NDA Mathematics 6 Sep 2020 Official Paper (Previous Year Paper)

Calculation:

Given: n(Z) = 90 and the number of elements in Y and Z are in the ratio 4 : 5

To find: b = ?

n(Y) = 16 + 18 + 17 + b

n(Z) = 12 + 18 + 17 + c = 90 

 

$\rm \frac{n(y)}{n(z)}=\frac{16+18+17+b}{12+18+17+c}=\frac{4}{5}\ \Rightarrow \frac{51+b}{90}=\frac{4}{5}\ \Rightarrow (51+b)\times 5=360 \Rightarrow255+5b=360\ \Rightarrow5b=105\ \Rightarrow b=21$

Hence, option (3) is correct.

Calculation:

Given: n(Z) = 90, 

From above diagram, we get

n(X) = a + 46, n(Y) = b + 51, n(X ∩ Y) = 16 + 18 = 34,  n(Y ∩ Z)  = 17 + 18 =35,  n(X ∩ Z) = 12 + 18 = 30,  n(X ∩ Y ∩ Z) = 18

n(X) + n(Y) + n(Z) - n(X ∩ Y) - n(Y ∩ Z) - n(X ∩ Z) + n(X ∩ Y ∩ Z) = 

a + 46 + b + 51 + 90 - 34 - 35 - 30 + 18

= a + b + 106

Hence, option (4) is correct.

Concept:

Complement of X = U - X, where U is a universal set.

 

Calculation:

We know, n(X) + n(Y) + n(Z) - n(X ∩ Y) - n(Y ∩ Z) - n(X ∩ Z) + n(X ∩ Y ∩ Z) = n(X ∪ Y ∪ Z)

Here U - n(X ∪ Y  ∪ Z) = p, whre U is a universal set 

⇒ U - (a + b + 106) = p

⇒ U = p + (a + b + 106) 

Now the number of elements in the complement of X, U - n(X) = p + (a + b + 106) - (a + 16 + 18 + 12)

= p + b + 60

Hence, option (1) is correct.

Concept:

Let $\rm \dfrac p q = \rm \dfrac r s$ then componendo and dividend: $\rm\frac{p+q}{p-q}=\rm\frac{r+s}{r-s}$⇒$\rm\frac{p-q}{p+q}=\rm\frac{r-s}{r+s}$

$\rm \sin(A+B)=sinAcosB+cosAsinB \ and \rm \sin(A-B)=sinAcosB-cosAsinB$

$\rm \sec^2A=1+tan^2A$

$\rm cos2A=cos^2A-sin^2A$

 

Calculation:

Given: $\rm \frac {\tan 3A}{\tan A} = K,$

$⇒ \rm \frac{\frac{\sin 3 A}{\cos 3 A}}{\frac{\sin A}{\cos A}}=k\ ⇒ \frac{\sin 3 A \cdot \cos A}{\cos 3 A \cdot \sin A}=\frac{K}{1}$

Applying componendo and dividend, 

$\rm \frac{\sin 3 A \cdot \cos A-cos 3 A \cdot \sin A}{\sin 3 A \cdot \cos A+cos 3 A \cdot \sin A}=\frac{k-1}{k+1}\ \Rightarrow \frac{sin(3A-A)}{sin(3A+A)}=\frac{k-1}{k+1}\ \Rightarrow \frac{sin(2A)}{sin(4A)}=\frac{k-1}{k+1}\ \Rightarrow \frac{2sin(A)cosA}{2sin(2A)cos2A}=\frac{k-1}{k+1}\ \Rightarrow \frac{sin(A)cosA}{2sin(A)cosAcos2A}=\frac{k-1}{k+1}\$

$\begin{aligned} &\rm \Rightarrow \frac{1}{2 \cos 2 A}=\frac{k-1}{k+1}\ &\rm \Rightarrow \frac{1}{2\left(\cos ^{2} A-\sin ^{2} A\right)}=\frac{k-1}{k+1}\ &\rm \Rightarrow \frac{\sec ^{2} A}{2\left(1-\tan ^{2} A\right)}=\frac{k-1}{k+1}\ &\rm \Rightarrow \frac{1+\tan ^{2} A}{1-\tan ^{2} A}=\frac{2 k-2}{k+1}\ &\rm \Rightarrow \frac{1+\tan ^{2} A-1+\tan ^{2} A}{1+\tan ^{2} A+1-\tan ^{2} A}=\frac{2 k-2-k-1}{2 k-2+k+1}\ &\rm \Rightarrow \frac{2 \tan ^{2} A}{2}=\frac{k-3}{3 k-1}\ &\rm \Rightarrow \tan ^{2} A=\frac{k-3}{3 k-1} \end{aligned}$

Hence, option (2) is correct.

