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Maths Maths NCERT Exemplar Solutions Class 12th Chapter Three

Total number of possible matrices of order $3 * 3$ with each entry 2 or 0 is

(A) 9

(B) 27

(C) 81

(D) 512

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f p o s s i b l e m a t r i c e s o f o r d e r 3 * 3 w i t h e a c h e n t r y 0 o r 2 = 2^{3 * 3} = 2^{9} = 512 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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Maths Exemplar Solution

The matrix $P = [\begin{matrix} 0 & 0 & 4 \\ 0 & 4 & 0 \\ 4 & 0 & 0 \end{matrix}]$ is a

(A) Square matrix

(B) Diagonal matrix

(C) Unit matrix

(D) None

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t A = [\begin{matrix} 0 & 0 & 4 \\ 0 & 4 & 0 \\ 4 & 0 & 0 \end{matrix}] \\ H e r e, n u m b e r o f c o l u m n s a n d t h e n u m b e r o f r o w s a r e e q u a l i . e ., 3 . S o, A i s a s q u a r e m a t r i x . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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Maths Maths NCERT Exemplar Solutions Class 11th Chapter Five

(i) For any two complex numbers $z_{1}, z_{2}$ and any real numbers $a, b$ , $? a^{2} z_{1} + b^{2} z_{2} ? - ? b^{2} z_{1} + a^{2} z_{2} ? = . . . . .$

(ii) The value of $\sqrt[]{- 25} \times \sqrt[]{- 9}$ is .

(iii) The number $\frac{{(1 - i)}^{3}}{1 - i}$ is equal to .

(iv) The sum of the series $i + i^{2} + i^{3} + \dots$ up to 1000 terms is .

(v) Multiplicative inverse of $1 + i$ is .

(vi) If $z_{1}$ and $z_{2}$ are complex numbers such that $z_{1} + z_{2}$ is a real number, then $z_{2} = . . . .$

(vii) $(z) + z (z \neq 0)$ is .

(viii) If $? z + 4 ? \leq 3$ , then the greatest and least values of $? z + 1 ?$ are . and .

(ix) If $\frac{z - 2}{z + 6} = \frac{π}{6},$ then the locus of $z$ is .

(x) If $? z ? = 4$ and $(z) = \frac{5 π}{6}$ , then $z = \dots \dots \dots \dots$

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4 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Five

(i) For any two complex numbers $z_{1}, z_{2}$ and any real numbers $a, b$ , $? a^{2} z_{1} + b^{2} z_{2} ? - ? b^{2} z_{1} + a^{2} z_{2} ? = . . . . .$

(ii) The value of $\sqrt[]{- 25} \times \sqrt[]{- 9}$ is .

(iii) The number $\frac{{(1 - i)}^{3}}{1 - i}$ is equal to .

(iv) The sum of the series $i + i^{2} + i^{3} + \dots$ up to 1000 terms is .

(v) Multiplicative inverse of $1 + i$ is .

(vi) If $z_{1}$ and $z_{2}$ are complex numbers such that $z_{1} + z_{2}$ is a real number, then $z_{2} = . . . .$

(vii) $arg (z) + arg z (z \neq 0)$ is .

(viii) If $? z + 4 ? \leq 3$ , then the greatest and least values of $? z + 1 ?$ are . and .

(ix) If $\frac{z - 2}{z + 6} = \frac{π}{6},$ then the locus of $z$ is .

(x) If $? z ? = 4$ and $(z) = \frac{5 π}{6}$ , then $z = \dots \dots \dots \dots$

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4 months ago

Maths Sequences and Series

If a₁, a₂, a₃…. and b₁, b₂, b₃…. are A.P., and a₁ = 2, a₁₀ = 3, a₁b₁ = 1 – a₁₀b₁₀, then a₄b₄ is equal to…….

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Payal Gupta

Contributor-Level 10

$a_{1}, a_{2}, a_{3} . . . . . .$ are in A.P. (Let common difference is d₁)

$b_{1}, b_{2}, b_{3} . . . .$ are in A.P. (Let common difference is d₂)

and a₁2, a₁₀ = 3, a₁b₁ = 1 = a₁₀b₁₀

$? a_{1} b_{1} = 1$

$∴ b_{1} = \frac{1}{2}$

$a_{10} b_{10} = 1$

$∴ b_{10} = \frac{1}{3}$

Now,

$a_{10} = a_{1} + 9 d_{1} \Rightarrow d_{1} = \frac{1}{9}$

$b_{10} = b_{1} + 9 d_{2} \Rightarrow d_{2} = \frac{1}{9} [\frac{1}{3} - \frac{1}{2}] = - \frac{1}{54}$

Now

$a_{4} = 2 + \frac{3}{9} = \frac{7}{3}$

$b_{4} = \frac{1}{2} - \frac{3}{54} = \frac{4}{9}$

$∴ a_{4} b_{4} = \frac{28}{27}$

...more

$a_{1}, a_{2}, a_{3} . . . . . .$ are in A.P. (Let common difference is d₁)

$b_{1}, b_{2}, b_{3} . . . .$ are in A.P. (Let common difference is d₂)

and a₁2, a₁₀ = 3, a₁b₁ = 1 = a₁₀b₁₀

$? a_{1} b_{1} = 1$

$∴ b_{1} = \frac{1}{2}$

$a_{10} b_{10} = 1$

$∴ b_{10} = \frac{1}{3}$

Now,

$a_{10} = a_{1} + 9 d_{1} \Rightarrow d_{1} = \frac{1}{9}$

$b_{10} = b_{1} + 9 d_{2} \Rightarrow d_{2} = \frac{1}{9} [\frac{1}{3} - \frac{1}{2}] = - \frac{1}{54}$

Now

$a_{4} = 2 + \frac{3}{9} = \frac{7}{3}$

$b_{4} = \frac{1}{2} - \frac{3}{54} = \frac{4}{9}$

$∴ a_{4} b_{4} = \frac{28}{27}$

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Maths Sets

Let S = $2 + \frac{6}{7} + \frac{12}{7^{2}} + \frac{20}{7^{3}} + \frac{30}{7^{4}} + . . . .$ Then 4S is equal to

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Payal Gupta

Contributor-Level 10

$S = 2 + \frac{6}{7} + \frac{12}{7^{2}} + \frac{20}{7^{3}} + \frac{30}{7^{4}} + . . . . . . . . (i)$

$\frac{1}{7} S = \frac{2}{7} + \frac{6}{7^{2}} + \frac{12}{7^{3}} + \frac{20}{7^{4}} + . . . . . . . . (i i)$

(i) - (ii)

$\frac{6}{7} S = 2 + \frac{4}{7} = \frac{6}{7^{2}} + \frac{8}{7^{3}} + . . . . . . . . . . . (i i i)$

$\frac{6}{7^{2}} S = \frac{2}{7} + \frac{4}{7^{2}} + \frac{6}{7^{3}} + . . . . . . . . (i v)$

(iii) – (iv)

$= 2 + [\frac{1}{1 - \frac{1}{7}}] = 2 (\frac{7}{6})$

$∴ 4 S = 8 {(\frac{7}{6})}^{3} = {(\frac{7}{3})}^{3}$

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4 months ago

Maths Conic Sections

The number of points of intersection of $| z - (4 + 3 i) | = 2 a n d | z | + | z - 4 | = 6, z \in C,$ is

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Payal Gupta

Contributor-Level 10

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