NDA Mathematics 14 Nov 2021 Official Paper (Previous Year Paper)

Given:

x2 + x + 1 = 0    ----(i)

Calculation:

We have x2 + x + 1 = 0

We have to find the value of x199 + x200 + x201

⇒ x199  (1 + x + x2) 

From (i), we get

⇒ x199 × 0

⇒ 0

∴ The value of x199 + x200 + x201  is 0.

Concept:

If a, b and c are three terms in GP then b² = ac and vice-versa

If three terms a, b, c are in HP, then b = $\rm \frac{2ac}{a + c}$ and vice-versa

When three terms are in AP, the middle one is called the arithmetic mean of the other two.

If a, b and c are three terms in AP then b = $\rm \frac{a + c}{2}$ and vice-versa

Calculation:

Given x, y, z are in G.P

y² = xz      ....(1)

On multiply both sides by 9, we get

⇒ (3y)2 = 9xz

Taking log on both sides, we get

⇒ 2In 3y = In 3x + In 3z

⇒ ln(3x), ln(3y), ln(3z) are in A.P

So, ln(3x), ln(3y), ln(3z) are in A.P

Statement 1 is correct.

Again taking eq(1)

y2 = xz 

Taking log on both sides, we get

⇒ 2In y = In x + In z

On adding 2xyz on both the sides, we get

⇒ 2(xyz + In(y)) = (xyz + In(x)) + (xyz + In(z))

⇒ xyz + ln(x), xyz + ln(y), xyz + ln(z) are in A.P

$\rm \frac{1}{xyz + In(x)}, \frac{1}{xyz + In(y)}, \frac{1}{xyz + In(z)}$ are in H.P

It means  xyz + ln(x), xyz + ln(y), xyz + ln(z) are not in HP.

Statement 2 is not correct.

Formula used:

• If a, b and c are three terms in AP then b = $\rm \frac{a + c}{2}$
• log_amⁿ = n log_am
• log_a [mn] = log_a m + log_a n
• If logam = logan then m = n

Calculation:

Given, log102,  log10(2x - 1), log10(2x + 3) are in AP

According to the question

2 log10(2x - 1) =  log102 + log10(2x + 3)

⇒ log10(2x - 1)² =  log10 2(2x + 3)

⇒ log₁₀​(2^2x + 1 − 2^x+1) = log₁₀​(2x+1 + 6) 

From the formula used

⇒ 22x + 1− 2x+1 = 2x+1 + 6

⇒ 22x + 1− 2 × 2x+1 = 6

⇒ 2^2x − 2^x+2 − 5 = 0 

Take, 2^x = y > 0 

y² − 4y − 5 = 0 

⇒ (y − 5)(y + 1) = 0 

⇒ y = 5,−1 

∵ y > 0

⇒ y = 5 

⇒ 2^x = 5 

⇒ x = log₂​5

∴ The value of x is log2​5.

Explanation:

We can solve this by filling the places according to the question 

Step 1:

Since the Number should be less than 500 

So, there are only 3 possibilities at 1 st place i.e. (2, 3, 4) 

Step 2:

According to the question

Repetition is not allowed we have to fix one number (let 2) in the 1^st place 

Now, At 2^nd place, there are a total of 6 ways (3, 4, 5, 6, 7, 9)

Step 3:

According to the question

Repetition is not allowed we have to fix one number (let 3) in the 2^nd place 

Now, At 3^rd place, there are a total of 5 ways (4, 5, 6, 7, 9)

∴ The total number of different 3-digit numbers less than 500 = 3 × 6 × 5 = 90.

Formula used:

sum of first n terms of GP formula

S_n = $\rm \frac{a(r^{n} - 1)}{r - 1}$, $\rm r\neq 1$ and r > 1

Calculation:

11 = 1 + 10¹

111 = 1 + 10 + 10²

1111 = 1 + 10 + 10² + 10³

(1111 ..... n-digit) = 1 + 10 + 10² + 10³ + ...... + 10^n-1

 1 + 10 + 102 + 103 + ...... + 10n-1  are in G.P

let P = 1 + 10 + 102 + 103 + ...... + 10n-1

⇒ p = $\rm \frac{a(r^{n} -1)}{r - 1} = \frac{10^{n} -1}{9}$

Now, we have to find the value of (9p2 + p)

⇒ 9 × $\rm \frac{(10^{n} -1)^{2}}{81} + \frac{10^{n} -1}{9}$

⇒$\rm \frac{(10^{n} -1)^{2}}{9} + \frac{10^{n} -1}{9}$

⇒ $\rm \frac{(10^{2n} \space+\space 1 -\space 2\space ×\space 10^{n} \space+\space 10^{n}\space-1\space)}{9}$

⇒ $\rm \frac{(10^{2n} - 10^{n})}{9}$

⇒ $\rm \frac{10^{n} (10^{n}- 1)}{9}$

⇒ 10ⁿ p

∴ The value of 9p2 + p is 10n p.