Calculation:

We have, $\rm \tan^2A=\frac {k - 3}{3k - 1}$

For real values of tan² A:

$\rm \frac {K - 3}{3K - 1}$ ≥ 0

At k = 1/3 and 3, $\rm \frac {K - 3}{3K - 1}$<0

∴For real values of tan A, K cannot lie between $\rm \frac 1 3;and;3$

Hence, option (2) is correct.

Concept:

In triangle ABC, with sides a, b, and c sine rule is given by, $\rm \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

sin (180 - θ) = sin θ 

Calculation:

1. In triangle ABD, using sine rule

sin θ /AB = sin α / AD

⇒ AD sin θ = AB sin α

So ,this is correct statement.

In triangle ABD, using sine rule

sin θ /AB = sin (180-(θ +α )) / BD

⇒ BD sin θ = AB sin (α + θ)

So ,this is also correct statement.

Hence, option (3) is correct.

Concept:

In triangle ABC, with sides a, b, and c sine rule is given by, $\rm \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

sin (A + B) = sinA.cosB + cosA.sin B

Calculation:

$\begin{array}{l} \rm In \ \Delta B C D \ \rm B D^{2}=B C^{2}+C D^{2} \ \rm B D^{2}=P^{2}+q^{2} \ \rm B D=\sqrt{P^{2}+q^{2}} \end{array}$

In triangle ABD, using sine rule,

$\rm \frac{AB}{\sin \theta} =\frac{B D}{\sin (180-(\theta+\alpha))} \ \Rightarrow\rm \frac{AB}{\sin \theta} =\frac{B D}{\sin (\theta+\alpha)} \ \Rightarrow A B =\frac{B D \sin \theta}{\sin (\theta+\alpha)} \ = \frac{B D^{2} \sin \theta}{{B D}(\sin (\theta+\alpha)}$

$\begin{array}{l} \rm= \frac{B D^{2} \sin \theta}{{3 D(\sin \theta \cos \alpha+\cos \theta \sin \alpha)}} \ \rm =\frac{\left(p^{2}+q^{2}\right) \sin \theta}{\sqrt{p^{2}+q^{2}}\left(\sin \theta \frac{q}{\sqrt{p^{2}+q^{2}}}+\cos\theta \frac{p}{\sqrt{p^{2}+q^{2}}}\right)} \end{array}$

= $\rm \frac {\left( p^2 + q^2 \right) \sin \theta}{p \cos \theta + q \sin \theta}$

Hence, option (1) is correct.

Concept:

tan (A - B) = $\rm \frac{\tan A-\tan B}{1+\tan A \tan B}$

 

Calculation:

$\rm \tan θ = \frac {\cos 17^\circ - \sin 17^\circ}{\cos 17^\circ + \sin 17^\circ},$

RHS = $\frac {\cos 17^\circ - \sin 17^\circ}{\cos 17^\circ + \sin 17^\circ}$

Divide by cos 17°

$=\frac{\frac{\cos 17}{\cos 17}-\frac{\sin 17}{\cos 17}}{\frac{\cos 17}{\cos 17}+\frac{\sin 17}{\cos 17}}$

$\begin{aligned} &= \frac{1-\tan 17}{1+\tan 17} \& = \frac{\tan 45^{\circ}-\tan 17}{1+1 \times \tan 17^{\circ}} \ & = \frac{\tan 45-\tan 17}{1+\tan 45 \tan 17} \end{aligned}$

= tan (45 - 17) = tan (28°) 

Now,

⇒ tan θ = tan (28°) 

∴ θ = 28° 

Hence, option (2) is correct.

Concept:

cos (A + B) = cosA. cosB - sinA. sin B

Calculation:

Given: cos 2B = 3 sin² A   .... (1)

and  3sin 2A = 2 sin 2B     .... (2)

Divide (1) by (2),

$\frac{cos(2B)}{2sin(2B)}=\frac{3sin^2(A)}{3sin(2A)}$ .... (3)

We know that,

sin 2A = 2 sin A∙ cos A

Hence, equation (3) becomes,

$\frac{cos(2B)}{2sin(2B)}=\frac{sin^2(A)}{2sin(A)cos(A)}$

⇒ $\frac{cos(2B)}{sin(2B)}=\frac{sin(A)}{cos(A)}$

⇒ cos (2B) cos (A) = sin (2B) sin (A)

⇒ cos (2B) cos (A) - sin (2B) sin (A) = 0

From above concept,

cos (A + 2B) = 0

⇒ (A + 2B) = cos^-1 (0)

⇒ (A + 2B) =  π/2

Concept:

$\begin{array}{l} \rm \sin 3 x=3 \sin x-4 \sin ^{3} x \ \rm \cos 3 x=4 \cos ^{3} x-3 \cos x \end{array}$

 

Calculation:

sin 3x + cos 3x + 4 sin³ x - 3 sin x + 3 cos x - 4 cos³ x

= 3 sin x - 4 sin³ x + 4 cos³ x - 3 cos x  + 4 sin³ x - 3 sin x + 3 cos x - 4 cos³ x

= 0

Hence, option (1) is correct.

