NDA Mathematics 4 Sep 2022 Official Paper (Previous Year Paper)

Calculation:

Odd digits - 1, 3, 5, 7, 9

A four-digit number with all digits odd has 5 choices in the first position, and since digits can be repeated, there are 5 choices for the second, third, and fourth positions also.

So, the number of four-digit natural numbers such that all of the digits are odd = 5 × 5 × 5 × 5 = 625

Concept:

Binomial expansion of (x + y)ⁿ is given by

(x + y)n = ⁿC₀xⁿ + nC1xn-1y + nC2xn-2y²+.....+ nCn-1xy^n-1 + nCnyn  

Calculation:

Given,  $\displaystyle\sum_{r=0}^n$2r C(n, r)

Expanding the expression,

⇒ nC020 + nC1 2¹ + nC222+.....+ nCn-12n-1 + nCn2n  

⇒ nC01n 20 + nC1 1n-1 2¹ + nC21n-2 22+.....+ nCn-1 2n-1 + nCn2n  

Comparing with binomial expansion x = 1 and y = 2

⇒  $\displaystyle\sum_{r=0}^n$2r C(n, r) = (1 + 2)ⁿ

⇒  $\displaystyle\sum_{r=0}^n$2r C(n, r) = 3n

∴ The correct option is (2).

Concept:

The number of ways of arranging n objects out of which p are of one type, q are of the second type is $n! \over p!q!$

Calculation:

Letters in the word 'MATHEMATICS' in alphabetical order are

AACEHIMMTTS

So in the list, the words that start with A only lie before the words start with C.

Now the number of words start with A are: 

{fix A at first position}

= $10! \over 2! 2!$ = 907200

∴ The correct answer is option (3).

Concept:

A relation f from set A to set B is said to be a function if every element of set A has a unique image in set B.

Calculation:

1. If f is the subset of Z × Z defined by f = {(xy, x − y); x, y ∈ Z},

let x = 2, y = 3 → (2 × 3, 2 - 3) = (6, -1) ∈ f

and for x = -2, y = -3 → (-2 × (-3), -2 + 3) = (6, 1) ∈ f

So, the element 6 ∈ Z has two different images 1 and -1 in Z 

⇒ f is not a function.

2. If f is the subset of N × N defined by f = {(xy, x + y); x, y ∈ N},

 let x = 3, y = 4→ (3 × 4, 3 + 4) = (12, 7) ∈ f

and for x = 2, y = 6 → (2 × (6), 2 + 6) = (12, 8) ∈ f

So, the element 12 ∈ N has two different images 7 and 8 in N

⇒ f is not a function.

∴ The correct option is (4).

Explanation:

Given:

Δ = \(\left|\begin{array}{lll}\text{a}{11} & \text{a}{12} & \text{a}{13} \ \text{a}{21} & \text{a}{22} & \text{a}{23} \ \text{a}{31} & \text{a}{32} & \text{a}_{33}\end{array}\right|\)

 a13 = yz, a23 = zx, a33 = xy and M₁₃ = (z − y), M₂₃ = (z − x), M₃₃ = (y − x)

Expanding the determinant along Column 3,

⇒ Δ = a13 (a₂₁a₃₂ - a₂₂a₃₁) - a23 (a₁₁a₃₂ - a₁₂a₃₁) + a33 (a11a2 - a12a1)

⇒ Δ = a₁₃M₁₃ - a₂₃M₂₃ + a₃₃M₃₃

where M_ij denotes the minor of a_ij element.

⇒ Δ = yz(z - y) - zx(z - x) + xy(y - x)

⇒ Δ = yz(z - y) - z²x + zx² + xy² - x²y

⇒ Δ = yz(z - y) + (xy2 - z2x) + (zx2 - x2y)

⇒ Δ = yz(z - y) + x(y2 - z2) + x²(z - y)

⇒ Δ = (z - y)[yz - x(y + z) + x2]

⇒ Δ = (z - y)[yz - xy - xz + x2]

⇒ Δ = (z - y)[y(z - x) - x(z - x)]

⇒ Δ = (z - y)[(z - x)(y - x)]

∴ The correct answer is option (1).

Concept:

If a matrix A = \(\left(\begin{array}{lll}\text{a}{11} & \text{a}{12} & \text{a}{13} \ \text{a}{21} & \text{a}{22} & \text{a}{23} \ \text{a}{31} & \text{a}{32} & \text{a}{33}\end{array}\right)\) then AT = \(\left(\begin{array}{lll}\text{a}{11} & \text{a}{21} & \text{a}{31} \ \text{a}{12} & \text{a}{22} & \text{a}{32} \ \text{a}{13} & \text{a}{32} & \text{a}{33}\end{array}\right)\)

Explanation:

If A = $\left(\begin{array}{ccc}1 & 0 & 0 \ 0 & \cos θ & \sin θ \ 0 & \sin θ & −\cos θ\end{array}\right)$, 

the cofactor matrix of A is given by,

cofactor(A) = $\left(\begin{array}{ccc}-1 & 0 & 0 \ 0 & -\cos θ & -\sin θ \ 0 & -\sin θ & \cos θ\end{array}\right)$

So, adj A = $\left(\begin{array}{ccc}-1 & 0 & 0 \ 0 & -\cos θ & -\sin θ \ 0 & -\sin θ & \cos θ\end{array}\right)$ (Transpose of cofactor matrix)

⇒ adj A = - A __(i)

⇒ adj A + A = 0 is a null matrix

So, statement 1 is correct.

Now, expanding A along Row 1,

|A| = 1(-cos²θ - sin²θ) = -1

⇒ A^-1 = $1 \over |A|$adj A = -adj A  

⇒ A-1 + adj A = 0 __(ii)

So, statement 2 is correct.