Concept:

Quadratic equation:

ax² + bx + c = 0

roots of equal magnitude and opposite sign then

b = 0, a ≠ 0, c ≠ 0

Calculation:

The given equation 3x2 - (k2 + 5k)x + 3k2 - 5k = 0

Here, a = 3, b = -(k2 + 5k) and c = 3k2 - 5k

According to the concept used

Case I. b = 0

⇒ (k² + 5k) = 0

⇒ k(k + 5) = 0

⇒ k = 0, -5

Case II. c ≠ 0

3k2 - 5k ≠ 0

⇒ k(3k - 5) ≠ 0

⇒ k ≠ 0, k ≠ 5/3.

From Case I and II, we get k = -5

But the question says roots are real

So, k = -5 is not possible.  [at k = -5, D < 0]

∴ The option 4^th is the correct option.

Formula used:

n! = n × (n - 1) × (n - 2) × ....... × 3 × 2 × 1

Calculation:

a_n = n (n)!

a_n = (n + 1 - n)n!

a_n = (n + 1)n! - n!

a_n = (n + 1)! - n!

Hence, 

a₁ = 2! - 1! 

a₂ = 3! - 2! 

a₁₀ = 11! - 10!

Now,

 a1 + a2 + a3 +...+ a10 

2! - 1! + 3! - 2! + ......11! - 10! 

= 11! - 1

∴ The value of a1 + a2 + a3 +...+ a10 is 11! - 1.

Formula used:

For any quadratic equation ax² + bx + c = 0,

sum of the roots = $\rm \frac{-b}{a}$

product of the roots = $\rm \frac{c}{a}$

Calculation:

Given equation 

x² + px + q = 0

Also, given p, q are the roots of the equation.

Sum of roots = -p/1

⇒ p + q = −p  

⇒ 2p + q  = 0   .....(1)

And product of roots = q/1

⇒ pq = q

⇒ p = 1

putting the value of p in eq(1), we get

2(1) + q = 0

⇒ q = -2

∴ q has only one value.

Concept::

• The interchange of any two rows (or columns) of the determinant changes its sign.
• The determinant remains unaltered if its rows are changed into columns and the columns into rows.

Calculation:

$\Delta = \begin{vmatrix} a & b & c\ d & e & f\ g & h & i \end{vmatrix}$

$A = \begin{vmatrix} 3d + 5g & 4a + 7g & 6g\ 3e + 5h & 4b + 7h & 6h\ 3f + 5i & 4c + 7i & 6i \end{vmatrix}$

$⇒ A = 6\begin{vmatrix} 3d + 5g & 4a + 7g & g\ 3e + 5h & 4b + 7h & h\ 3f + 5i & 4c + 7i & i \end{vmatrix}$

C₁ → C1 - 5C₃, C₂ → C2 - 7C₃

 $A = 6\begin{vmatrix} 3d & 4a & g\ 3e & 4b & h\ 3f & 4c& i \end{vmatrix}$

$A =6\times 3\times4 \begin{vmatrix} d & a & g\ e & b & h\ f & c& i \end{vmatrix}$

Using the property of determinants

$A =-72 \begin{vmatrix} a & b & c\ d & e & f\ g & h& i \end{vmatrix}$

∴  A = -72Δ

Concept:

• If a, b and c are three terms in GP then b2 = ab and vice-versa
• If three terms a, b, c are in HP, then b = $\rm \frac{2ac}{a + c}$  and vice-versa
• When three quantities are in AP, the middle one is called as the arithmetic mean of the other two.
• If a, b and c are three terms in AP then b = $\rm \frac{a + c}{2}$  and vice-versa
• If a, b and c are in A.P then 1/a, 1/b, 1/c are in H.P and vice-versa

Calculation:

$\frac{1}{b+c}, \frac{1}{c+a},\frac{1}{a+b}$ are in HP

(b + c), (c + a) and (a + b) are in A.P

⇒ 2(c + a) = b + c + a + b    

⇒ 2c + 2a = 2b + a + c     -----(i)

⇒ 2c + 2a - 2b - a - c = 0

⇒ 2b = c + a

So, a, b, c are in AP

Now, Let (b + c)2, (c + a)2, (a + b)2 are in GP

(c + a)² = √[(b + c).(a + b)]²

⇒ c² + a² + 2ac = (b + c).(a + b)

⇒ c2 + a2 + 2ac = ab + b² + ac + bc

⇒ c2 + a2 - b2 + ac - ab - bc = 0

From here we are unable to check the relation between a, b and c 

So, Our assumption was wrong 

So, b + c)2, (c + a)2, (a + b)2 are not in GP.

∴ Only the 1^st option is correct.