Concept:

 −1 ≤ cosx ≤ 1

 

Calculation:

Here, y = 2 + cos x

We know,  −1 ≤ cosx ≤ 1

Adding 2, we get

⇒ 2 −1 ≤ 2 + cosx ≤ 2 + 1

⇒ 1 ≤ y ≤ 3

So the value of y-coordinate lies between y = 1 to y = 3

Hence, option (4) is correct.

Concept:

2 sin x. cos x = sin 2x

Calculation:

Let, x = 8 cos 10° ⋅ cos 20° ⋅ cos 40°

Now, multiplying both sides by sin 10°

$\begin{array}{l} ⇒ x \sin 10=4 \times 2 \sin 10 . \cos 10 . \cos 20 . \cos 40 \ ⇒ x \sin 10=2 \times 2 \sin 20 . \cos 20 . \cos 40 \ ⇒ x \sin 10=2 \times \sin 40 . \cos 40 \ ⇒ x \sin 10=\sin 80 \ ⇒ x \sin 10=\sin (90-10)=\cos 10 \end{array}$

⇒ x = cos 10/ sin 10

⇒ x = cot 10° 

Hence, option (2) is correct.

Concept:

$\rm \cos (A)-\cos (B)=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$

$\cos 3θ =4 \cos ^{3} θ-3 \cos θ$

 

Calculation:

cos 48° - cos 12°  

 $= \rm-2 \sin \left(\frac{48+12}{2}\right) \sin \left(\frac{48-12}{2}\right)\ =-2\sin 30\sin18\ =-2\times \frac{1}{2}sin 18\ =-\sin 18$

sin 18°

 $\begin{array}{l} \Rightarrow θ=18^{\circ} \ \Rightarrow 2 θ=36^{\circ} \ \Rightarrow \sin 2 θ=\sin 36^{\circ} \ \Rightarrow \sin 2 θ=\sin (90-54) =\cos 54\ \Rightarrow \sin 2 θ=\cos 3θ \ \Rightarrow 2 \sin θ \cos θ=4 \cos ^{3} θ-3 \cos θ \ \Rightarrow 2 \sin θ \cos θ=\cos θ\left(4 \cos ^{2} θ-3\right) \end{array}$

$\Rightarrow 2 \sin θ= 4(1-\sin ^{2} θ)-3 \ \Rightarrow 4-4 \sin ^{2} θ-3-2 \sin θ=0\ \Rightarrow -4 \sin ^{2} θ-2 \sin θ+1=0\ \Rightarrow 4 \sin ^{2} θ+2 \sin θ-1=0$

$\Rightarrow \sin θ= \frac{-2 \pm \sqrt{4+16}}{8} \ =\frac{-2 \pm 2 \sqrt{5}}{8} \ \sin θ =\frac{-1 \pm \sqrt{5}}{4}$

Now since θ is in I quadrant 

sin 18° =  $\frac{-1 + \sqrt{5}}{4}$

cos 48° - cos 12°  = - sin 18° = $\frac{1 - \sqrt{5}}{4}$

= $\frac{1 - \sqrt{5}}{4}$

Hence, option (2) is correct.

Concept:

Pythagoras theorem: In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.

Sine Rule in triangle ABC, having sides a, b, c:

$\rm \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{K}$

sin A - cos B = $\rm 2\cos(\frac{A +B}{2})\sin(\frac{A -B}{2})$

Calculation:

1. We have,

sin $\rm B = \frac 1 3=\frac{P}{H}$

Hwence, Right angle triangle can be drawn as,

Here, P = 1

H = 3

B = x

Using Pythagoras theorem,

3² = 1² + x²

x² = 8

$\therefore x=2\sqrt{2}$

Now , Cosec C = $\frac{3}{2\sqrt{2}}$

Hence Ststement (1) is not correct.

2. Given,

b cos B = c cos C

Using the sine rule,

$\rm \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{K}$

⇒ b = K sin B & c = K sin C

⇒ 2 sin B cos B = 2 sin C cos C

Assume,

K = 2

⇒ 2 sin B cos B = 2 sin C cos C

⇒ sin 2B = sin 2C  [2 sin x cos x = sin 2x]

⇒ sin 2B - sin 2C = 0

Using Formula:

sin C - sin D = 2 cos $(\frac{C+D}{2})$ × sin $(\frac{C-D}{2})$

⇒ 2 cos (B + C) sin (B - C) = 0

In this case,

Either,

cos (B + C) = 0

Hence, (B + C) = 90° .... (1)

Or,

Or sin (B -C) = 0

B - C = 0

⇒ B = C  .... (2)

From equation (1) & (2),

B = C = 45°

So ABC must be issosceles 

Hence, option (2) is correct.

