]$

$gof(4) = g(f(4))=g(1) = 3$ $\left [ f(4)=1 \, and\, g(1)=3 \right ]$

Hence, $gof$ = $\left \{ (1,3),(3,1),(4,3) \right \}$

Question:2 Let $f$ , $g$ and $h$ be functions from $R$ to $R$ . Show that $\\(f + g) o h = foh + goh\\ (f \cdot g) o h = (foh) \cdot (goh)$

Answer:

To prove : $\\(f + g) o h = foh + goh$

$((f + g) o h)(x)$

$=(f + g) ( h(x) )$

$=f ( h(x) ) +g(h(x))$

$=(f o h)(x) +(goh)(x)$

$=\left \{ (f o h) +(goh) \right \}(x)$ $x\forall R$

Hence, $\\(f + g) o h = foh + goh$

To prove: $(f \cdot g) o h = (foh) \cdot (goh)$

$((f . g) o h)(x)$

$=(f . g) ( h(x) )$

$=f ( h(x) ) . g(h(x))$

$=(f o h)(x) . (goh)(x)$

$=\left \{ (f o h) .(goh) \right \}(x)$ $x\forall R$

$\therefore$ $(f \cdot g) o h = (foh) \cdot (goh)$

Hence, $(f \cdot g) o h = (foh) \cdot (goh)$

Question:3(i) Find $gof$ and $fog$ , if

(i) $f (x) = | x |$ and $g(x) = \left | 5x-2 \right |$

Answer:

$f (x) = | x |$ and $g(x) = \left | 5x-2 \right |$

$gof$ $= g(f(x))$

$= g( | x |)$

$= |5 | x |-2|$

$fog$ $= f(g(x))$

$=f( \left | 5x-2 \right |)$

$=\left \| 5x-2 \right \|$

$=\left | 5x-2 \right |$

Question:3(ii) Find gof and fog, if

(ii) $f (x) = 8x^{3}$ and $g(x) = x^{\frac{1}{3}}$

Answer:

The solution is as follows

(ii) $f (x) = 8x^{3}$ and $g(x) = x^{\frac{1}{3}}$

$gof$ $= g(f(x))$

$= g( 8x^{3})$

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

$=2x$

$fog$ $= f(g(x))$

$=f(x^{\frac{1}{3}} )$

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

$=8x$

Question:4 If $f(x) = \frac{4x + 3}{6x - 4}, x \neq \frac{2}{3}$ show that $fof (x) = x$ , for all $x \neq\frac{2}{3}$ . What is the inverse of $f$ ?

Answer:

$f(x) = \frac{4x + 3}{6x - 4}, x \neq \frac{2}{3}$

$fof (x) = x$

$(fof) (x) = f(f(x))$

$=f( \frac{4x + 3}{6x - 4})$

$=\frac{4( \frac{4x + 3}{6x - 4}) +3}{6( \frac{4x + 3}{6x - 4}) -4}$

$= \frac{16x+12+18x-12}{24x+1824x+16}$

$= \frac{34x}{34}$

$\therefore fof(x) = x$ , for all $x \neq \frac{2}{3}$

$\Rightarrow fof=Ix$

Hence,the given function $f$ is invertible and the inverse of $f$ is $f$ itself.

Question:5(i) State with reason whether following functions have inverse

(i) $f : \{1, 2, 3, 4\} \rightarrow\{10\}$

with $f = \{(1, 10), (2, 10), (3, 10), (4, 10)\}$

Answer:

(i) $f : \{1, 2, 3, 4\} \rightarrow\{10\}$ with
$f = \{(1, 10), (2, 10), (3, 10), (4, 10)\}$

From the given definition,we have:

$f\left ( 1 \right )=f\left ( 2 \right )=f\left ( 3 \right )=f(4)=10$

$\therefore$ f is not one-one.

Hence, f do not have an inverse function.

Question:5(ii) State with reason whether following functions have inverse

(ii) $g : \{5, 6, 7, 8\} \rightarrow \{1, 2, 3, 4\}$ with
$g = \{(5, 4), (6, 3), (7, 4), (8, 2)\}$

Answer:

(ii) $g : \{5, 6, 7, 8\} \rightarrow \{1, 2, 3, 4\}$ with
$g = \{(5, 4), (6, 3), (7, 4), (8, 2)\}$

From the definition, we can conclude :

$g(5)=g(7)=4$

$\therefore$ g is not one-one.