Putting the value of (i) in (ii),

⇒ A-1 - A = 0  

or A - A^-1 = 0

So, statement 3 is correct.

∴ The correct option is (4).

Concept:

If A is a matrix of order m × n and B is a matrix of order p × q, then AB is a matrix of order m × q with mq number of elements.

and AB is defined if and only if n = p

Calculation:

If X is a matrix of order 3 × 3, Y is a matrix of order 2 × 3 and Z is a matrix of order 3 × 2,

1.  (ZY)X is a square matrix having 9 entries.

Order of Z = 3 × 2, order of Y = 2 × 3,

⇒ Order of (ZY) = 3 × 3

And order of (ZY) = 3 × 3, order of X = 3 × 3,

⇒ order of (ZY)X = 3 × 3

⇒ (ZY)X is a square matrix having 3 × 3 = 9 entries.

So, statement 1 is true

2.  Y(XZ) is a square matrix having 4 entries.

Order of X = 3 × 3, order of Z = 3 × 2,

⇒ order of (XZ) = 3 × 2

And order of Y = 2 × 3, order of (XZ) = 3 × 2, 

⇒ order of Y(XZ) = 2 × 2

⇒ Y(XZ) is a square matrix having 2 × 2 = 4 entries.

So, statement 2 is true

3. X(YZ) is not defined.

Order of Y = 2 × 3, order of Z = 3 × 2,

⇒ order of YZ = 2 × 2

order of X = 3 × 3, order of YZ = 2 × 2

and 3 ≠ 2

⇒  X(YZ) is not defined.

So, statement 3 is true

∴ The correct answer is option (4).

Concept:

For a quadratic equation ax² + bx + c = 0

Sum of roots = -b/a

and the product of roots = c/a

Calculation:

If the sum of roots = product of roots

⇒ $\frac{-b}{a}=\frac ca$

⇒ -b = c

So, there can be infinitely many quadratic equations with -b = c and a ≠ 0

∴ The correct option is (4).

Concept:

Between any two real numbers, there are infinitely many rationals and irrational.

{The numbers which are not rational are irrational}

Real = Rational ∪ Irrational

Explanation:

1. The set of all irrational numbers between $\sqrt{2}$ and $\sqrt{5}$ is an infinite set.

There are infinitely many rationals and irrational between any two real numbers.

 $\sqrt{2}$ and $\sqrt{5}$ are two irrational numbers and there are infinitely many rationals and irrationals between them.

⇒ The set of all irrational numbers between $\sqrt{2}$ and $\sqrt{5}$ is an infinite set.

So, statement 1 is true.

2. The set of all odd integers less than 100 is a finite set.

The set of all odd integers less than 100 is given by

{......,-5 ,-3 ,-1 ,1 ,3 ,5 ,.....,95, 97, 99}

this set is bounded above but not bounded below

So, It is an infinite set, and statement 2 is false.

∴ The correct option is (1).

Concept:

• The sum of first n natural numbers = 1 + 2 + 3 + .. + n = $n(n+1) \over 2$
• The series of form a, a + d, a + 2d, ....., a + (n - 1)d is an AP and the sum of the first n terms of an A.P. is ${n \over 2}(a + a_n)$ where a_n is the n^th term.
• A number is prime if it has no factors except 1 and itself

Calculation:

1. 2 + 4 + 6 + ........ + 2n = n2 + n

 2 + 4 + 6 + ........ + 2n is an A.P.

with first term 2, number of terms n and last term 2n 

⇒ Sum of A.P. = ${n \over 2}(2 + 2n)$ = n(n + 1) = n2 + n

So statement 1 is true

2. The expression n2 + n + 41 always gives a prime number for every natural number n

If n = 41

n2 + n + 41 = 412 + 41 + 41= 41(41 + 1 + 1) = 41 × 43

As n2 + n + 41  has factors 41 and 43

⇒ n2 + n + 41 is not prime for n = 41 ∈ N

So, statement 2 is false

∴ The correct answer is option (1)

Concept:

If α and β are the roots of the equation ax² + bx + c = 0,

then sum of roots = α + β = -$b \over a$

and product of roots = αβ = $c \over a$

Calculation:

If p, q(p > q) are the roots of the quadratic equation x2 + bx + c = 0 

then, p + q = -b  __(i)

and pq = c   __(ii)

Squaring equation (i),

⇒ (p + q)² = (-b)²

⇒ p² + q² + 2pq = b²

⇒ p2 + q2 + 2c = b2  (From (i))

⇒ p2 + q2 = b2 - 2c  __(iii)

Take equation (iii) - 11(ii),

⇒ p2 + q2 − 11pq = b2 - 2c - 11c = 0

⇒ b² = 13c  __(iv)

Now (p - q)² =  p2 + q2 - 2pq

Put values from (ii) and (iii),

⇒ (p - q)2 =  b2 - 2c  - 2c = b2 - 4c

⇒ (p - q)2 =  13c - 4c {from (iv)}

⇒ (p - q)2 =  9c

⇒ p - q = 3√c

∴ The correct answer is option (1).