Concept:

• Consider complex number z = x + iy,
• $\text { Modulus of } z=|z|=\sqrt{x^{2}+y^{2}}$
• Division of modulus of two complex numbers $\rm |\frac{z_1}{z_2}|= \frac{|z_1|}{|z_2|}$
• sin²θ + cos²θ = 1

 

Calculation:

We have to find the value of $|\frac {\cos θ + i \sin θ}{\cos θ - i \sin θ}|$

Now, According to the concept of division of modulus

$⇒ \frac {|\cos θ + i \sin θ|}{|\cos θ - i \sin θ|}$

$⇒ \frac{\cos^2θ +sin^2θ }{cos^2θ +(-sin θ )^2}$

⇒ 1/1 = 1

Hence, option (2) is correct.

Alternate Method We have given complex numbers,

z = $\rm \frac {\cos θ + i \sin θ}{\cos θ - i \sin θ},$

Multiply in the numerator and denominator by (cos θ + i sin θ)

⇒ z = $\rm \frac {\cos θ + i \sin θ}{\cos θ - i \sin θ}\times \rm \frac {\cos θ + i \sin θ}{\cos θ + i \sin θ}$

⇒ z = $\rm \frac {(\cos θ + i \sin θ)^2}{\cos^2 θ - i^2 \sin^2 θ}$

We know that,

sin²A + cos²A = 1

and, i² = -1

Hence, ⇒ z = $\rm \frac {(\cos^2 θ + i^2 \sin^2 θ+2i\ cosθ \ sinθ )}{\cos^2 θ + \sin^2 θ}$

or, z = (cos²θ - sin²θ) + 2i sin θ cos θ .... (1)

Using formula,

cos 2θ = (cos2θ - sin2θ)

and, sin 2θ = 2sin θ cos θ

Hence, equation (1) becomes,

z = cos 2θ + i sin 2θ

Hence, |z| = $\sqrt{(cos^22\theta+sin^22\theta)}$ = 1

Concept:

A proper subset is one that contains few elements of the original set.

A superset is one which contain all the element including elements of the original set.

 

Calculation:

The proper subsets of {1, 2, 3, 4} and supperset of set {3} are

{1, 3}, {2, 3}, {3, 4}, {1, 2, 3}, {2, 3, 4}, {1, 3, 4}, {1, 2, 3, 4}

Hence, option (3) is correct.

Concept:

If A is the arithmetic mean of numbers a and b and is given by ⇔ $\rm AM = \dfrac{a+b}{2}$

If G is the geometric mean of the numbers a and b and is given by  ⇔ $\rm GM = \sqrt{ab}$

If H is the Harmonic mean of numbers a and b and is given by ⇔ $\rm HM = \dfrac{2ab}{a+b}$

Relation between AM, GM and HM

G² = AH

AM ≥ GM ≥ HM

 

Calculation: 

\rm D = \left| {\begin{array}{*{20}{c}} \rm p&\rm q&\rm r\ \rm q&\rm r&\rm p\ \rm r&\rm p&\rm q \end{array}} \right|,

$\begin{array}{l} =\rm p\left(q r-p^{2}\right)-q\left(q^{2}-p r\right)+r\left(p q-r^{2}\right) \ =\rm p q r-p^{3}-q^{3}+p q r+p q r-r^{3} \ =\rm3 p q r-\left(p^{3}+q^{3}+r^{3}\right) \end{array}$

As we know AM ≥ GM

Now, AM of p³, q³ and r³ = $\rm \dfrac{p^3+q^3+r^3}{3}$

Now, GM of p³, q³ and r³ = $\rm ({p^3\times q^3 \times r^3})^{\frac{1}{3}} = pqr$

Therefore, $\rm \dfrac{p^3+q^3+r^3}{3}$ > pqr

⇒ $\rm p^{3}+q^{3}+r^{3} > 3pqr$

$\Rightarrow \rm3 p q r-(p^{3}+q^{3}+r^{3}) < 0$

∴ D < 0

Hence, option (2) is correct.