Formula used:

A = $\rm \begin{bmatrix} a & b\ c & d \end{bmatrix}$

det(A) = ad - cb

Identity matrix: 

An identity matrix is a given square matrix of any order which contains on its main diagonal elements with a value of one, while the rest of the matrix elements are equal to zero.

$\rm I=\begin{bmatrix} 1 & 0\ 0 & 1 \end{bmatrix}$

Calculation:

Given that,

$\rm A^2=\begin{bmatrix} 1 & 2a\ 0 & 1 \end{bmatrix}$

$\rm A^3 = A^2\times A=\begin{bmatrix} 1 & 3a\ 0 & 1 \end{bmatrix}$

Therefore, the value of A100 - A50 - 2A25

⇒ $\begin{bmatrix} 1 & 100a \ 0 & 1 \end{bmatrix}-\begin{bmatrix} 1 & 50a \ 0 & 1 \end{bmatrix}-2\begin{bmatrix} 1 & 25a \ 0 & 1 \end{bmatrix}$

⇒ $\begin{bmatrix} 1 & 100a \ 0 & 1 \end{bmatrix}-\begin{bmatrix} 1 & 50a \ 0 & 1 \end{bmatrix}-2\begin{bmatrix} 1 & 25a \ 0 & 1 \end{bmatrix}$

⇒ $\begin{bmatrix} 1 & 50a \ 0 & 1 \end{bmatrix}-\begin{bmatrix} 2 & 50a \ 0 & 2 \end{bmatrix}$

⇒ $-2\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$

∴ A100 - A50 - 2A25  = -2I

Alternate Method\()​

$\rm A=\begin{bmatrix} 1 & a \ 0 & 1 \end{bmatrix}$​​​⇒ |A|  = 1 - 0 = 1

⇒ A100 - A50 - 2A25

⇒  (1)100 - 150 - 2(1)25

⇒  1 - 1 - 2

⇒  -2

⇒  -2I

Formula used:

$\rm A = \begin{bmatrix} a & b & c\ d & e & f\ g & h & i \end{bmatrix}$

det(A) = a(ei - hf) - b(di - gf) + c(dh - eg)

Calculation:

$\begin{vmatrix} a & -b & a - b - c\ -a & b & -a + b - c\ -a & -b & -a - b + c \end{vmatrix} = kabc$

R₂ → R₂ + R₁

$\begin{vmatrix} a & -b & a - b - c\ 0 & 0 & - 2c\ -a & -b & -a - b + c \end{vmatrix} = kabc$

⇒ a[-(-2c)(-b)] - (-b)[-(-2c)(-a)]

⇒ - 2abc - 2abc + (a - b - c ) × 0 = kabc

⇒ - 4abc = kabc

⇒ k = - 4

∴ The value of k is - 4

Formula used:

Powers of iota.

i² = -1, i³ = -i, i⁴ = 1,

i⁵ = ( i . i4 ) = i,

i⁶ = i² . i4  = -1,

i⁷ = -i...... and so on.

 ⇒ i1 + i2  + i3 + i4 = i -1 - i + 1 = 0

Calculation:

$\rm \sum\limits_{n=1}^{8n +7} i^n$

⇒ i¹ + i²  + i³ + i⁴ + i⁵ + i⁶ + i⁷ + ..... + i⁸ⁿ⁺⁷ 

⇒  i1 + i2  + i3 + i4 + i5 + i6 + i7 + ..... +  i8n +  i8n+1 + i8n+2 + .... +  i8n+6  + i8n+7  

⇒ (i -1 - i + 1) +  (i -1 - i + 1) + .... +  (i8n+1 + i8n+2 +  i8n+3 + i8n+4 ) +  (i8n+5 +  i8n+6  + i8n+7)

⇒ 0 + i8n+5 +  i8n+6  + i8n+7 = i1 + i2  + i3 =  i -1 - i = -1

∴ The value of  $\rm \sum\limits_{n=1}^{8n +7} i^n$ is - 1.

Formula used:

z = a + ib

$\rm \overline{z} = a - ib$

Equation of circle: (x - h)² + (y - k)² = r²

Where (h,k) is the coordinates of center of the circle and r is the radius. 

Calculation:

we have $z \overline{z} + |z|^2 + 4(z + \overline{z})-48 = 0$

(x + iy)(x - iy) + x² + y² + 4(x + iy + x - iy) - 48 = 0

⇒ x² + y² +  x2 + y2 + 4 × 2x - 48 = 0

⇒ 2(x² + y²) + 8x - 48 = 0

⇒ x² + y² + 4x - 24 = 0

⇒ (x + 2)² + (y - 0)² = $\rm (\sqrt{28})^{2}$

Since, the above equation is similar to (x - h)2 + (y - k)2 = r2

∴ The given equation represent a circle.