Concept:

Sine Rule:

In a triangle Δ ABC, where a is the side opposite to A, b is the side opposite to B, c is the side opposite to C and where R is the circum-radius:

$\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R$

The right-angled triangle follow the Pythagoras theorem,

Hypotenuse2 = Perpendicular2 + Base2

The hypotenuse is the greatest side of the right-angled triangle

Solution:

Given, Length of sides of polygon a = 1

Number of sides of polygon n = 12

 

The right-angled triangle follow the Pythagoras theorem,

$\frac{sin 15}{0.5}=\frac{sin 75}{R}$

$\Rightarrow \frac{\sqrt3-1}{(2\sqrt2)0.5}=\frac{\sqrt6+\sqrt2}{4R}$

$\Rightarrow \frac{2(\sqrt3-1)}{(2\sqrt2)}=\frac{\sqrt6+\sqrt2}{4R}$

$\Rightarrow \frac{(\sqrt3-1)}{(\sqrt2)}=\frac{\sqrt6+\sqrt2}{4R}$

$\Rightarrow R=\frac{\sqrt6+\sqrt2}{4}\times \frac{\sqrt2}{\sqrt3-1}$

$\Rightarrow D = 2R=\frac{\sqrt6+\sqrt2}{4}\times \frac{2\sqrt2}{\sqrt3-1}$ ...... (Diameter D = 2R)

$\Rightarrow D =\frac{\sqrt6+\sqrt2}{2}\times \frac{\sqrt2}{\sqrt3-1}$

$\Rightarrow D =\frac{\sqrt6+\sqrt2}{\sqrt2 \sqrt2}\times \frac{\sqrt2}{\sqrt3-1}$

$\Rightarrow D =\frac{\sqrt6+\sqrt2}{\sqrt2}\times \frac{1}{\sqrt3-1}$

$\Rightarrow D =\frac{\sqrt6+\sqrt2}{\sqrt6-\sqrt2}$

$\Rightarrow D =\frac{\sqrt6+\sqrt2}{\sqrt6-\sqrt2} \times \frac{\sqrt6+\sqrt2}{\sqrt6+\sqrt2}$

$\Rightarrow D =\frac{6+2\sqrt{12}+2}{4}$

$\Rightarrow D =\frac{8+2\sqrt{12}}{4}$

$\Rightarrow D =\frac{8+2\sqrt{4 \times3}}{4}$

$\Rightarrow D =2+\sqrt3$

Concept:

The range of a function is the set of all possible outputs of the function.

Calculation:

Given  A = {7, 8, 9, 10, 11, 12, 13; 14, 15, 16} 

and f ∶ A → N be defined by f(x) = the highest prime factor of x.

Range of f = f(A)

⇒ Range of f = {f(7), f(8), f(9), f(10), f(11), f(12), f(13); f(14), f(15), f(16)}

⇒ Range of f = {f(7), f(2³), f(3²), f(2 × 5), f(11), f(2² × 3), f(13); f(2 × 7), f(3 × 5), f(2⁴)}  

⇒ Range of f = {7, 2, 3, 5, 11, 3, 13; 7, 5, 2}  

⇒ Range of f = { 2, 3, 5, 7,11,13}

⇒ |Range of f| = 6

∴ The correct answer is option (3).

Explanation:

Given, R = {(x, y): x, y ∈ N and x2 = y3}.

1. (x, x) ∈ R for all x ∈ N

 If (x,x) ∈ R ⇒ x2 = x3

⇒ x²(x - 1) = 0 

⇒ (x, x) ∈ R for x = 0, 1 only

So, statement 1 is not correct.

2. (x, y) ∈ R ⇒ (y, x) ∈ R

 let x = 8 and y = 4

⇒ x² = 64 = 4³ = y³

⇒ (x,y) ∈ R 

but y² = 16 ≠ 8³ ⇒ (y,x) ∉ R

So, statement 2 is not correct.

3. (x, y) ∈ R and (y, z) ∈ R ⇒ (x, z) ∈ R

 If (x, y) ∈ R and (y, z) ∈ R

⇒ x2 = y3 and y2 = z3}

⇒ x2 = (z^3/2)3 

⇒ x2 = (z3/2)3 ≠ z^9/2

So, statement 3 is not correct.

∴ The correct answer is option (4).

Concept:

A ∪ B = {x| x ∈ A or x ∈ B}

A ∩ B = {x| x ∈ A and x ∈ B}

Explanation:

1. A ∩ B = A ∩ C ⇒ B = C

 If A = {1,2,3,4,5}, B = {1,2,p,q} and C = {1,2,m,n}

Here, A ∩ B = A ∩ C = {1,2}

But B ≠ C

So statement 1 is false.

2. A ∪ B = A ∪ C ⇒ B = C

 If A = {1,2,3,4}, B = {1,2,p,q} and C = {p,q}

Here,A ∪ B = A ∪ C = {1,2,3,4,p,q}

But B ≠ C

So statement 2 is false.

∴ The correct option is (4).

Concept:

If z = x + iy 

⇒ Modulus of z = $|z| = \sqrt {x^2 +y^2}$

Calculation:

Given, 

z = $\frac{1+\text{i}\sin θ}{1−\text{i}\sin θ}$ where i = $\sqrt{−1}$

Multiply numerator and denominator by (1 + i sin θ),

$⇒ z =\frac{1+\text{i}\sin θ}{1−\text{i}\sin θ} \times \frac{1+\text{i}\sin θ}{1+\text{i}\sin θ}$,

$⇒ z =\frac{(1+\text{i}\sin θ)^2}{1^2−(\text{i}\sin θ)^2}$

$⇒ z =\frac{(1+2{i}\sin θ - \sin^2\theta )}{1+\sin^2 θ}$

$⇒ z =\frac{(1 - \sin^2\theta )}{1+\sin^2 θ} +{i}\frac{2\sin θ}{1+\sin^2 θ}$

$⇒ z = \frac{ \cos^2\theta }{1+\sin^2 θ} +{i}\frac{2\sin θ}{1+\sin^2 θ}$

$⇒ |z| =\sqrt{(\frac{ \cos^2\theta }{1+\sin^2 θ})^2 +(\frac{2\sin θ}{1+\sin^2 θ})^2}$