Concept:

(1 + x)ⁿ  = ⁿΣ_r-0 nC_r . x^r = [C₀ + C₁ x + C₂ x² + … C_n x_n]

$\large{\rm n_{C_{n-r}}=n_{C_{r}}}$

Calculation:

\(\begin{aligned} &(1+\mathrm{x})^{9}={ }^{9} \mathrm{C}{0}+{ }^{9} \mathrm{C}{1} \mathrm{x}+{ }^{9} \mathrm{C}{2} \mathrm{x}^{2}+\ldots . .^{9} \mathrm{C}{9} \mathrm{x}^{9}\ &\text { Substituting } \mathrm{x}=1, \text { we get }\ &2^{9}={ }^{9} \mathrm{C}{0}+{ }^{9} \mathrm{C}{1}+{ }^{9} \mathrm{C}{2}+\ldots .^{9} \mathrm{C}{9}----(1)\end{aligned} \)

As we know, ⁿC_r = ⁿC_n-r

So,

\({ }^{9} \mathrm{C}{0}+{ }^{9} \mathrm{C}{1}+ +{ }^{9} \mathrm{C}{2} +{ }^{9} \mathrm{C}{3}+ ^{9} \mathrm{C}{4} ={ }^{9} \mathrm{C}{9}+{ }^{9} \mathrm{C}{8}+{ }^{9} \mathrm{C}{7}+{ }^{9} \mathrm{C}{6}+{ }^{9} \mathrm{C}{5} \)     ----(2)

Now, From (1) and (2), we get,

\(2^{9}=2\left[{ }^{9} \mathrm{C}{9}+{ }^{9} \mathrm{C}{8}+{ }^{9} \mathrm{C}{7}+9 \mathrm{C}{6}+{ }^{9} \mathrm{C}_{5}\right]\)

⇒ \(\left[{ }^{9} \mathrm{C}{9}+{ }^{9} \mathrm{C}{8}+{ }^{9} \mathrm{C}{7}+9 \mathrm{C}{6}+{ }^{9} \mathrm{C}_{5}\right] = 2^{8}\)

⇒ 256

Hence, option (1) is correct.

Concept:

If A is non-singular then A^-1 exist.

$\rm A^{-1} = \dfrac{1}{|A|}.adj.A$

 

Calculations:

Given: the matrix A is a non singular matrix of order 3.

Since A is non-singular A^-1 exist.

$\rm A^{-1} = \dfrac{1}{|A|}.adj.A$                      ....(1)

Pre-multiply equation (1) by A

$\rm AA^{-1} = \dfrac{A}{|A|}.adj.A$

⇒ I |A| = A.(adj.A)

⇒A.(Adj. A) = |A|I                       ....(2)

Now, post multiply equation (1) by A

$\rm A^{-1}A =\dfrac{1}{|A|} (adj.A)A$

⇒ I|A| = (adj.A)A

⇒ (adj.A)A = |A|I                     ....(3)

From (2) and (3), we have

The statement A (adj A) = (adj A) A is correct.

 

Also, A is a non-singular matrix.

Hence, A^-1 exist

AA^-1= I

A ($\rm \dfrac{1}{|A|}.adj.A$) = I

A (adj A) = |A| I

take determinants Both side

| A (adj A) | = |  |A| I

We know that if A be an n-rowed square matrix  and ' k'  be any scalar then| K A | = Kⁿ |A|

Now

| A | | adj A| = | A |ⁿ | I_n |

| A | | adj A| = | A |ⁿ

| adj A| = | A |^n-1

The statement |adj A| = |A| is not correct.

Concept:

The equation of line is:

y = mx + c

where m is the slope of the line.

In the square, the diagonals are perpendicular to each other.

Calculations:

Given,

3x + 2y = 5 is the equation of one diagonal of the square.

Consider m₁ be the slope of one diagonal. and m₂ be the slope of other diagonal.

⇒ 2y = - 3x +5

⇒ $\rm y = \dfrac{-3}{2} x+ \dfrac{5}{2}$

⇒ m₁ = $\dfrac{-3}{2}$

In square, the diagonals are perpendicular to each other.

⇒ m₁.m₂ = -1

⇒m₂ = $\rm\dfrac{2}{3}$

Equation of other diagonal is 

 y = m₂x + c

⇒ y = $\rm\dfrac{2}{3}$ x + c....(1)

Since, the other diagonal is passing through the point (1, -1). It must satisfy the equation of line.

Put y = -1 and x = 1 in above equation

⇒ -1 = $\rm\dfrac{2}{3}$ + c

⇒ c = $\rm\dfrac{-5}{3}$

 Put the value of c in equation (1).

⇒y = $\rm\dfrac{2}{3}$ x - $\rm\dfrac{5}{3}$ 

⇒ 3y = 2x - 5

⇒ 2x - 3y = 5 is the equation of other diagonal.

The point (1, -1) is one of the vertices of a square. If 3x + 2y = 5 is the equation of one diagonal of the square, then the equation of the other diagonal is 2x - 3y = 5

Shortcut TrickSince We have to take any point on the given ellipse 

Take y = 0 

Form the equation of ellipse 25x2 + 16y2 = 400

25x2 = 400

x2 = 16

We get x = ±4

So we take P as (4, 0)

Now, Applying the distance formula we get 

PQ + PR = $\sqrt{[(4 - 0)^2 + (0 - 3)^2] }+ \sqrt{[(4 - 0)^2 + (0 + 3)^2] }$

⇒ PQ + PR = 5 + 5 = 10.