Hence, function g does not have inverse function.

Question:5(iii) State with reason whether following functions have inverse

(iii) $h : \{2, 3, 4, 5\}\rightarrow \{7, 9, 11, 13\}$ with
$h = \{(2, 7), (3, 9), (4, 11), (5, 13)\}$

Answer:

(iii) $h : \{2, 3, 4, 5\}\rightarrow \{7, 9, 11, 13\}$ with
$h = \{(2, 7), (3, 9), (4, 11), (5, 13)\}$

From the definition, we can see the set $\left \{ 2,3,4,5 \right \}$ have distant values under h.

$\therefore$ h is one-one.

For every element y of set $\left \{ 7,9,11,13 \right \}$ ,there exists an element x in $\left \{ 2,3,4,5 \right \}$ such that $h(x)=y$

$\therefore$ h is onto

Thus, h is one-one and onto so h has an inverse function.

Question:6 Show that $f : [-1, 1] \rightarrow R$ , given by $f(x) = \frac{x}{(x + 2)}$ is one-one. Find the inverse of the function $f : [-1, 1] \rightarrow Range f$

Answer:

$f : [-1, 1] \rightarrow R$

$f(x) = \frac{x}{(x + 2)}$

One -one:

$f(x)=f(y)$

$\frac{x}{x+2}=\frac{y}{y+2}$

$x(y+2)=y(x+2)$

$xy+2x=xy+2y$

$2x=2y$

$x=y$

$\therefore$ f is one-one.

It is clear that $f : [-1, 1] \rightarrow Range f$ is onto.

Thus,f is one-one and onto so inverse of f exists.

Let g be inverse function of f in $Range f\rightarrow [-1, 1]$

$g: Range f\rightarrow [-1, 1]$

let y be an arbitrary element of range f

Since, $f : [-1, 1] \rightarrow R$ is onto, so

$y=f(x)$ for $x \in \left [ -1,1 \right ]$

$y=\frac{x}{x+2}$

$xy+2y=x$

$2y=x-xy$

$2y=x(1-y)$

$x = \frac{2y}{1-y}$ , $y\neq 1$

$g(y) = \frac{2y}{1-y}$

$f^{-1}=\frac{2y}{1-y},y\neq 1$

Question:7 Consider $f : R \rightarrow R$ given by $f (x) = 4x + 3$ . Show that f is invertible. Find the inverse of $f$ .

Answer:

$f : R \rightarrow R$ is given by $f (x) = 4x + 3$

One-one :

Let $f(x)=f(y)$

$4x + 3 = 4y+3$

$4x=4y$

$x=y$

$\therefore$ f is one-one function.

Onto:

$y=4x+3\, \, \, , y \in R$

$\Rightarrow x=\frac{y-3}{4} \in R$

So, for $y \in R$ there is $x=\frac{y-3}{4} \in R$ ,such that

$f(x)=f(\frac{y-3}{4})=4(\frac{y-3}{4})+3$

$= y-3+3$

$= y$

$\therefore$ f is onto.

Thus, f is one-one and onto so $f^{-1}$ exists.

Let, $g:R\rightarrow R$ by $g(x)=\frac{y-3}{4}$

Now,

$(gof)(x)= g(f(x))= g(4x+3)$

$=\frac{(4x+3)-3}{4}$

$=\frac{4x}{4}$

$=x$

$(fog)(x)= f(g(x))= f(\frac{y-3}{4})$

$= 4\times \frac{y-3}{4}+3$

$= y-3+3$

$= y$

$(gof)(x)= x$ and $(fog)(x)= y$

Hence, function f is invertible and inverse of f is $g(y)=\frac{y-3}{4}$ .

Question:8 Consider f : R+ → [4, ∞) given by $f(x) = x^2+4$ . Show that f is invertible with the inverse $f^{-1}$ of f given by $f^{-1}(y)= \sqrt{y-4}$ , where R+ is the set of all non-negative real numbers.