$⇒ |z| ={1 \over 1+ \sin^2 \theta }\sqrt{\cos^4 \theta + 4\sin^2 \theta}$

$⇒ |z| = {1 \over 1+ \sin^2 \theta }\sqrt{(1 -\sin^2 \theta)^2 + 4\sin^2 \theta}$

$⇒ |z| ={1 \over 1+ \sin^2 \theta }\sqrt{1 -2\sin^2 \theta + \sin^4 \theta + 4\sin^2 \theta}$

$⇒ |z| ={1 \over 1+ \sin^2 \theta }\sqrt{1 +2\sin^2 \theta + \sin^4 \theta }$

$⇒ |z| ={1+ \sin^2 \theta \over 1+ \sin^2 \theta } = 1$

∴ The correct answer is option (1).

Concept:

If z = x + iy is purely real, then y = 0

Calculation:

Given, 

z = $\frac{1+\text{i}\sin θ}{1−\text{i}\sin θ}$ where i = $\sqrt{−1}$

Multiply numerator and denominator by (1 + i sin θ) 

$⇒ z =\frac{1+\text{i}\sin θ}{1−\text{i}\sin θ} \times \frac{1+\text{i}\sin θ}{1+\text{i}\sin θ}$

$⇒ z =\frac{(1+\text{i}\sin θ)^2}{1^2−(\text{i}\sin θ)^2}$

$⇒ z =\frac{(1+2{i}\sin θ - \sin^2\theta )}{1+\sin^2 θ}$

$⇒ z =\frac{(1 - \sin^2\theta )}{1+\sin^2 θ} +{i}\frac{2\sin θ}{1+\sin^2 θ}$

Since z is purely real,

⇒ $\frac{2\sin θ}{1+\sin^2 θ} = 0$

⇒ sin θ = 0

⇒ θ = nπ 

∴ The correct answer is option (3).

Concept:

If z = x + iy is purely imaginary, then x = 0

Calculation:

Given, 

z = $\frac{1+\text{i}\sin θ}{1−\text{i}\sin θ}$ where i = $\sqrt{−1}$

Multiply numerator and denominator by (1 + i sin θ), 

$⇒ z =\frac{1+\text{i}\sin θ}{1−\text{i}\sin θ} \times \frac{1+\text{i}\sin θ}{1+\text{i}\sin θ}$,

$⇒ z =\frac{(1+\text{i}\sin θ)^2}{1^2−(\text{i}\sin θ)^2}$

$⇒ z =\frac{(1+2{i}\sin θ - \sin^2\theta )}{1+\sin^2 θ}$

$⇒ z =\frac{(1 - \sin^2\theta )}{1+\sin^2 θ} +{i}\frac{2\sin θ}{1+\sin^2 θ}$

$⇒ z = \frac{ \cos^2\theta }{1+\sin^2 θ} +{i}\frac{2\sin θ}{1+\sin^2 θ}$

Since z is purely imaginary,

⇒ $\frac{\cos^2 θ}{1+\sin^2 θ} = 0$

⇒ cos²θ = 0

⇒ θ =  $\frac{(2n+1)\pi}{2}$

∴ The correct answer is option (2).

Solution:

There are two AP with the different first terms and common difference.

For the first AP A:

Let first term be a and common difference = d

Sum of n terms P = $\frac{n}{2}[2a+(n-1)d]$

and n^th term a_n = a + (n - 1)d

For the second AP B:

Let first term be M and common difference = D

Sum of n terms Q = $\frac{n}{2}[2M+(n-1)D]$

and n^th term M_n = M + (n - 1)D

It is given that

$\frac{Sum\ of\ n\ terms\ of\ first\ A.P}{Sum\ of\ n\ terms\ of\ second\ A.P }= \frac{5n+4}{9n+6}$

⇒ $\frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2M+(n-1)D] }= \frac{5n+4}{9n+6}$

⇒ $\frac{[2a+(n-1)d]}{[2M+(n-1)D] }= \frac{5n+4}{9n+6}$

⇒ $\frac{a+\left ( \frac{n-1}{2} \right )d}{M+\left ( \frac{n-1}{2} \right )D}= \frac{5n+4}{9n+6}$.....(1)

Now, we need to find the ratio of the first term of A and B

$\frac{1^{st}\ term\ of\ A}{1^{st}\ term\ of\ B}$

= $\frac{a}{M}$

Hence, $\frac{n-1}{2}=0\Rightarrow n=1$

Putting n = 1 in equation (1), we get

$\frac{a}{M}=\frac{5(1)+4}{9(1)+6}$

⇒ $\frac{a}{M}=\frac{9}{15}$

⇒ $\frac{a}{M}=\frac{3}{5}$

⇒ $\frac{1^{st}\ term\ of\ A}{1^{st}\ term\ of\ B} =\frac{3}{5}$

Hence the ratio of the first term of A and B = $\frac{3}{5}$

∴ The correct option is (4)

Solution:

There are two AP with the different first terms and common difference.