Formula used:

(a + b)² = a² + b² + 2ab

(a + ib) (a - ib) = a² - i²b² = a² + b²

i² = -1

Concept Used:

if y = x² then we can say that x is the square root of y

Calculation:

We have the algebric expression $\rm 2a+2\sqrt{a^2 + b^2}$

⇒ {(a + ib) + (a - ib)} + $\rm 2\sqrt{(a + ib)(a - ib)}$

⇒ $\rm {(\sqrt{(a + ib)})^{2} + ((\sqrt{(a - ib)})^{2} + 2\sqrt{(a + ib)(a - ib)}}$

⇒ $\rm (\sqrt{a + ib} + \sqrt{a - ib})^{2}$

∴ The square root of  $\rm 2a+2\sqrt{a^2 + b^2}$ is  $\rm (\sqrt{a + ib} + \sqrt{a - ib})$.

Formula used:

For any quadratic equation ax2 + bx + c = 0,

when α and β are the roots then

α + β = $\rm \frac{-b}{a}$

αβ = $\rm \frac{c}{a}$

(sinθ + cos θ)2 = sin2θ + cos2θ + 2 sin θ cos θ

sin2θ + cos2θ = 1

Calculation:

According to the question

sin θ and cos θ are the roots of the equation ax² + bx + c = 0,

sin θ + cos θ = $\rm \frac{-b}{a}$     ----(i)

sin θ . cos θ = $\rm \frac{c}{a}$     ----(ii)

(sinθ + cos θ)² = sin²θ + cos²θ + 2 sin θ cos θ    ----(iii)

Form (i), (ii) and (iii), we get 

$\rm \frac{b^{2}}{a^{2}} = 1 + \frac{2c} {a}$

⇒ b² = a² + 2ac

⇒ a² - b² + 2ac = 0

∴ The correct relation between a, b and c is a2 - b2 + 2ac = 0.

Formula used:

• C(n,r) = $\rm \frac{n!}{r!(n-r)!}$
• If a, b and c are in A.P then 2b = a + c

Calculation:

According to the question C(n, 4), C(n, 5) and C(n, 6) are in AP

2 C(n, 5) = C(n, 4) + C(n, 6) 

⇒ 2 × $\rm \frac{n!}{5!(n-5)!}$ = $\rm \frac{n!}{4!(n-4)!}$ + $\rm \frac{n!}{6!(n-6)!}$

⇒ $\rm \frac{1}{60(n-5)(n - 6)!} = \frac{1}{24(n-4)(n - 5)(n - 6)!} + \frac{1}{720(n - 6)!}$

Multiply by 24(n - 5)(n - 6) on both side

⇒ $\rm \frac{1}{5} - \frac{1}{2(n-4)} = \frac{n - 5}{60}$

Again, multiply 60(n - 4) on both side

⇒ 12(n - 4) = 30 + (n - 5)(n - 4)

⇒ 12n - 48 = 30 + n² - 9n + 20

⇒ n² - 21n + 98 = 0

⇒ n = 14, 7

∴ The value of n is 7.

Concept:

If n is a positive integer and r is a whole number, such that r < n, then P(n, r) represents the number of all possible arrangements or permutations of n distinct objects taken r at a time.

It can be represented as ⁿP_r = $\frac{n!}{(n-r)!}$.

The combination is defined as “An arrangement of objects where the order in which the objects are selected does not matter.” 

ⁿC_r = $\frac{n!}{r!(n-r)!}$ ,  when n < r

Where n = distinct object to choose from

C = Combination

r = spaces to fill

Calculation:

Vowels = 2

Consonants = 5

Total Alphabets = 7

Since 4 letter words must include 2 vowels, we don't need to select them, and the rest of the 2 letters will be taken from 5 consonants.

Number of ways of selecting 2 letters from 5 consonants = ⁵C₂ = 10

Arrangement of all 4 letters will be given by 4! = 24 ways

Total number of arrangements = 5C2 × 4! = 10 × 24 = 240 ways

∴ The total number of words that can be formed is 240.

Explanation:

We only need 2 points to form straight lines

We have 20 points so the number of straight lines drawn is ²⁰C₂ ​= 190

We only need 3 points to form a triangle

We have 20 points, so number of triangles drawn is ²⁰C₃ = 1140

∴ Statement 1 is incorrect and Statement 2 is correct.

Formula used:

If a and b are real numbers and n is a positive integer, then

(a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ an - 1 b¹ + .........  ⁿCn bⁿ  for $\rm 0 \leq r \leq n$

Total number of terms in (a + b)ⁿ = n + 1

Calculation:

We have to find the total number of terms in $\rm \left(\frac{a^2}{b^2}+\frac{b^2}{a^2}+2\right)^{21}$ where a ≠ 0, b ≠ 0

⇒ $[{\rm \left(\frac{a}{b}+\frac{b}{a}\right)^{2}}]^{21}$

⇒ ${\rm \left(\frac{a}{b}+\frac{b}{a}\right)^{42}}$

∴ The total number of terms in the expression is 42 + 1 = 43.