 

Alternate Method Concept:

Ellipse: The locus of a point that moves such that its distance from a fixed point and a fixed straight line is always less than one,

(Eccentricity = e < 1) ⇔ $\frac{{{\rm{;}}{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} + \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1$ (b > a)

Sum of the focal distance: The sum of the focal distance of any point on an ellipse is constant and equal to the length of the major axis of the ellipse. 

 

Sum of the focal distance = PS + PS’ = 2a for a > b

Sum of the focal distance = PS + PS’ = 2b for b > a

Focus: (0, ±be) for b > a

Eccentricity: ${\rm{e}} = {\rm{;}}\sqrt{ {1 - {\rm{;}}\frac{{{{\rm{a}}^2}}}{{{{\rm{b}}^2}}}} }$ for b > a

Calculation:

Given: 25x2 + 16y2 = 400

 $\frac{{{{\rm{x}}^2}}}{{16}} + \frac{{{{\rm{y}}^2}}}{25} = 1$

Here,

a2 = 16,

b2 = 25

⇒ a = 4 and b = 5 (a < b)

Eccentricity:

${\rm{e}} = {\rm{;}}\sqrt {{1 - {\rm{;}}\frac{{{{\rm{a}}^2}}}{{{{\rm{b}}^2}}}}} = {\rm{;}\sqrt{ {1 - {\rm{;}}\frac{16}{{25}}}}} = {\rm{;}}\sqrt{\frac{{9}}{{25}}} = {\rm{;}}\frac{3}{5}$ 

Focus = (0, ± be) = (0, ± 3)

So, focus are (0, 3) and (0, -3)

Hence given point Q(0, 3) and R(0, -3) are the focus of the ellipse

PQ + PR = Sum of the focal distance = 2b = 2 × 5 = 10

We have

Circumcentre of Δ is (-1, -1)

Consider,

Equation (1) ⇒ x + 2 = 0

Equation (2) ⇒ y + 2 = 0

Equation (3) ⇒ kx + y + 2 = 0

From equation (1) & (2),

(x, y) = (-2, -2) = C1

From equation (2) & (3),

y = - 2

And, kx + (-2) + 2 = 0

x = 0

(x, y) = (0, 2) = C2

From equation (1) & (3),

x + 2 = 0 ⇒ x = -2

kx + y + 2 = 0

or -2k + y + 2 = 0

or y = (2k - 2)

∴ (x, y) = (-2, 2k - 2) = C3

From the given coordinate, the graph can be drawn as,

Since the given diagram shows that the Δ acts as a right-angle triangle, and for a right-angle triangle, the circumcentre lies at the midpoint of the hypotenuse.

Hence,

(-1, -1) = $(\frac{0-2}{2},\ \frac{-2-2K+2}{2})$

⇒ -1 = - K

Hence, K = 1

Alternate Method

it is clear that the Δ acts as a right-angle triangle and in this Δ the circumcentre lies at the midpoint of the hypotenuse.

We have

Kx + y + 2 = 0

at x = y = -1

⇒ K(-1) + (-1) + 2 = 0

⇒ K = 1

Concept:

The coordinates of the point where the chord cut the parabola satisfIes  the equation of a parabola.

Calculation:

 

 

Given:

The equation of a parabola is y² = x.

The angle made by Chord OA with x-axis is θ

Let the length of the chord OA of the parabola is L 

So, Length of AM = L sinθ 

and Length of OM = L cosθ 

So, The coordinate of A = (L cos θ, L sin θ)

And this point will satisfy the equation of parabola y2 = x.

⇒ (Lsin θ)² = L cos θ

⇒L² sin² θ = L cos θ

⇒ L = cos θ. cosec2 θ

∴ The required length of chord is cos θ. cosec2 θ.

Concept:

Let A = (x1 ,y1) , B = (x2, y2), C = (x3 ,y3) be the points of triangle.

Area of $\rm\triangle ABC$ = $\rm\dfrac{1}{2}\begin{vmatrix} x_1 &y_1 &1 \ x_2 &y_2 &1 \ x_3&y_3 &1 \end{vmatrix}$

If A, B, C be the points of a triangle and colinear then Area of $\rm\triangle ABC$ = 0

Calculations:

Let A = (x1 ,y1) , B = (x2, y2), C = (x3 ,y3) be the points of triangle.

Area of $\rm\triangle ABC$ = $\rm\dfrac{1}{2}\begin{vmatrix} x_1 &y_1 &1 \ x_2 &y_2 &1 \ x_3&y_3 &1 \end{vmatrix}$

Let A = (a, b), B = (c, d) and C = (a - c, b - d) be the points of triangle. They are collinear.