Answer:

It is given that
$f : R^+ \rightarrow [4,\infty)$ , $f(x) = x^2+4$ and

Now, Let f(x) = f(y)

⇒ x ² + 4 = y ² + 4

⇒ x ² = y ²

⇒ x = y

⇒ f is one-one function.

Now, for y $\epsilon$ [4, ∞), let y = x ² + 4.

⇒ x ² = y -4 ≥ 0

⇒ for any y $\epsilon$ R, there exists x = $\epsilon$ R such that

= y -4 + 4 = y.

⇒ f is onto function.

Therefore, f is one–one and onto function, so f-1 exists.

Now, let us define g: [4, ∞) → R+ by,

g(y) =

Now, gof(x) = g(f(x)) = g(x ² + 4) =

And, fog(y) = f(g(y)) = =

Therefore, gof = gof = I _R .

Therefore, f is invertible and the inverse of f is given by

f-1(y) = g(y) =

Question:9 Consider $f : R_+ \rightarrow [- 5, \infty)$ given by $f (x) = 9x^2 + 6x - 5$ . Show that $f$ is invertible with $f^{-1} (y) = \left (\frac{(\sqrt{y + 6}) - 1}{3} \right )$

Answer:

$f : R_+ \rightarrow [- 5, \infty)$

$f (x) = 9x^2 + 6x - 5$

One- one:

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

$9x^{2}+6x-5=9y^{2}+6y-5$

$9x^{2}+6x=9y^{2}+6y$

$\Rightarrow$ $9(x^{2}-y^{2})=6(y-x)$

$9(x+y)(x-y)+6(x-y)= 0$

$(x-y)(9(x+y)+6)=0$

Since, x and y are positive.

$(9(x+y)+6)> 0$

$\therefore x=y$

$\therefore$ f is one-one.

Onto:

Let for $y \in [-5,\infty)$ , $y=9x^{2}+6x-5$

$\Rightarrow$ $y=(3x+1)^{2}-1-5$

$\Rightarrow$ $y=(3x+1)^{2}-6$

$\Rightarrow$ $y+6=(3x+1)^{2}$

$(3x+1)=\sqrt{y+6}$

$x = \frac{\sqrt{y+6}-1}{3}$

$\therefore$ f is onto and range is $y \in [-5,\infty)$ .

Since f is one-one and onto so it is invertible.

Let $g : [-5,\infty)\rightarrow R_+$ by $g(y) = \frac{\sqrt{y+6}-1}{3}$

$(gof)(x)=g(f(x))=g(9x^{2}+6x)-5=g((3x+1)^{2}-6)\\=\sqrt{ { (3x+1)^{2} }-6+6} -1$

$(gof)(x)=\frac{3x+1-1}{3}=\frac{3x}{3}= x$

$(fog)(x)=f(g(x))=f(\frac{\sqrt{y+6}-1}{3})$

$=[3(\frac{\sqrt{y+6}-1}{3})+1]^{2}-6$

$=(\sqrt{y+6})^{2}-6$

$=y+6-6$

$=y$

$\therefore gof=fog=I_R$

Hence, $f$ is invertible with the inverse $f^{-1}$ of $f$ given by $f^{-1} (y) = \left (\frac{(\sqrt{y + 6}) - 1}{3} \right )$

Question:10 Let $f : X \rightarrow Y$ be an invertible function. Show that f has a unique inverse. (Hint: suppose $g_1$ and $g_2$ are two inverses of $f$ . Then for all $y \in Y$ ,
$fog_1 (y) = I_Y (y) = fog_2 (y)$ . Use one-one ness of f).

Answer:

Let $f : X \rightarrow Y$ be an invertible function

Also, suppose f has two inverse $g_1 and g_2$

For $y \in Y$ , we have

$fog_1(y) = I_y(y)=fog_2(y)$

$\Rightarrow$ $f(g_1(y))=f(g_2(y))$

$\Rightarrow$ $g_1(y)=g_2(y)$ [f is invertible implies f is one - one]

$\Rightarrow$ $g_1=g_2$ [g is one-one]

Thus,f has a unique inverse.

Question:11 Consider $f : \{1, 2, 3\} \rightarrow \{a, b, c\}$ given by $f (1) = a$ , $f (2) = b$ and $f (3) = c$ . Find $f^{-1}$ and show that $(f^{-1})^{-1} = f$ .