Concept:

Consider three linear eqaution in two variable:

a₁x + b₁y + c_{1 = 0}

a₂x + b₂y + c₂ = 0

a₃x + b₃y + c₃ = 0

Condition for the consistency of three simultaneous linear equations in 2 variables:

​​​​ \left| {\begin{array}{*{20}{c}} a_1&b_1&c_1\ a_2&b_2&c_2\ a_3&b_3&c_3 \end{array}} \right|=0

Formula for Quadratic equation:

ax² + bx + c = 0

x = $\rm \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

Calculation:

2k2x + 3y - 1 = 0      ....(1)

7x - 2y + 3 = 0      ....(2)

6kx + y + 1 = 0      ....(3)

For consistency of given simultaneous equation,

\left| {\begin{array}{*{20}{c}} 2k^2&3&-1\ 7&-2&3\ 6k&1&1 \end{array}} \right|=0

⇒ 2k²(-2- 3) - 3(7 - 18k) - 1(7 + 12k) = 0

⇒ -10k² - 21 + 54k - 7 - 12k = 0

⇒ -10k² + 42k - 28 =  0

⇒ 5k2 - 21k + 14 =  0

By using the formula,

$x=\rm \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

$\Rightarrow k=\rm \frac{-(-21) \pm \sqrt{(-21)^{2} - 4(5)(14)}}{2\times 5}$

$\therefore k=\rm \frac{21 \pm \sqrt{161}}{10}$

Concept:

AA^-1 = I

Calculation:

A^-1 = $\rm \begin{bmatrix} -2 & 1 \ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}$

Let A = \rm \begin{bmatrix} a & b \c & d \end{bmatrix}

We know that AA^-1 = I

∴ \rm \begin{bmatrix} a & b \c & d \end{bmatrix}(\rm \begin{bmatrix} -2 & 1 \ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}\) = $\rm \begin{bmatrix} 1 & 0\0 & 1 \end{bmatrix}$

⇒ $\rm \begin{bmatrix} -2a +\frac{3}{2}b & a -\frac{b}{2} \-2c +\frac{3d}{2}& c-\frac{d}{2} \end{bmatrix}$= $\rm \begin{bmatrix} 1 & 0\0 & 1 \end{bmatrix}$

Two matrices are equal if their order and corresponding elements are equal.

Therefore, - 2a + $\frac{3}{2}b = 1$  ---(1)

and a - b/2 = 0 ⇒ a = b/2 --- (2)

From equations (1) and (2),

⇒ $-2(\frac{b}{2})+\frac{3}{2}b = 1$

⇒ $-b +\frac{3b}{2} =1$

⇒ $\frac{b}{2} = 1$

∴ b = 2 

From equation (2), a = 1

Similarly, $-2c +\frac{3d}{2} = 0$  --- (3)

and $c -\frac{d}{2} = 1$ --- (4)

Solving equation (3) and (4), we get c = 3 and d = 4

∴ Matrix A = $\rm \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$

Formula used:

cos2θ = 1 - 2sin²θ 

Calculation:

f(x) = ln(2 + sin2x)

⇒ f(x) = ln[2 + (1 - cos2x)/2]

⇒ f(x) = ln[(5 - cos2x)]/2      ----(i)

If we put (x + π) in place of x, we get

⇒ f(x + π) = ln[(5 - cos2(x + π)/2]

⇒ f(x + π) = ln[5 - cos(2π + 2x)]/2 

⇒ f(x + π) = ln[(5 - cos2x)]/2      ----(ii) 

From (i) and (ii), we get 

f(x + π) = f(x)

∴ The period of the function f(x) = ln(2 + sin2x) is π.

Formula used:

sin $\rm \frac{\pi}{2}$ = 1

sin $\rm \frac{\pi}{6}$ = $\rm \frac{1}{2}$

Calculation:

Given

sin(A + B) = 1 

⇒ sin(A + B) = sin $\rm \frac{\pi}{2}$

⇒ A + B = $\rm \frac{\pi}{2}$     ----(1)

2 sin(A - B) = 1

⇒ sin(A - B) = $\rm \frac{1}{2}$

⇒ sin(A - B) = sin $\rm \frac{\pi}{6}$ 

⇒ A - B = $\rm \frac{\pi}{6}$     ----(2)

On solving (1) and (2), we get

A = $\rm \frac{\pi}{3}$, B = $\rm \frac{\pi}{6}$

Now, We have to find the value of "tan A ∶ tan B"

⇒ $\rm \frac{tan \frac{\pi}{3}}{tan \frac{\pi}{6}}$ = $\rm \frac{\sqrt{3}}{\frac{1}{\sqrt{3}}}$ = 3 : 1

∴ The value of "tan A ∶ tan B" equal to 3 : 1.

Formula used:

Triangle having no side common with those of the polygon = 

$\rm \frac{n(n-4)(n-5)}{6}$

n is the sides of the polygon.