Area of $\rm\triangle ABC$ = 0 

= $\rm\dfrac{1}{2}\begin{vmatrix} x_1 &y_1 &1 \ x_2 &y_2 &1 \ x_3&y_3 &1 \end{vmatrix}$ = 0

⇒$\rm\begin{vmatrix} x_1 &y_1 &1 \ x_2 &y_2 &1 \ x_3&y_3 &1 \end{vmatrix}$ = 0

⇒$\rm\begin{vmatrix} a &b &1 \ c &d &1 \ a - c& b - d &1 \end{vmatrix}$ = 0

⇒ a(d - b + d) - b(bc - dc - ad + cd) + 1(bc - dc - ad + cd) = 0

⇒ ad - bc = 0

⇒ ad = bc

Concept:

The midpoint theorem states that “The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.”

Calculations:

Given, ABC be a triangle. If D(2, 5) and E(5, 9) are the mid-points of the sides AB and AC respectively.

Hence, DE = $\rm\sqrt{(2-5)^2+(5 -9)^2}$

⇒ DE = $\rm\sqrt{9+16}$

⇒ DE = 5

The midpoint theorem states that “The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.”

Since, D(2, 5) and E(5, 9) are the mid-points of the sides AB and AC then their length is half of the third side.

⇒ BC = 2(DE)

⇒ BC = 2(5)

⇒ BC = 10

Concept:

If two nonvertical lines are perpendicular, then the product of their slopes is −1.

The slope of a line passing through the distinct points (x₁, y₁) and (x₂, y₂) is $\rm \frac{y_2-y_1}{x_2-x_1}$

 

Calculation:

 

 

Slope of line passing through points (0, k) and (3, 1)

 $=\rm \frac{1-k}{3-0} \=\frac{1-k}{3}$

3x - 4y - 5 = 0 

⇒4y = 3x - 5

⇒ y = $\frac{3}{4}\rm x-\frac{5}{4}$ 

So, the slope of line 3x - 4y - 5 = 0 is 3/4

Now since line OP and 3x - 4y - 5 = 0 are perepndicular 

$\rm \frac{1-k}{3}\times \frac{3}{4}=-1.....(\text {Product of slopes of perpendicular lines} )\ \Rightarrow 1-k=-4\ \Rightarrow k =5$

Hence, option (3) is correct.

Concept:

The angle between the lines whose slopes are m1 and m2 be given by,

$\rm\tan;\theta = |\dfrac{m_1-m_2}{1+m_1m_2}|$

 

Calculations:

The angle between the lines whose slopes are m1 and m2 be given by,

$\rm\tan;\theta = |\dfrac{m_1-m_2}{1+m_1m_2}|$

Given, m1 = 2 - √3 and m2 = 2 + √3

$\rm\tan;\theta = |\dfrac{(2-\sqrt3)-(2+\sqrt3)}{1+(2-\sqrt3)(2+\sqrt3)}|$

⇒ $\rm\tan;\theta = |\dfrac{-2\sqrt3}{2}|$

⇒ $\rm\tan;\theta = |-\sqrt3|$

⇒ $\rm\tan;\theta =\pm \sqrt3$

⇒ $\rm\theta = \dfrac{\pi}{3}, \dfrac {2\pi} {3}$

Concept:

The distance between the lines ax + by + c₁ = 0 and ax + by + c₂ = 0  is $\rm \frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$

Area of square = (side)² 

 

Calculation:

Here, 3x - 4y - 5 = 0 and 3x - 4y + 15 = 0 are the equations of a pair of opposite sides of a square.

Distance between them =

 =\frac{|-5-15|}{\sqrt {3^2+4^2}}\= \frac{20}{\sqrt 25}\ =\frac{20}{5} \=4

Now area of square = 4² sq. units 

= 16 sq. units 

Hence, option (3) is correct.

Concept:

A hermitian matrix is a square matrix, which is equal to its conjugate transpose matrix.

The non-diagonal elements of a hermitian matrix are all complex numbers.

The complex numbers in a hermitian matrix are such that the element of the i^th row and j^th column is the complex conjugate of the element of the j^th row and i^th column.