Answer:

It is given that
$f : \left \{ 1,2,3 \right \}\rightarrow \left \{ a,b,c \right \}$

$f(1) = a, f(2) = b \ and \ f(3) = c$

Now,, lets define a function g :
$\left \{ a,b,c \right \}\rightarrow \left \{ 1,2,3 \right \}$ such that

$g(a) = 1, g(b) = 2 \ and \ g(c) = 3$
Now,

$(fog)(a) = f(g(a)) = f(1) = a$
Similarly,

$(fog)(b) = f(g(b)) = f(2) = b$

$(fog)(c) = f(g(c)) = f(3) = c$

And

$(gof)(1) = g(f(1)) = g(a) = 1$

$(gof)(2) = g(f(2)) = g(b) = 2$

$(gof)(3) = g(f(3)) = g(c) = 3$

Hence, $gof = I_X$ and $fog = I_Y$ , where $X = \left \{ 1,2,3 \right \}$ and $Y = \left \{ a,b,c \right \}$

Therefore, the inverse of f exists and $f^{-1} = g$

Now,
$f^{-1} : \left \{ a,b,c \right \}\rightarrow \left \{ 1,2,3 \right \}$ is given by

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

Now, we need to find the inverse of $f^{-1}$ ,

Therefore, lets define $h: \left \{ 1,2,3 \right \}\rightarrow \left \{ a,b,c \right \}$ such that

$h(1) = a, h(2) = b \ and \ h(3) = c$

Now,

$(goh)(1) = g(h(1)) = g(a) = 1$

$(goh)(2) = g(h(2)) = g(b) = 2$

$(goh)(3) = g(h(3)) = g(c) = 3$

Similarly,

$(hog)(a) = h(g(a)) = h(1) = a$

$(hog)(b) = h(g(b)) = h(2) = b$

$(hog)(c) = h(g(c)) = h(3) = c$

Hence, $goh = I_X$ and $hog = I_Y$ , where $X = \left \{ 1,2,3 \right \}$ and $Y = \left \{ a,b,c \right \}$

Therefore, inverse of $g^{-1} = (f^{-1})^{-1}$ exists and $g^{-1} = (f^{-1})^{-1} = h$

$\Rightarrow h = f$

Therefore, $(f^{-1})^{-1} = f$

Hence proved

Question:12 Let $f : X \rightarrow Y$ be an invertible function. Show that the inverse of $f^{-1}$ is $f$ , i.e., $(f^{-1})^{-1} = f$

Answer:

$f : X \rightarrow Y$

To prove: $(f^{-1})^{-1} = f$

Let $f:X\rightarrow Y$ be a invertible function.

Then there is $g:Y\rightarrow X$ such that $gof =I_x$ and $fog=I_y$

Also, $f^{-1}= g$

$gof =I_x$ and $fog=I_y$

$\Rightarrow$ $f^{-1}of = I_x$ and $fof^{-1} = I_y$

Hence, $f^{-1}:Y\rightarrow X$ is invertible function and f is inverse of $f^{-1}$ .

i.e. $(f^{-1})^{-1} = f$

Question:13 If $f : R \rightarrow R$ be given by $f(x) = (3 - x^3)^{\frac{1}{3}}$ , then $fof(x)$ is

(A) $x^{\frac{1}{3}}$

(B) $x^3$

(D) $(3 - x^3)$

Answer:

$f(x) = (3 - x^3)^{\frac{1}{3}}$

$fof(x)$ $=f(f(x))=f((3-x^{3})^{\frac{1}{3}})$

$=[3- ((3-x^{3})^{\frac{1}{3}})^{3}]^{\frac{1}{3}}$

$=([3- ((3-x^{3})]^{\frac{1}{3}})$

$= (x^{3})^{\frac{1}{3}}$

$=x$

Thus, $fof(x)$ is x.

Hence, option c is correct answer.

Question:14 Let $f: R - \left\{-\frac{4}{3}\right\} \rightarrow R$ be a function defined as $f(x) = \frac{4x}{3x + 4}$ . The inverse of $f$ is the map $g : Range\;f \rightarrow R - \left \{-\frac{4}{3} \right \}$ given by

(A) $g(y) = \frac{3y}{3 -4y}$

(B) $g(y) = \frac{4y}{4 -3y}$

(D) $g(y) = \frac{3y}{3 -4y}$

Answer:

$f: R - \left\{-\frac{4}{3}\right\} \rightarrow R$

$f(x) = \frac{4x}{3x + 4}$

Let f inverse $g : Range\;f \rightarrow R - \left \{-\frac{4}{3} \right \}$

Let y be the element of range f.