Calculation:

Here n = 10

Triangle having no side common with those of the polygon

= $\rm \frac{10(10-4)(10-5)}{6}$

= $\rm \frac{10\times 6\times 5}{6}$

= 50

∴ The number of triangles is 50.

Formula used:

Quadratic Equation formula: 

A quadratic equation can be written in a standard form as

ax² + bx + c = 0

$\rm x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

Concept Used:

Algebraically, the absolute value of a number equals the nonnegative square root of its square

i.e Modulus of that number.

Calculation:

Given equation:

x4 - 10x2 + 9 = 0.

⇒ x² = $\rm \frac{10 ± \sqrt{100 - 4\times 9}}{2}$

⇒ x2 = $\rm \frac{10 ± \sqrt{64}}{2}$

⇒ x² = $\rm \frac{10 ± 8}{2}$

Case I. Take positive sign

x² = 9

⇒ x = ± 3

Case II. Take negative sign

x² = 1

⇒ x = ± 1

∴ The sum of the absolute values of the roots = 3 + 1 + |-3| + |-1| = 8.

Formula used:

This binomial expansion formula gives the expansion of (x + y)ⁿ where 'n' is a natural number. The expansion of (x + y)ⁿ has (n + 1) terms. 

(x + y)ⁿ = ⁿC ₀ xⁿ y⁰ + ⁿC ₁ x^{n - 1} y¹ + ⁿC ₂ x^n-2 y² + ^_nC ₃ x^{n - 3} y³ + ... + ⁿC _n−1 x y^{n - 1} + ⁿC _n x⁰yⁿ

Calculation:

(1 + x)ⁿ = 1 + n x + ...  + n.x^n-1+ xⁿ

From above equation we can write 

p = 1, q = n

r = n, s = 1

Now, We have to find the value of (ps + qr)

⇒  1 + n²

∴ (ps + qr) equal to (1 + n2).

Formula used:

sin^-1 x = $\rm \frac{\pi}{2}$

x = sin ($\rm \frac{\pi}{2}$)

sin ($\rm \frac{\pi}{2}$) = 1

Calculation:

sin-1 x + sin-1 y + sin-1 z = $\frac{3\pi}{2}$ 

This is possible only when

sin-1 x = sin-1 y = sin-1 z = $\rm \frac{\pi}{2}$ 

⇒ sin-1 x = $\rm \frac{\pi}{2}$

⇒ x = 1    ----(i)

⇒ sin-1 y = $\rm \frac{\pi}{2}$

⇒ y = 1      ----(ii)

⇒ sin-1 z = $\rm \frac{\pi}{2}$

⇒ z = 1      ----(iii)

Now, we have to find the value of 

x1000 + y1001 + z1002

Now, from (i), (ii), and (iii), we get

⇒ (1)1000 + (1)1001 + (1)1002

⇒ 1 + 1 + 1

⇒ 3

∴ The value of x1000 + y1001 + z1002 is 3.

Formula used:

sin x + sin y $=2\ sin\ {\left(\frac{x+y}{2}\right)}\ cos\ {\left(\frac{x-y}{2}\right)}$

cos x + cos y $=2\ cos\ {\left(\frac{x+y}{2}\right)}\ cos\ {\left(\frac{x-y}{2}\right)}$

tan θ = $\rm \frac{sin : θ}{cos : θ}$

Calculation:

Given, sin x + sin y = cos x + cos y 

We know that

sin x + sin y $=2\ sin\ {\left(\frac{x+y}{2}\right)}\ cos\ {\left(\frac{x-y}{2}\right)}$

cos x + cos y $=2\ cos\ {\left(\frac{x+y}{2}\right)}\ cos\ {\left(\frac{x-y}{2}\right)}$

sin x + sin y = cos x + cos y

$\Rightarrow 2\ sin\ {\left(\frac{x+y}{2}\right)}\ cos\ {\left(\frac{x-y}{2}\right)}=2\ cos\ {\left(\frac{x+y}{2}\right)}\ cos\ {\left(\frac{x-y}{2}\right)}$

$\Rightarrow \ sin\ {\left(\frac{x+y}{2}\right)}=\ cos\ {\left(\frac{x+y}{2}\right)}$

$\frac{\ sin\ \frac{x+y}{2}}{ cos\frac{x+y}{2}}=1$................($\frac{sinx}{cosx}=tanx$)

$tan (\frac{x+y}{2})=1$

$tan \left(\frac{x}{2}+\frac{y}{2}\right)=1$

∴ The value of $\rm tan \left(\frac{x}{2}+\frac{y}{2}\right)$ is equal to 1.