Example:

\left[ {\begin{array}{*{20}{c}} {3}&{\rm{3+2i}}\ { {\rm{3-2i}}}&{2} \end{array}} \right]

Calculations:

Given, matrix {\rm{A}} = \left[ {\begin{array}{*{20}{c}} {1 - {\rm{i}}}&{\rm{i}}\ { - {\rm{i}}}&{1 - {\rm{i}}} \end{array}} \right] where $\rm i = \sqrt {-1},$

⇒      {\rm{\bar A}} = \left[ {\begin{array}{*{20}{c}} {1 + {\rm{i}}}&{\rm{-i}}\ { {\rm{i}}}&{1 + {\rm{i}}} \end{array}} \right]

⇒  {\rm({\bar A})^T} = \left[ {\begin{array}{*{20}{c}} {1 + {\rm{i}}}&{\rm{i}}\ { {\rm{-i}}}&{1 + {\rm{i}}} \end{array}} \right]

 \({\rm({\bar A})^T}+ A = \left[ {\begin{array}{{20}{c}} {1 + {\rm{i}}}&{\rm{i}}\ { {\rm{-i}}}&{1 + {\rm{i}}} \end{array}} \right]+\left[ {\begin{array}{{20}{c}} {1 - {\rm{i}}}&{\rm{i}}\ { {\rm{-i}}}&{1 - {\rm{i}}} \end{array}} \right]\)

⇒ {\rm({\bar A})^T}+ A = \left[ {\begin{array}{*{20}{c}} {2}&{\rm{2i}}\ { {\rm{-2i}}}&{2} \end{array}} \right]

Since, the diagonal elements of (A̅)T + A is real.

(A̅)^T + A is hermitian is correct statement.

Concept:

(x + y)ⁿ = ⁿΣ_r=0 nC_r x^{n – r} · y^r 

In independent term: Power of variables will be zero.

 

Calculation:

 $\rm \left( \frac {2}{x^2} - \sqrt x \right)^{10}=\sum 10C_r(\frac{2}{x^2})^{10-r}(-x^{1/2})^r\ ⇒\sum 10C_r({2}{x^{-2}})^{10-r}(-x^{1/2})^r\ ⇒\sum 10C_r({2}^{10-r}{x^{-20+2r}})(-x^{\frac{r}{2}})\ ⇒\sum 10C_r({2}^{10-r}{x^{-20+2r+\frac{r}{2}}})(-1^{\frac{r}{2}})\$

Now, for independent term, -20 + 2r + r/2 = 0

⇒(4r + r)/2 = 20

⇒ 5r = 40

⇒ r = 8

So, the independent term=

 $\rm 10 C_r\times2^{10-r}\times (-1)^{r/2}\ ⇒ 10 C_8\times2^{10-8}\times (-1)^{8/2}\ ⇒ \frac{10!}{8!2!}\times2^{2}\times 1\ ⇒\frac{10\times 9}{2}\times 4\ ⇒180$

Hence, option (1) is correct.

Calculation:

(1 + 2x - x²)⁶ = a₀ + a₁x + a₂x² + ... + a₁₂x¹²

On putting x = -1 in the above expression, we get

(1 - 2 - 1)⁶ = $\rm a_0-a_1+a_2-a_3...+a_{12}$

⇒  $\rm a_0+a_1+....+a_{12}= 64$ 

Hence, option (2) is correct.

Concept:

C(n, r) = $\rm\dfrac{n!}{r!(n - r)!}$

n! = n. (n - 1)!

Calculations:

Given, C(20, n + 2) = C(20, n - 2)

$\rm\frac{20!}{(n+2)!(18-n)!} = \rm\frac{20!}{(n-2)!(22-n)!}$

$\rm\frac{1}{(n+2)!(18-n)!} = \rm\frac{1}{(n-2)!(22-n)!}$

we know that

(n + 2)! =(n - 2)! (n - 1)n(n + 1)(n+2)

(22 - n)! = (18 - n)! (19 - n)(20 - n)(21 - n)(22 - n)

$\rm\frac{1}{((n - 2)! (n - 1)n(n + 1)(n+2)(18-n)!} = \rm\frac{1}{(n-2)!(18 - n)! (19 - n)(20 - n)(21 - n)(22 - n))}$

$\rm\frac{1}{(n - 1)n(n + 1)(n+2)} = \rm\frac{1}{ (19 - n)(20 - n)(21 - n)(22 - n))}$

$\rm {(n - 1)n(n + 1)(n+2)} = \rm (19 - n)(20 - n)(21 - n)(22 - n)$

Now , put n = 10 

L.H.S = (n - 1)n(n + 1)(n+2) = (9)(10)(11)(12) 

R.H.S = (19 - n)(20 - n)(21 - n)(22 - n) = (9)(10)(11)(12) 

L.H.S. = R.H.S.

Hence, if C(20, n + 2) = C(20, n - 2), then n = 10

Alternate MethodWe have,

C(20, n + 2) = C(20, n - 2)

⇒ $^{20}C_{(n+2)}=^{20}C_{(n−2)}$

Consider,

n = 20

x = (n + 2)

y = (n - 2)

x ≠ y,

∵ n = 20 and, n ≠ 0

x + y = n (Possible)

In that case,

(n + 2) + (n - 2) = 20

or, 2n = 20

Hence, n = 10