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Concept:

Matrix Multiplication:

• Matrix multiplication is defined only when the number of columns in the first matrix is equal to the number of rows in the second matrix.
• For two matrices \( A_{m \times n} \) and \( B_{n \times p} \), their product \( AB \) will result in a matrix \( C_{m \times p} \).
• In general, the product of \( X \) and \( Y \) is defined if and only if the number of columns in \( X \) equals the number of rows in \( Y \).

Calculation:

We have the following matrix operations:

\( A_{m \times n} B_{n \times p} = (AB)_{m \times p} \)

Now, consider the matrix multiplications:

\([Z_{3 \times 2} . Y_{2\times3}].X_{3 \times 3}] = [ZYX]_{3 \times 3}\)

\( Y_{2 \times 3} [X_{3 \times 3} Z_{3 \times 2}] = [YXZ]_{2 \times 2} \)

Finally, we observe that:

\( X_{3 \times 3} [Y_{2 \times 3} Z_{3 \times 2}] = X_{3 \times 3} [YZ]_{2 \times 2} \)

Conclusion:

\(\text{No. of columns in X} \neq \text{No. of rows in (YZ)}\)

Hence, X(YZ) is not defined.

∴ the Correct answer is Option 4

Concept:

Infinite and Finite Sets:

• A finite set contains a limited number of elements that can be counted to an end.
• An infinite set contains endlessly many elements; you cannot finish counting them.
• Between any two different real numbers there are infinitely many irrational numbers.
• A set described without a lower bound (for integers) typically includes infinitely many negative members.

Calculation:

Given statements:

I. All irrational numbers between √12 and √15

II. All odd integers less than 1000

Step wise check:

⇒ For I: The interval between √12 and √15 is a normal real interval.

⇒ For I: Any real interval contains infinitely many irrational numbers.

⇒ So I is true.

⇒ For II: "Less than 1000" gives no lower limit.

⇒ For II: This includes ..., -5, -3, -1, 1, 3, ..., 999.

⇒ For II: We can continue to more negative odd integers without end.

⇒ So II is not finite (it is infinite).

∴ Only statement I is correct.

Concept:

Counting 4-digit Numbers with All Odd Digits:

• Each digit in a 4-digit number can be one of the odd digits: 1, 3, 5, 7, or 9.
• There are 5 choices for each digit since there are 5 odd digits available.
• For a 4-digit number, we need to select a digit for each of the 4 positions independently.
• The total number of 4-digit numbers with all odd digits is the product of choices for each digit.

Calculation:

Given,

Choices for each digit = 5 (as the digits are 1, 3, 5, 7, 9)

Number of digits, n = 4

The total number of 4-digit numbers with all odd digits is,

\(5^4\) =  5 × 5 ×5 × 5 = 625

∴ 625 4-digit numbers have all digits as odd.5×5×5×5 5×5×5×\there

∴ The total number of 4-digit numbers with all odd digits is 625.

Explanation:

(1+ω -ω²)¹⁰⁰+ (1-ω +ω2)100

= (– ω² – ω² )¹⁰⁰ + (–ω – ω)¹⁰⁰  {∵ 1 + ω  + ω² = 0}

= 2¹⁰⁰(ω²⁰⁰ + ω¹⁰⁰)

= 2¹⁰⁰(ω² + ω)     (∵ ω³ = 1)

= –2¹⁰⁰ 

∴ Option (d) is correct

Explanation:

Given AB = O

Let  A = \(\left[\begin{array}{lll} a & b \ c & d \end{array}\right]\)

B = \(\left[\begin{array}{lll} p & q \ r & s \end{array}\right]\)

Now AB = \(\left[\begin{array}{lll} a & b \ c & d \end{array}\right] \left[\begin{array}{lll} p & q \ r & s \end{array}\right] = \left[\begin{array}{lll} 0 & 0 \ 0 & 0 \end{array}\right]\)