For the first AP A:

Let first term be a and common difference = d

Sum of n terms P = $\frac{n}{2}[2a+(n-1)d]$

and nth term an = a + (n - 1)d

For the second AP B:

Let first term be M and common difference = D

Sum of n terms Q = $\frac{n}{2}[2M+(n-1)D]$

and nth term Mn = M + (n - 1)D

It is given that

$\frac{Sum\ of\ n\ terms\ of\ first\ A.P}{Sum\ of\ n\ terms\ of\ second\ A.P }= \frac{5n+4}{9n+6}$

⇒ $\frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2M+(n-1)D] }= \frac{5n+4}{9n+6}$

⇒ $\frac{[2a+(n-1)d]}{[2M+(n-1)D] }= \frac{5n+4}{9n+6}$

⇒ $\frac{a+\left ( \frac{n-1}{2} \right )d}{M+\left ( \frac{n-1}{2} \right )D}= \frac{5n+4}{9n+6}$.....(1)

Now, we need to find the ratio of the 10^th term of A and B

$\frac{10^{th}\ term\ of\ A}{10^{th}\ term\ of\ B}$

= $\frac{a_{10}\ of\ first\ A.P}{M_{10}\ of\ second\ A.P}$

= $\frac{a+(10-1)d}{M+(10-1)D}$

= $\frac{a+9d}{M+9D}$                 ---(2)

From equation (1) & (2)

Hence, $\frac{n-1}{2}=9\Rightarrow n=19$

Putting n = 19 in equation (1), we get

$\frac{a+9d}{M+9D}=\frac{5(19)+4}{9(19)+6}$

 

⇒ $\frac{a+9d}{M+9D}=\frac{99}{177}$

⇒ $\frac{a+9d}{M+9D}=\frac{33}{59}$

⇒ $\frac{10^{th}\ term\ of\ A}{10^{th}\ term\ of\ B} =\frac{33}{59}$

Hence the ratio of the 10^th term of A and B = $\frac{33}{59}$

∴ The correct option is (3)

Solution:

There are two AP with the different first terms and common difference.

For the first AP A:

Let the first term be a and common difference = d

Sum of n terms P = $\frac{n}{2}[2a+(n-1)d]$

and nth term an = a + (n - 1)d

For the second AP B:

Let the first term be M and common difference = D

Sum of n terms Q = $\frac{n}{2}[2M+(n-1)D]$

and nth term Mn = M + (n - 1)D

It is given that

$\frac{Sum\ of\ n\ terms\ of\ first\ A.P}{Sum\ of\ n\ terms\ of\ second\ A.P }= \frac{5n+4}{9n+6}$

⇒ $\frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2M+(n-1)D] }= \frac{5n+4}{9n+6}$

⇒ $\frac{[2a+(n-1)d]}{[2M+(n-1)D] }= \frac{5n+4}{9n+6}$

⇒ $\frac{a+\left ( \frac{n-1}{2} \right )d}{M+\left ( \frac{n-1}{2} \right )D}= \frac{5n+4}{9n+6}$.....(1)

Now, we need to find the ratio of the first term of A and B

$\frac{1^{st}\ term\ of\ A}{1^{st}\ term\ of\ B}$

= $\frac{a}{M}$

Hence, $\frac{n-1}{2}=0\Rightarrow n=1$

Putting n = 1 in equation (1), we get

$\frac{a}{M}=\frac{5(1)+4}{9(1)+6}$

⇒ $\frac{a}{M}=\frac{9}{15}$

⇒ $\frac{a}{M}=\frac{3}{5}$

⇒ $\frac{1^{st}\ term\ of\ A}{1^{st}\ term\ of\ B} =\frac{3}{5}$

Now, we need to find the ratio of the second term of A and B

$\frac{2^{nd}\ term\ of\ A}{2^{nd}\ term\ of\ B}$

= $\frac{a+(2-1)d}{M+(2-1)D}$

= $\frac{a+d}{M+D}$

Hence, $\frac{n-1}{2}=1\Rightarrow n=3$

Putting n = 3 in equation (1), we get

$\frac{a+d}{M+D}=\frac{5(3)+4}{9(3)+6}$

⇒ $\frac{a+d}{M+D}=\frac{19}{33}$

⇒ $\frac{2^{nd}\ term\ of\ A}{2^{nd}\ term\ of\ B} =\frac{19}{33}$

Let the first term of A = 3 and the second term of A = 19

then common difference d = 19 - 3 = 16

Let the first term of B = 5 and the second term of B = 33

then common difference D = 33 - 5 = 28

Hence, D > d

∴ The correct option is (1)

Concept:

General term of binomial expansion:

The general term in the expansion of (x + y)ⁿ is given by:

 T_r+1 = ⁿC_r x^r y^n-r

Calculation:

In the binomial expansion of (p + qx)9 :

the general term in the expansion of (p + qx)9 is given by:

 Tr+1 = 9Cr p9 (qx)9-r

The term with x³ when 9 - r = 3  __(i)

⇒ r = 9 - 3 = 6

⇒ The term with x³ = T7 = $^9C_6 \ p^6 (qx)^{9-6}$ = ${9! \over 6!(9-6)!}$p⁶q³x³

⇒ The term with x3 =  84 p6q3x3

⇒ Coefficient of x³ = 84p6q3 __(i)

The term with x6 when 9 - r = 6

⇒ r = 9 - 6 = 3

⇒ The term with x6 = T4 = $^9C_3 \ p^3 (qx)^{9-3}$ 

⇒ The term with x6 = ${9! \over 3!(9-3)!}$p3q6x6

⇒ Coefficient of x6 = 84p3q6 __(ii)

If the coefficients of x3 and x6 are equal

Equating equation (i) and (ii),

⇒  84p6q3 = 84p3q6

⇒ p³ = q³

⇒  p = q

∴ The correct option is (1).

Concept:

General term of binomial expansion:

The general term in the expansion of (x + y)n is given by:

 Tr+1 = nCr xr yn-r

Middle term in binomial expansion: The middle terms in the expansion of (x + y)n depend on the value of n.

