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New question posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Five

(i) For any two complex numbers <math><mrow><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub></mrow></math> and any real numbers <math><mrow><mi>a</mi><mo>,</mo><mi>b</mi></mrow></math> , <math> <mrow> <mo>?</mo> <msup> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msub> <mrow> <mi>z</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msub> <mrow> <mi>z</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>?</mo> <mo>−</mo> <mo>?</mo> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msub> <mrow> <mi>z</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msup> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msub> <mrow> <mi>z</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>?</mo> <mo>=</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> </mrow> </math> (ii) The value of <math><mrow><mroot><mrow><mo>−</mo><mn>2</mn><mn>5</mn></mrow><mrow></mrow></mroot><mo>×</mo><mroot><mrow><mo>−</mo><mn>9</mn></mrow><mrow></mrow></mroot></mrow></math> is . (iii) The number <math><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo><mrow><mn>1</mn><mo>−</mo><mi>i</mi></mrow><mo>)</mo></mrow></mrow><mrow><mn>3</mn></mrow></msup></mrow><mrow><mn>1</mn><mo>−</mo><mi>i</mi></mrow></mfrac></mrow></math> is equal to . (iv) The sum of the series <math><mrow><mi>i</mi><mo>+</mo><msup><mrow><mi>i</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><msup><mrow><mi>i</mi></mrow><mrow><mn>3</mn></mrow></msup><mo>+</mo><mo>…</mo></mrow></math> up to 1000 terms is . (v) Multiplicative inverse of <math><mrow><mn>1</mn><mo>+</mo><mi>i</mi></mrow></math> is . (vi) If <math><mrow><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub></mrow></math> and <math><mrow><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub></mrow></math> are complex numbers such that <math><mrow><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>+</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub></mrow></math> is a real number, then <math><mrow><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>=</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo></mrow></math> (vii)  <math> <mrow>arg <mrow><mo>(</mo><mrow><mi>z</mi></mrow><mo>)</mo></mrow><mo>+</mo>arg <mi>z</mi><mrow><mo>(</mo><mrow><mi>z</mi><mo>≠</mo><mn>0</mn></mrow><mo>)</mo></mrow></mrow> </math> is . (viii) If <math><mrow><mo>?</mo><mi>z</mi><mo>+</mo><mn>4</mn><mo>?</mo><mo>≤</mo><mn>3</mn></mrow></math> , then the greatest and least values of <math><mrow><mo>?</mo><mi>z</mi><mo>+</mo><mn>1</mn><mo>?</mo></mrow></math> are . and . (ix) If <math><mrow><mfrac><mrow><mi>z</mi><mo>−</mo><mn>2</mn></mrow><mrow><mi>z</mi><mo>+</mo><mn>6</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>π</mi></mrow><mrow><mn>6</mn></mrow></mfrac><mo>,</mo></mrow></math> then the locus of <math><mrow><mi>z</mi></mrow></math> is . (x) If <math><mrow><mo>?</mo><mi>z</mi><mo>?</mo><mo>=</mo><mn>4</mn></mrow></math> and <math><mrow><mrow><mo>(</mo><mrow><mi>z</mi></mrow><mo>)</mo></mrow><mo>=</mo><mfrac><mrow><mn>5</mn><mi>π</mi></mrow><mrow><mn>6</mn></mrow></mfrac></mrow></math> , then <math><mrow><mi>z</mi><mo>=</mo><mo>…</mo><mo>…</mo><mo>…</mo><mo>…</mo></mrow></math>

4 Views|Posted a year ago

Asked by Payal

New answer posted•a year 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

9 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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•a year 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 $\frac{d y}{d x} =_________.$

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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•a year 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 = __________.

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

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

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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•a year 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

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year 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

9 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year 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) $\frac{1}{2}$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

If $f (x) = 1 + x + \frac{x^{2}}{2} + \dots + \frac{x^{100}}{100}$ , then $f^{'} (1)$ is equal to

(a) $\frac{1}{100}$
(b) 100
(c) Does not exist
(d) 0

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = 1 + x + \frac{x^{2}}{2} + \dots + \frac{x^{100}}{100} \\ f' (x) = 1 + \frac{2 x}{2} + \dots + \frac{100 x^{99}}{100} \\ ∴ f' (1) = 1 + 1 + 1 + \dots + 1 (100 t i m e s) = 100 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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