\(\left[\begin{array}{lll} ap+br & aq+bs \ cp+dr & cq+ds \end{array}\right]= \left[\begin{array}{lll} 0 & 0 \ 0 & 0 \end{array}\right]\)

Comparing both sides, we get

⇒ ap + br = 0...(1)

⇒ aq + bs = 0...(2)

⇒ cp + dr = 0...(3)

⇒ cq + ds = 0...(4)

From Eqs. (i) and (iii), we get:

⇒ (ad – bc)p = 0 and (ad – bc)r = 0 

From Eqs. (ii) and (iv), we get:

⇒ (ad – bc) q = 0 and (ad – bc)s = 0 

Now, if A is non-singular

ad – bc ≠ 0

So p = q = r = s = 0.

Then,  B is the null matrix.

∴ Option (c) is correct.

7Concept:

Binomial Expansion:

• The binomial theorem is used to expand expressions of the form \((1 + x)^n\).
• The general term in the expansion of \((1 + x)^n\) is given by \(T_k = C(n, k) \cdot x^k\), where ⁿC_k is the binomial coefficient.
• To find the coefficient of x^3, we identify the corresponding terms from the expansion and set the coefficient equal to 35.

Calculation:

Given the expansion of \((1 + x)^p \cdot (1 + x)^q\), we have:

The coefficient of x³ in  the expansion is 35.

We use the binomial expansion formula for the term x³

⇒ (p+ q)_c3 = 35 = ⁷C₃

⇒ p+q =7

∴ the correct answer is Option C

Explanation:

Let 'a' and 'd' be the first term and common difference of AP

(p + q)^th Term = T_p+q 

= a + (p + q - 1)......(1)

Given

p. T_p = q.T_p

⇒ p.[ a + (p -1) d] = q. [ a + (q - 1) d]

⇒ (p - q) [ a + (p + q -1) d] =0

⇒ [ a + (p + q -1) d] =0           (∵ p ≠ q)

 ⇒Tp+q = 0 ......(From i)

∴ Option (a) is correct

Concept:

Arithmetic Progression (AP):

• The terms are in AP if the difference between consecutive terms is constant.
• The common difference is given by: \(d = q - p = r - q\)

Geometric Progression (GP):

• The terms are in GP if the ratio between consecutive terms is constant.
• The common ratio is given by: \(r = q \times p\)

Calculation:

Given

p = ln(x),

q = ln(x3)  = 3 lnx

and r = ln(x5) = 5 lnx

Clearly, p - q  = r - q = \(2lnx\)

⇒ p, q, r are in AP

Also \(\frac{q}{p} ≠ \frac{r}{p} \because \frac{q}{p} = 3 , \frac{r}{p} = \frac{5}{3}\)

∴ p, q, r can never be in GP

∴ Option (c) is correct.

Explanation:

Given

 \(\rm z=\frac{1}{3}\begin{vmatrix}i&2i&1\\ 2i&3i&2\\ 3&1&3\end{vmatrix}\) 

\(Z = \frac{1}{3}[[ (i9 – 2) – 2i (6i – 6) 1(2i – 9i)]\)

\(Z = \frac{1}{3}[3+3i] =1 + i \)

|Z| = \(√{1^2+1^2} =\) √2 

∴ Option (b) is correct.

Explanation:

Let I =  \(\rm \Sigma_{n=1}^{20}(i^{n-1}+i^n+i^{n+1})\)

⇒ I = \(\rm \Sigma_{n=1}^{20}(i^n (\frac{1}{i} +1+ i)\)

⇒ I = \(\rm \Sigma_{n=1}^{20}(i^n (1)\)

⇒ I = \(i^1+ i^2+i^3+i^4i^5+........+ i^{20}\)

⇒ I = \(5.(i^1+ i^2+i^3+i^4) = 0\) ( {∵  i + i² + i³ + i⁴ = 0} 

∴ Option (b) is correct.