• If n is even, there will be n + 1 term in the expansion. So, there is only one middle term and the middle term is ($n \over 2$ + 1)^th term.
• If n is odd, there will be n +1 term in the expansion. So, there will be two middle terms which are  ($n+1 \over 2$)th and ($n+3 \over 2$)th term.

 

Calculation:

In the binomial expansion of (p + qx)9, there are 9 + 1 = 10 terms.

Since n is odd, there will be two middle terms.

⇒The middle terms are ($9+1 \over 2$)th and ($9+3 \over 2$)th term, that is, 5^th and 6^th terms.

In the expansion of (p + qx)9,  

5^th term = T₅ = T₄₊₁  

⇒ T4+1 = $^9C_4 \ p^4 (qx)^{9-4}$ 

⇒ T4+1 = ${9! \over 4!(9-4)!} \ p^4 (qx)^{5}$ = 126p4q5x5

⇒ Coefficient of 5^th term = 126p4q5 __(i)

6^th term = T6 = T5+1  

⇒ T5+1 = $^9C_5 \ p^5 (qx)^{9-5}$ 

⇒ T5+1 = ${9! \over 5!(9-5)!} \ p^5 (qx)^{4}$ = 126p5q4x4

⇒ Coefficient of 6^th term = 126p5q4 __(ii)

The ratio of the coefficients of middle terms in the expansion

= Coefficient of 5^th term: Coefficient of 6^th term 

Putting values from equations (i) and (ii),

The ratio of the coefficients of middle terms in the expansion = 126p5q4 : 126p4q5

⇒ The ratio of the coefficients of middle terms in the expansion = p : q

∴ The correct option is (2).

Concept:

General term of binomial expansion:

The general term in the expansion of (x + y)n is given by:

 Tr+1 = nCr xr yn-r

Calculation:

In the binomial expansion of (p + qx)9 :

the general term in the expansion of (p + qx)9 is given by:

 Tr+1 = 9Cr pr (qx)9-r

The term with x2 when 9 - r = 2  __(i)

⇒ r = 9 - 2 = 7

⇒ The term with x2 = T8 = $^9C_7 \ p^7 (qx)^{9-7}$ 

⇒ The term with x2 = ${9! \over 7!(9-7)!}$p7q2x2

⇒ The term with x2 =  36p7q2x2

⇒ Coefficient of x2 = 36p7q2__(i)

And the term with x4 when 9 - r = 4

⇒ r = 9 - 4 = 5

⇒ The term with x4 = T5+1 = $^9C_5 \ p^5 (qx)^{9-5}$ 

⇒ The term with x4 = ${9! \over 5!(9-5)!}$p5q4x4

⇒ The term with x4 = 126p5q4x4

⇒ Coefficient of x4 = 126p5q4 __(ii)

If the coefficients of x2 and x4 are equal

From (i) and (ii),

⇒  36p7q2 = 126p5q4

⇒ 2p2 = 7q2

⇒  p2 ∶ q2 = 7 ∶ 2

∴ The correct option is (2).

Concept:

The number of ways of choosing m objects out of n different objects = ⁿC_m

The number of ways of arranging n objects = n!

Explanation:

Given word 'QUESTION' which has 8 different letters

{4 vowels (U,E,I,O) and 4 consonants(Q,S,T,N)}

The number of ways of choosing two vowels from 4 vowels = 4C2 = 6

And the number of ways of choosing two vowels from 4 vowels = 4C2 = 6

⇒ The number of ways of choosing two vowels and two consonants from 4 vowels and 4 consonants = ⁴C₂ × 4C2 = 36

And the number of ways of arranging these four different letters = 4! = 24

⇒ Number of 4-letter words each of two vowels and two consonants with or without meaning = 36 × 24 = 864

∴ The correct option is (4).

Concept:

The number of ways of choosing m objects out of n different objects = nCm

The number of ways of arranging n objects = n!

Explanation:

Given word 'QUESTION' which has 8 different letters

{4 vowels (U,E,I,O) and 4 consonants(Q,S,T,N)}

If vowels occupy odd positions and consonants occupy even positions,

then 4 consonants are to be arranged in 4 positions and 4 vowels are to be arranged in 4 positions

As the number of ways of arranging 4 objects in 4 positions = 4! = 24

⇒ The number of ways arranging 4 vowels in 4 positions and 4 consonants in 4 positions = 24 × 24 = 576

Similarly, for the case when vowels occupy even positions and consonants occupy odd positions

So, the number of 8-letter words that can be formed such that consonants and vowels occupy alternate positions = 576 × 2 = 1152

∴ The correct answer is option (3).

Concept:

The number of ways of choosing m objects out of n different objects = nCm

The number of ways of arranging n objects = n!

Explanation:

Given word 'QUESTION' which has 8 different letters

{4 vowels (U,E,I,O) and 4 consonants(Q,S,T,N)}

If all consonants are together consider them as one string, which can be rearranged internally in 4! ways 

⇒ Number of ways of rearranging 4 consonants = 24

Now for forming a word, we have to arrange 4 vowels and one string of consonants 

The number of ways of arranging 5 objects = 5! = 120

So, the total number of words formed = 120 × 24 = 2880

∴ The correct answer is option (2).

Concept:

Multiplying the elements of any row with their respective cofactors gives the determinant of the matrix.

Calculation:

Given matrix A = \(\left(\begin{array}{lll}{a}{11}&{a}{12}&{a}{13}\{a}{21}&{a}{22}&{a}{23}\{a}{31}&{a}{32}&{a}_{33}\end{array}\right)\) 

Expanding along row 1,

|A| = Δ = a₁₁(a₂₂a₃₃ - a₂₃a₃₂) + a12(a23a31 - a21a33) + a

13(a21a32 - a22a31)

 Here,C11 = a22a33 - a23a32, C12 = a23a31 - a21a33 ,C13 = a21a32 - a22a31

where, C11, C12, C13 are the cofactors of a11, a12, a13 respectively.

