Maths NCERT Exemplar Solutions Class 11th Chapter Two

The word is 'LETTER'.
Consonants are L, T, R.
Vowels are E, E.
Total number of words (with or without meaning) from the letters of the word 'LETTER' is:
6! / (2! 2!) = 720 / 4 = 180.
Total number of words (with or without meaning) from the letters of the word 'LETTER' if vowels are together:
Treat (EE) as a single unit. We now arrange {L, T, R, (EE)}. This is 5 units.
Number of arrangements = 5! / 2! (for the two T's) = 120 / 2 = 60.
∴ The number of words where vowels are not together = Total words - Words with vowels together
Required = 180 - 60 = 120.

|x + y|² = |x|²
(x+y)· (x+y) = x·x
|x|² + 2x·y + |y|² = |x|²
|y|² + 2x·y = 0 (1)
and (2x + λy)·y = 0
2x·y + λ|y|² = 0 (2)
From (1), 2x·y = -|y|².
Substitute into (2):
-|y|² + λ|y|² = 0
(λ-1)|y|² = 0
Assuming y is a non-zero vector, |y|² ≠ 0, therefore λ=1.

|λ-1 3λ+1 2λ|
|λ-1 4λ-2 λ+3| = 0
|2 3λ+1 3 (λ-1)|
R? → R? - R? and R? → R? - R? (from a similar matrix setup, applying operations to simplify)
The provided solution uses a slightly different matrix but let's follow the subsequent steps.
A different matrix from the image is used in the calculation:
|λ-1 3λ+1 2λ|
|0 λ-3 -λ+3|
|3-λ 0 λ-3 |
C? → C? + C?
|3λ-1 3λ+1 2λ |
|3-λ λ-3-λ | = 0
|0 λ-3 |
⇒ (λ-3) [ (3λ-1) (λ-3) - (3λ+1) (3-λ)] = 0
⇒ (λ-3) [ (λ-3) (3λ-1) + (λ-3) (3λ+1)] = 0
⇒ (λ-3)² [3λ-1 + 3λ+1] = 0
⇒ (λ-3)² [6λ] = 0 ⇒ λ = 0, 3
Sum of values of λ = 3

Given f (1) = a = 3, and assuming the function form is f (x) = a?
So f (x) = 3?
∑? f (i) = 363
⇒ 3 + 3² + . + 3? = 363
This is a geometric progression. The sum is S? = a (r? -1)/ (r-1).
3 (3? -1)/ (3-1) = 363
3 (3? -1)/2 = 363
3? - 1 = 242
3? = 243
3? = 3? ⇒ n = 5

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:A=\left\{a,b\right\},B=\left\{x,y\right\}\\ \therefore A*B=\left\{\left(a,x\right),\left(a,y\right),\left(b,x\right),\left(b,y\right)\right\}\\ Hence,thestatementis\text{'}True\text{'}.\end{array}$

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:\left(x-2,y+5\right)=\left(-2,\frac{1}{3}\right)\\ \Rightarrow x-2=-2\Rightarrow x=0\\ andy+5=\frac{1}{3}\Rightarrow y=\frac{1}{3}-5=-\frac{14}{3}\\ Hence,thestatementis\text{'}False\text{'}.\end{array}$

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:A=\left\{1,2,3\right\},B=\left\{3,4\right\}andC=\left\{4,5,6\right\}\\ \therefore A*B=\left\{\left(1,3\right),\left(1,4\right),\left(2,3\right),\left(2,4\right),\left(3,3\right),\left(3,4\right)\right\}\\ andA*C=\left\{\left(1,4\right),\left(1,5\right),\left(1,6\right),\left(2,4\right),\left(2,5\right),\left(2,6\right),\left(3,4\right),\left(3,5\right),\left(3,6\right)\right\}\\ \left(A*B\right)\cup \left(A*C\right)=\left\{\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right)\right\}\\ Hence,thestatementis\text{'}True\text{'}.\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:f\left(x\right)=3x^2-1andg\left(x\right)=3+x\\ f\left(x\right)=g\left(x\right)\\ \Rightarrow 3x^2-1=3+x\\ \Rightarrow 3x^2-x-4=0\Rightarrow 3x^2-4x+3x-4=0\\ \Rightarrow x\left(3x-4\right)+1\left(3x-4\right)=0\Rightarrow \left(x+1\right)\left(3x-4\right)=0\\ \Rightarrow x+1=0or3x-4=0\\ \Rightarrow x=-1orx=\frac{4}{3}\\ \therefore Domain=\left\{-1,\frac{4}{3}\right\}\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:f\left(x\right)=2-\left|x-5\right|\\ f\left(x\right)isdefinedforx\in R\\ \therefore Domainoff\left(x\right)=R\\ Now,\left|x-5\right|\ge 0\Rightarrow -\left|x-5\right|\le 0\\ \Rightarrow 2-\left|x-5\right|\le 2\Rightarrow f\left(x\right)\le 2\\ \therefore Rangeoff\left(x\right)=\left(-\infty ,2\right]\\ Hence,thecorrectoptionis\left(b\right).\end{array}$