Then there is $x \in R - \left\{-\frac{4}{3}\right\}$ such that

$y=f(x)$

$y=\frac{4x}{3x+4}$

$y(3x+4)=4x$

$3xy+4y=4x$

$3xy-4x+4y=0$

$x(3y-4)+4y=0$

$x= \frac{-4y}{3y-4}$

$x= \frac{4y}{4-3y}$

Now , define $g : Range\;f \rightarrow R - \left \{-\frac{4}{3} \right \}$ as $g(y)= \frac{4y}{4-3y}$

$gof(x)= g(f(x))= g(\frac{4x}{3x+4})$

$= \frac{4(\frac{4x}{3x+4})}{4-3(\frac{4x}{3x+4})}$

$=\frac{16x}{12x+16-12x}$

$=\frac{16x}{16}$

$=x$

$fog(y)=f(g(y))=f(\frac{4y}{4-3y})$

$= \frac{4(\frac{4y}{4-3y})}{3(\frac{4y}{4-3y}) + 4}$

$=\frac{16y}{12y+16-12y}=\frac{16y}{16}$

$=y$

Hence, g is inverse of f and $f^{-1}=g$

The inverse of f is given by $g(y)= \frac{4y}{4-3y}$ .

The correct option is B.

Benefits of NCERT Solutions for class 12 maths chapter 1 exercise 1.3

The Class 12th maths chapter 1 exercise is provided above in detail which is solved by subject matter experts according to the NCERT syllabus .
Students are recommended to practice Exercise 1.3 Class 12 Maths to prepare for topics like inverse functions etc., direct questions are asked in Board exams.
These Class 12 Maths NCERT book chapter 1 exercise 1.3 solutions can be referred by students to revise just before the exam.
NCERT Solutions for Class 12 Maths chapter 1 exercise 1.3 can be used to prepare inverse function topics of physics also.

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Key Features Of NCERT Solutions for Exercise 1.3 Class 12 Maths Chapter 1

Comprehensive Coverage: The solutions encompass all the topics covered in ex 1.3 class 12, ensuring a thorough understanding of the concepts.
Step-by-Step Solutions: In this class 12 maths ex 1.3, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
Accuracy and Clarity: Solutions for class 12 ex 1.3 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
Conceptual Clarity: In this 12th class maths exercise 1.3 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
Inclusive Approach: Solutions for ex 1.3 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
Relevance to Curriculum: The solutions for class 12 maths ex 1.3 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

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Frequently Asked Questions (FAQs)

1. Which concepts are covered in Exercise 1.3 Class 12 Maths?

Concepts related to inverse function, imposition of one function on other eg. fog(x) or gof(x) etc., are discussed in the Exercise 1.3 Class 12 Maths

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 of the chapter relations and functions for CBSE board exam ?

Two chapters 'relation and function' and 'inverse trigonometry' combined has 10 % weightage in the CBSE final board exam.

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

As CBSE board exam paper is designed entirely based on NCERT textbooks and most of the questions in CBSE board exam are directly asked from NCERT textbook, students must know the NCERT very well to perform well in the exam. NCERT solutions are not only important when you are stuck while solving the problems but students will get how to answer in the board exam in order to get good marks in the board exam.

5. What are relations in Class 12 Maths?

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

6. What are some types of relations discussed in Exercise 1.3 Class 12 Maths

Questions related to inverse function, imposition of one function on other eg. fog(x) or gof(x) etc are discussed in the Exercise 1.3 Class 12 Maths

7. How many questions are covered in Exercise 1.3 Class 12 Maths ?

There are 14 questions in Exercise 1.3 Class 12 Maths NCERT syllabus.

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

There are 5 exercises including a miscellaneous exercise in the NCERT book 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!

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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 Exercise 1.3 Class 12 Maths Chapter 1 - Relations and Functions

NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.3

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Assess NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.3

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions: Exercise 1.3

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