Putting the values,

⇒ a11C11 + a12C12 + a13C13 = Δ

∴ The correct answer is option (3).

Concept:

Multiplying the elements of any row with the cofactors of another row is 0.

Calculation:

If C₁₁, C₁₂, C₁₃ be the cofactors of a₁₁, a₁₂, a₁₃ respectively,

⇒ C₁₁ = a₂₂a₃₃ - a₂₃a₃₂, C₁₂ = a₂₃a₃₁ - a₂₁a₃₃ ,C₁₃ = a₂₁a₃₂ - a₂₂a₃₁,

Putting the values in a₂₁C₁₁ + a₂₂C₁₂ + a₂₃C₁₃

⇒ a₂₁C₁₁ + a₂₂C₁₂ + a₂₃C₁₃ = a₂₁(a₂₂a₃₃ - a₂₃a₃₂) + a₂₂(a₂₃a₃₁ - a₂₁a₃₃) + a₂₃(a₂₁a₃₂ - a₂₂a₃₁)

⇒ a₂₁C₁₁ + a₂₂C₁₂ + a₂₃C₁₃ = a₂₁a₂₂a₃₃ - a₂₁a₂₃a₃₂ + a₂₂a₂₃a₃₁ - a₂₁a₂₂a₃₃ + a₂₃a₂₁a₃₂ - a₂₂a₂₃a₃₁

⇒ a₂₁C₁₁ + a₂₂C₁₂ + a₂₃C₁₃ = 0

∴ The correct answer is option (1).

Concept:

• By interchanging two rows or columns determinant changes to its negative value.
• The determinant of the transpose of a matrix is same as the determinant of A.

Calculation:

If  Δ = |A| = \(\left|\begin{array}{lll}\text{a}{11} & \text{a}{12} & \text{a}{13} \ \text{a}{21} & \text{a}{22} & \text{a}{23} \ \text{a}{31} & \text{a}{32} & \text{a}_{33}\end{array}\right|\),

Taking transpose of A,

⇒ |A^T| = \(\left|\begin{array}{lll}\text{a}{11} & \text{a}{21} & \text{a}{31} \ \text{a}{12} & \text{a}{22} & \text{a}{32} \ \text{a}{13} & \text{a}{23} & \text{a}_{33}\end{array}\right|\)  = Δ ,

Take C₁ ↔ C₃,

⇒  \(\left|\begin{array}{lll}\text{a}{31} & \text{a}{21} & \text{a}{11} \ \text{a}{32} & \text{a}{22} & \text{a}{12} \ \text{a}{33} & \text{a}{23} & \text{a}_{13}\end{array}\right|\) = - Δ 

Take C1 ↔ C2,

⇒  \(\left|\begin{array}{lll}\text{a}{21} & \text{a}{31} & \text{a}{11} \ \text{a}{22} & \text{a}{32} & \text{a}{12} \ \text{a}{23} & \text{a}{33} & \text{a}_{13}\end{array}\right|\) = -(- Δ) = Δ

Take R2 ↔ R3,

⇒  \(\left|\begin{array}{lll}\text{a}{21} & \text{a}{31} & \text{a}{11} \text{a}{23} & \text{a}{33} & \text{a}{13} \text{a}{22} & \text{a}{32} & \text{a}_{12} \end{array}\right|\) = - Δ

∴ The correct option is (4).

Concept:

• If f(x) is a function such that  f(x + y) = f(x) f(y),

           then the function will be of the form f(x) = a^x, a ∈ R.

• For a G.P., with first term a and common ratio r,

           Sum of first n terms = $a({r^n - 1 \over r - 1})$, r > 1

Calculation:

Given, f(x) is a function satisfying f(x + y) = f(x) f(y) for all x, y ∈ N such that f(1) = 2 

Since,  f(x + y) = f(x) f(y) 

⇒ f(1 + 1) = f(1) f(1) =  2²

⇒ f(1 + 2) =  f(3)  = f(1) f(2) =  2 × 22 = 23

Similarly,

f(n) = 2n

So, $\displaystyle\sum_{x=2}^n$f(x) = ​​​​$\displaystyle\sum_{x=2}^n$ 2^x

⇒ $\displaystyle\sum_{x=2}^n$f(x) = ​​​​(2² + 2³ + 2⁴ + ... + 2ⁿ) = 2044

which is a G.P. with first term 2² and common ratio 2.

and number of terms = n - 1

⇒ $2^2({ 2^{n-1}-1 \over 2-1} )$ = 2044

⇒ 4(2^n-1 - 1) = 2044

⇒ 2^n-1 - 1 = 511

⇒ 2^n-1 = 512 = 2⁹

⇒ n -1 = 9

⇒ n = 10  

∴ The correct answer is option (3).

Concept:

• If f(x) is a function such that  f(x + y) = f(x) f(y),

           then the function will be of the form f(x) = ax, a ∈ R.​

Calculation:

Given, f(x) is a function satisfying f(x + y) = f(x) f(y) for all x, y ∈ N such that f(1) = 2 

Let f(x) = ax 

f(1) = a1 = 2 

⇒ f(x) = 2x

So, $\displaystyle\sum_{x=1}^5$f(2x - 1) = ​​​​$\displaystyle\sum_{x=1}^5$22x- 1

⇒ $\displaystyle\sum_{x=1}^5$f(x) = ​​​​(2 + 23 + 25 + 27 + 2⁹ )

⇒ $\displaystyle\sum_{x=1}^5$f(x) = 2 + ​​​​8 + 32 + 128 + 512

⇒ $\displaystyle\sum_{x=2}^5$f(x) = ​​​​682

∴ The correct answer is option (2).