Formula used:

Multiplication matrix: 

$\rm \begin{bmatrix} a & b\ c & d \end{bmatrix} \times \begin{bmatrix} w & x\ y & z \end{bmatrix} = \begin{bmatrix} aw + by & ax + bz\ cw + dy & cx + dz \end{bmatrix}$

Identity matrix: 

An identity matrix is a given square matrix of any order which contains on its main diagonal elements with a value of one, while the rest of the matrix elements are equal to zero.

I = $\rm \begin{bmatrix} 1 & 0\ 0 & 1 \end{bmatrix}$

Calculation:

$I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$

⇒  $I^2 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$

$A = \begin{bmatrix} 0 & 2 \ -2 & 0 \end{bmatrix}$

⇒  A² = $\begin{bmatrix} -4 & 0 \ 0 & -4 \end{bmatrix}$

According to question,

(mI + nA)2 = A

⇒ m²I² + n²A² + 2mnA = A   

⇒  m2I2 + n2A2 = A(1 - 2mn)

⇒  m² $\rm \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$ + n² $\rm \begin{bmatrix} -4 & 0 \ 0 & -4 \end{bmatrix}$ = (1 - 2mn)$\rm \begin{bmatrix} 0 & 2 \ -2 & 0 \end{bmatrix}$ $\rm \begin{bmatrix} m^{2} & 0\ 0 & m^{2} \end{bmatrix} + \begin{bmatrix} -4n^{2} & 0\ 0 & -4n^{2} \end{bmatrix} = \begin{bmatrix} 0 & 2(1-2mn)\ -2(1-2mn) & 0 \end{bmatrix}$

On equating on both sides 

⇒ m² - 4n² = 0

⇒ m = 2n     ----(1)

⇒ 2(1 - 2mn) = 0

⇒ 1 - 2mn = 0

⇒  mn = $\rm \frac{1}{2}$

From equation (1)

(2n)n = $\rm \frac{1}{2}$

⇒ n = $\rm ± \frac{1}{2}$

Taking positive sign 

When n =  $\rm \frac{1}{2}$ then, m = 1

∴ m + n = 1 + $\rm \frac{1}{2}$ = $\rm \frac{3}{2}$

Formula used:

cot (cot^–1 x) = x, x ϵ R

sin^–1 x + sin^–1  y = sin^–1 [x √(1 - y²) + y √(1 - x2)]

where x > 0, y > 0 & (x² + y²) ≤ 1

cot^-1x = tan–1 (1/x) 

Calculation:

$\rm cot \left[sin^{-1} \frac{3}{5}+cot^{-1}\frac{3}{2} \right]$    ----(i)

Let a = cot^-1(3/2)

⇒ cot a = 3/2

⇒ sin a = 2/√13

⇒ a = sin^-1(2/√13)

From (i)

$\Rightarrow \rm cot \left[sin^{-1} \frac{3}{5}+sin^{-1}\frac{2}{√13} \right]$

⇒ cot [sin^-1{(3/5) √{1 - (4/13)} + (2/√13) √{1 - (9/25)}]

⇒ cot [sin-1(17/5√13)]

⇒ cot [cot-1(6/17)]

⇒ 6/17

∴ The value of $\rm cot \left[sin^{-1} \frac{3}{5}+cot^{-1}\frac{3}{2} \right]$is 6/17.

Formula used:

sin 60^∘ = $\rm \frac{\sqrt{3}}{2}$, tan 180∘ = 0

If sin a = sin b then a = b

Calculation:

Given, 4sin2 x = 3

⇒ sin x = $\rm \frac{\sqrt{3}}{2}$

⇒ sin x = sin60° 

⇒ x = 60^∘ or 120°   [∵ 0 ≤ x ≤ π]

We have to find the value of tan 3x

⇒ tan 3(60^∘) or tan 3 × 120° 

⇒ tan 180^∘ or tan360° 

⇒ 0 or 0

∴ The value of tan3x is equal to 0.

Formula used:

The formula for Arithmetic Sequence:

an = a1 + (n – 1)d

Where,

• an = nth term that has to be found
• a1 = 1^st term in the sequence
• n = Number of terms
• d = Common difference
• Sn = Sum of n terms

Calculation:

According to the question

q = p + (3 - 1)d   

⇒ q = p + 2d    ----(1)

And 3 = p + (5 - 1)d 

⇒ 3 = p + 4d    ----(2)

Solving (1) and (2), we get 

q = 3 - 2d

and p = 3 - 4d      [from (2)]

According to the question pq is minimum 

⇒ (3 - 4d) (3 - 2d) is minimum  

⇒ (9 - 18d + 8d²) is minimum 

Differentiation the above equation with respect to d is 0

⇒ -18 + 16d = 0 

⇒ d = 9/8 

∴ The value of d is 9/8.

Concept:

Non-Collinear Points: The set of points that do not lie on the same line are called non-collinear points. We cannot draw a single straight line through these points.

