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3 months ago

Exemplar Solution 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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3 months ago

Exemplar Solution 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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3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$\underset{x \to 0}{l i m} \frac{{(1 + x)}^{n} - 1}{x}$ is

(a) $n$

(b) 1

(c) – $n$

(d) 0

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alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{{(1 + x)}^{n} - 1}{x} \\ = \underset{x \to 0}{l i m} \frac{{(1 + x)}^{n} - {(1)}^{n}}{(1 + x) - (1)} = \underset{1 + x \to 1}{l i m} \frac{{(1 + x)}^{n} - {(1)}^{n}}{(1 + x) - (1)} = n {(1)}^{n - 1} = n [? \underset{x \to a}{l i m} \frac{x^{n} - a^{n}}{x - a} = n . a^{n - 1}] \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$l i m_{x \to 3} \frac{x}{[x]} =_________$ .

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alok kumar singh

Contributor-Level 10

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n \underset{x \to 3^{+}}{l i m} \frac{x}{[x]} \\ = \underset{h \to 0}{l i m} \frac{(3 + h)}{[3 + h]} = 1 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 1 . \end{array}$

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New answer posted

3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

If $y = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . .$ , then $\frac{d y}{d x} =_________.$

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alok kumar singh

Contributor-Level 10

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t y = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots \\ \frac{d y}{d x} = 0 + \frac{1}{1!} + \frac{2 x}{2!} + \frac{3 x^{2}}{3!} + \dots \\ = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots = y \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s y . \end{array}$

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New answer posted

3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

$l i m_{x \to 0} s i n (m x) c o t$ x/ $\sqrt$ 3 = 2 then m = __________.

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} (s i n m x c o t \frac{x}{\sqrt[]{3}}) = 2 \\ \Rightarrow \underset{\begin{array}{l} x \to 0 \\ ∴ m x \to 0 \end{array}}{l i m} \frac{s i n m x}{m x} * m x \underset{x \to 0}{l i m} (c o t \frac{x}{\sqrt[]{3}}) = 2 \\ \Rightarrow 1 * m x \underset{x \to 0}{l i m} \frac{1}{t a n \frac{x}{\sqrt[]{3}}} = 2 \\ \Rightarrow \underset{x \to 0}{l i m} m x * \frac{\frac{x}{\sqrt[]{3}}}{\frac{x}{\sqrt[]{3}} . t a n \frac{x}{\sqrt[]{3}}} = 2 \\ \Rightarrow \frac{m x}{\frac{x}{\sqrt[]{3}}} * (1) = 2 \Rightarrow \sqrt[]{3} m = 2 \Rightarrow m = \frac{2}{\sqrt[]{3}} = \frac{2 \sqrt[]{3}}{3} . \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s \frac{2 \sqrt[]{3}}{3} . \end{array}$

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New answer posted

3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

If $f (x) = \frac{t a n x}{x - π}$ , then $l i m_{x \to π} f (x) =_________$ .

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n f (x) = \underset{x \to π}{l i m} \frac{- t a n (π - x)}{x - π} \\ = \underset{π - x \to 0}{l i m} \frac{- t a n (π - x)}{- (π - x)} = 1 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 1 . \end{array}$

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New answer posted

3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

If $f (x) = 1 - x + x^{2} - x^{3} + . . . - x^{99} + x^{100}$ , then $f^{'} (1)$ is equal to

(a) 150

(b) –50

(c) –150

(d) 50

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = 1 - x + x^{2} - x^{3} + \dots - x^{99} + x^{100} \\ ∴ f' (x) = - 1 + 2 x - 3 x^{2} + \dots - 99 x^{98} + 100 x^{99} \\ S o, f' (1) = - 1 + 2 - 3 + \dots - 99 + 100 \\ = (- 1 - 3 - 5 \dots - 99) + (2 + 4 + 6 + \dots + 100) \\ = \frac{50}{2} [2 * - 1 + (50 - 1) (- 2)] + \frac{50}{2} [2 * 2 + (50 - 1) (2)] \\ = 25 [- 2 - 98] + 25 [4 + 98] = 25 * - 100 + 25 * 102 \\ = 25 [- 100 + 102] = 25 * 2 = 50 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted

3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

If $f (x) = x^{100} + x^{99} + . . . + x + 1$ , then $f^{'} (1)$ is equal to

(a) 5050

(b) 5049

(c) 5051

(d) 5005

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = x^{100} + x^{99} + \dots + x + 1 \\ ∴ f' (x) = 100 x^{99} + 99 x^{98} + \dots + 1 \\ S o, f' (1) = 100 + 99 + 98 + \dots + 1 \\ = \frac{100}{2} [2 * 100 + (100 - 1) (- 1)] = 50 [200 - 99] = 50 * 101 = 5050 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

If $f (x) = \frac{x^{n} - a^{n}}{x - a}$ for some constant $a$ , then $f^{'} (a)$ is

(a) 1

(b) 0

(c) Does not exist

(d) $\frac{1}{2}$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = \frac{x^{n} - a^{n}}{x - a} \\ f' (x) = \frac{(x - a) (n . x^{n - 1}) - (x^{n} - a^{n}) . 1}{{(x - a)}^{2}} \\ ∴ f' (a) = \frac{(a - a) (n . a^{n - 1}) - (a^{n} - a^{n}) . 1}{{(a - a)}^{2}} \\ S o, f' (a) = \frac{0}{0} = d o e s n o t e x i s t \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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