Concept:

• If f(x) is a function such that  f(x + y) = f(x) f(y),

           then the function will be of the form f(x) = ax, a ∈ R.

• For a G.P., with first term a and common ratio r,

           Sum of first n terms = $a({r^n - 1 \over r - 1})$, r > 1

Calculation:

Given, f(x) is a function satisfying f(x + y) = f(x) f(y) for all x, y ∈ N such that f(1) = 2 

Let f(x) = ax 

f(1) = a1 = 2 

⇒ f(x) = 2x

So, $\displaystyle\sum_{x=1}^6$2x f(x) = ​​​​$\displaystyle\sum_{x=1}^6$2x 2x

⇒ $\displaystyle\sum_{x=1}^6$2^x f(x) =  ​​​​$\displaystyle\sum_{x=1}^6$22x = ​​​​(22 + 24 + 2⁶ + 28 + 2¹⁰ + 212)

which is a G.P. with first term 22 and common ratio 4 and n = 6.

​⇒ $\displaystyle\sum_{x=1}^6$ 2x f(x)  = $2^2({ 4^6-1 \over 4-1} )$ = $4(4096 - 1) \over 3$

⇒ $\displaystyle\sum_{x=1}^6$ 2^x f(x) = 4 × 1365 = 5460

∴ The correct answer is option (4).

Concept:

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)

Explanation:

Given: x students (x < 6) got medals in all three sports and the medals went to a total of 15x students. It awarded 5x medals in basketball, (4x + 15) medals in football, and (x + 25) medals in volleyball.

Let A = students who got medals in basketball

B = students who got medals in football

C = students who got medals in volleyball

From given data,

⇒ n(A) = 5x, n(B) = 4x + 15, n(C) = x + 25, n(A ∪ B ∪ C) = 15x, n(A ∩ B ∩ C) = x

Putting in the formula,

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)

⇒ 15x = 5x + 4x + 15 + x + 25 - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + x

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 11x - 15x + 40 

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 40 - 4x  __(i)

From Venn diagram,

n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = (p + s) + (r + s) + (q + r)

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = p + q + r + 3s

where (p + q + r) is the number of students who received medals in exactly two of the three sports.

and s = the number of students who received medals in all the three sports

⇒ The number of students who received medals in exactly two of the three sports

= n(A ∩ B) + n(B ∩ C) + n(C ∩ A) - 3x

Putting value from (i),

⇒ The number of students who received medals in exactly two of the three sports = 40 - 4x - 3x = 40 - 7x

∴ The correct option is (3).

Explanation: 

Given: x students (x < 6) got medals in all three sports and the medals went to a total of 15x students. It awarded 5x medals in basketball, (4x + 15) medals in football, and (x + 25) medals in volleyball.

Let A = students who got medals in basketball

B = students who got medals in football

C = students who got medals in volleyball

From given data,

⇒ n(A) = 5x, n(B) = 4x + 15, n(C) = x + 25, n(A ∪ B ∪ C) = 15x, n(A ∩ B ∩ C) = x

Putting in the formula,

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)

⇒ 15x = 5x + 4x + 15 + x + 25 - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + x

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 11x - 15x + 40 

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 40 - 4x  __(i)

From the Venn diagram,

n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = (p + s) + (r + s) + (q + r)

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = p + q + r + 3s  ___(ii)

where (p + q + r) is the number of students who received medals in exactly two of the three sports.

and s = the number of students who received medals in all the three sports = x

⇒ p + q + r + s = the number of students who received medals in atleast two sports.

Putting the value of (i) in (ii),

⇒ 40 - 4x = p + q + r + 3s

⇒ 40 - 4x = p + q + r + s + 2x

⇒ p + q + r + s = 40 - 6x

⇒ The number of students who received medals in atleast two sports = 40 - 6x

∴ The correct option is (4).

Explanation: 

Given: x students (x < 6) got medals in all three sports and the medals went to a total of 15x students. It awarded 5x medals in basketball, (4x + 15) medals in football, and (x + 25) medals in volleyball.

Let A = students who got medals in basketball

B = students who got medals in football

C = students who got medals in volleyball

From given data,

⇒ n(A) = 5x, n(B) = 4x + 15, n(C) = x + 25, n(A ∪ B ∪ C) = 15x, n(A ∩ B ∩ C) = x

Putting in the formula,

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)

⇒ 15x = 5x + 4x + 15 + x + 25 - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + x

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 11x - 15x + 40 

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 40 - 4x  __(i)

From the Venn diagram,

n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = (p + s) + (r + s) + (q + r)

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = p + q + r + 3s  ___(ii)

where (p + q + r) is the number of students who received medals in exactly two of the three sports.

and s = the number of students who received medals in all the three sports = x

⇒ p + q + r + s = the number of students who received medals in atleast two sports.

Putting the value of (i) in (ii),

⇒ 40 - 4x = p + q + r + 3s

⇒ 40 - 4x = p + q + r + s + 2x

⇒ p + q + r + s = 40 - 6x __(iii)

Now, from the vein diagram

the number of students who received medals in exactly one of three sports 

= n(A ∪ B ∪ C) - (p + q + r + s)

Putting the value from (iii)

the number of students who received medals in exactly one of three sports = 15x - (40 - 6x) = 21x - 40

∴ The correct option is (1).