Formula Used:

a³ - b³ = (a - b) (a² + ab + b²)

Calculation:

Given equation x³ - 8 = 0

⇒ x3 - 2³ = 0

⇒ (x - 2) (x² + 2x + 4) = 0

⇒ x - 2 = 0 or x2 + 2x + 4 = 0 

⇒ x = 2 or x = -1 + i√3 or x = -1 - i√3

We can take these three points on complex plane 

So, we can write 1^st point as 2 + i.0, -1 + i√3, -1 - i√3

The three points on complex plain is (2, 0), (-1,√3), (-1, -√3)

Case I. 

Equation of line passes through (-1,√3), (-1, -√3) is

(y - √3) = (2√3/0) (x + 1)

⇒ x + 1 = 0 

And this line does not passes through (2, 0).

∴ These points (roots) are not collinear.

Case 2. 

We know that these points are non-collinear it means they will definitely pass through a circle 

Let the equation of circle be (x - a)² + (y - b)² = r²

Now, According to our assumption 

(2 - a)2 + (0 - b)2 = r2     ----(i)

(-1 - a)2 + (√3 - b)2 = r2     ----(ii)

(-1 - a)2 + (-√3 - b)2 = r2     ----(iii)

On solving (i), (ii) and (iii), we get 

a = 0, b = 0 and r = 2

Hence the equation of circle is (x - 0)2 + (y - 0)2 = 22

∴ The roots represents a circle but radius of circle is 2.

Formula used:

sec x = $\rm \frac{1}{cos :x}$, cosec x = $\rm \frac{1}{sin :x}$

(sin x - cos x)2 = sin2 x + cos2 x - 2 sin x cos x

sin2 x + cos2 x = 1, tan x = $\rm \frac{sin : x}{cos : x}$

Calculation:

We have sec x × cosec x = p

⇒ $\rm \frac{1}{cos x}× \frac{1}{sin x}$ = p

⇒ sinx.cos x = $\rm \frac{1}{p}$     ----(i)

We know that, 

(sin x - cos x)2 = sin2 x + cos2 x - 2 sin x cos x

From (i), we get 

⇒ (sin x - cos x)2 = 1 - 2 × $\rm \frac{1}{p}$    ----(ii)

We know tha the square of any number is ≥ 0

(sin x - cos x)2 ≥ 0

From (ii), we get

$1 - 2 \times \rm \frac{1}{p} ≥ 0$

⇒ 1 ≥ 2/p

⇒ 1/1 ≤ p/2

⇒ p ≥ 2

∴ The smallest value of p is 2.

Formula used:

sin 2θ = 2 sinθ cosθ 

cos 2θ = 2cos²θ - 1

Concept:

f'(x) = 0 at a point x = a is 

• If f'(x) changes the sign from positive to negative while the graph of the function passes through x = a then x = a is said to be a point local maxima.
• If f''(a) < 0 then x = a is a point of local maxima.

Calculation:

Let f(θ) = sinθ + sinθ cosθ

⇒ f(θ) = sinθ + $\rm \frac{1}{2}$​ sin 2θ

⇒ f'(θ) = cosθ + $\rm \frac{1}{2}$​ cos 2θ × 2

⇒ f'(θ) =  cosθ + cos 2θ 

Put f'(θ) = 0

⇒ cosθ + 2cos2θ - 1 = 0

⇒ 2cos2θ + 2cosθ - cosθ  - 1 = 0

⇒ 2cosθ (cosθ + 1) - 1(cosθ + 1) = 0 

⇒ (cosθ + 1) (2cosθ - 1) = 0

⇒ cosθ = 1/2, cosθ = -1

⇒ θ = π/3, π 

According to the given range 0 < θ < $\frac{π}{2}$

θ = π/3

At this value of θ = π/3, the function can be maximum and minimum 

So, we have to check second derivatives

f"(θ) = - sinθ - 2sin2θ 

at θ = π/3 

f"(π/3 ) ≤ 0

So, we can say that θ = π/3 is a point of local maxima.

Concept:

Definition of De Morgan’s law: 

The complement of the union of two sets is equal to the intersection of

their complements and the complement of the intersection of two sets

is equal to the union of their complements.

Explanation:

**Statement:**1

The union over the intersection of sets is distributive.

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

Statement 1 is correct

**Statement:**2 

According to De Morgan law,

(A U B)' = A' ∩ B' 

Hence, statement 2 is correct.

Statement:3 

If A and B are two sets, then their difference is given by

A - B or B - A. 

If A = {2, 3, 4} and B = {2, 3, 4, 5, 6} 

A - B means elements of A that are not the elements of B i.e,

in the above example A - B = ϕ

But here, A and B are not equal sets so statement 3 is incorrect

Also, B - A = {5, 6} ≠ ϕ 

Additional Information 

In general, B - A = {x : x ∈ B, and x ∉ A} 

• If A - B is the empty set, then we can show that A is a subset of B
• If B - A is also the empty set, then we can also show that B is a subset of A