प्रश्नों के उत्तर लिखिए। (प्रत्येक प्रश्न 3 अंक का हे) - गणित कक्षा 11 साइन्स Questions - Vidyadip

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प्रश्नों के उत्तर लिखिए। (प्रत्येक प्रश्न 3 अंक का हे)

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15 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks

दिए गए $21x^2 - 28x + 10 = 0$ समीकरण को हल कीजिए।

Answer

$21x^2 - 28x + 10 = 0$
यहां $a = 21, b = -28, c = 10$
$\because x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\Rightarrow x=\frac{-(-28) \pm \sqrt{(-28)^{2}-4 \times 21 \times 10}}{2 \times 21}$
$\Rightarrow x=\frac{28 \pm \sqrt{784-840}}{42}$
$\Rightarrow x=\frac{28 \pm \sqrt{-56}}{42}=\frac{28 \pm 2 \sqrt{14} i}{42}$
$\Rightarrow x=\frac{2}{3} \pm \frac{\sqrt{14}}{21} i$

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Question 23 Marks

दिए गए $27x^2 - 10x + 1 = $0 समीकरण को हल कीजिए।

Answer

$27x^2 - 10x + 1 = 0$
यहां $a = 27, b = -10, c = 1$
$\because x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\Rightarrow x=\frac{-(-10) \pm \sqrt{(-10)^{2}-4 \times 27 \times 1}}{2 \times 27}$
$\Rightarrow x=\frac{10 \pm \sqrt{100-108}}{54}$
$\Rightarrow x=\frac{10 \pm \sqrt{-8}}{54}=\frac{10 \pm 2 \sqrt{2} i}{54} $
$\Rightarrow x=\frac{5}{27}+\frac{\sqrt{2}}{27} i$

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Question 33 Marks

दिए गए $x^2 - 2x + \frac 32 = 0 $ समीकरण को हल कीजिए।

Answer

$x^2 - 2x + \frac 32 = 0$
$\Rightarrow x^2 - 2x + 1 - 1 + \frac 32 = 0$
$\Rightarrow (x - 1)^2 + \frac 12 = 0$
$\Rightarrow \left(x-\frac{1}{2}\right)^{2}-\left(\frac{1}{\sqrt{2}} i\right)^{2} = 0$
$\Rightarrow \left(x-\frac{1}{2}+\frac{1}{\sqrt{2}} i\right)\left(x-\frac{1}{2}-\frac{1}{\sqrt{2}} i\right) = 0$
$\Rightarrow x = \frac{1}{2}-\frac{1}{\sqrt{2}} i, \frac{1}{2}+\frac{1}{\sqrt{2}} i$

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Question 43 Marks

दिए गए $3x^2 - 4x + \frac {20}3 = 0$ समीकरण को हल कीजिए।

Answer

$3x^2 - 4x + \frac {20}3 = 0$
यहां $a = 3, b = -4, c = \frac{20}{3}$
$\because x = \frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
$\Rightarrow x = \frac{-(-4) \pm \sqrt{(-4)^{2}-4 \times 3 \times \frac{20}{3}}}{2 \times 3}$
$\Rightarrow x = \frac{4 \pm \sqrt{16-80}}{6}$
$\Rightarrow x = \frac{4 \pm \sqrt{-64}}{6}=\frac{4 \pm 8 i}{6}$
$\Rightarrow x = \frac{2}{3} \pm \frac{4}{3} i$

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Question 53 Marks

ध्रुवीय रूप में परिवर्तित कीजिए: $\frac{1+3 i}{1-2 i}$

Answer

z = $\frac{1+3 i}{1-2 i}=\frac{1+3 i}{1-2 i} \times \frac{1+2 i}{1+2 i}$
z = $\frac{1+6 i^{2}+3 i+2 i}{1-4 i^{2}}=\frac{-5+5 i}{5}$
z = -1 + i = $\sqrt{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)$
$\therefore$ ध्रुवीय रूप
z = r(cos $\theta$ + i sin $\theta$)
= $\sqrt{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)$

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Question 63 Marks

ध्रुवीय रूप में परिवर्तित कीजिए: $\frac{1+7 i}{(2-i)^{2}}$

Answer

$z = \frac{1+7 i}{(2-i)^{2}}=\frac{1+7 i}{4+i^{2}-4 i}$
$\Rightarrow z =\frac{1+7 i}{4-1-4 i}=\frac{1+7 i}{3-4 i}$
$\Rightarrow z =\frac{1+7 i}{3-4 i} \times \frac{3+4 i}{3+4 i}$
$\Rightarrow z=\frac{3+4 i+21 i+28 i^{2}}{9-16 i^{2}}$
$z = \frac{3+25 i-28}{9+16}$
$= \frac{-25+25 i}{25}$
$\therefore z + -1 + i,$ इसका संगत बिन्दु $(-1, 1)$ है।
$\Rightarrow r = |z| = \sqrt{(-1)^{2}+1^{2}}=\sqrt{2}$
$\theta =$ कोणांक$ (z) = \tan^{-1}\left|\frac{1}{-1}\right| = \tan^{-1}(1)$
$\theta=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}$
$\therefore$ ध्रुवीय रूप
$z = r(\cos \theta + i \sin \theta)$
$= \sqrt{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)$

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Question 73 Marks

यदि $ x - iy = \sqrt{\frac{a-i b}{c-i d}}$, तो सिद्ध कीजिए कि $(x^2+ y^2) = \frac{a^{2}+b^{2}}{c^{2}+d^{2}}$

Answer

$x - iy = \sqrt{\frac{a-i b}{c-i d}} ...(i)$
दोनों पक्षों की संयुग्मी सम्मिश्र संख्या लेने पर,
$\overline{x-i y}=\sqrt{\left(\overline{\frac{a-i b}{c-i d}}\right)}=\sqrt{\frac{\overline{(a-i b)}}{(c-i d)}}$
$\Rightarrow x + iy = \sqrt{\frac{a+i b}{c+i d}} ...(ii)$
समीकरण $(i)$ और $(ii)$ का गुणा करने पर,
$(x - iy)(x + iy) = \sqrt{\frac{a-i b}{c-i d}} \times \sqrt{\frac{a+i b}{c+i d}}$
$\Rightarrow x^2- i^2y^2= \sqrt{\frac{a^{2}-i^{2} b^{2}}{c^{2}-i^{2} d^{2}}}$
$\Rightarrow x^2+ y^2= \sqrt{\frac{a^{2}+b^{2}}{c^{2}+d^{2}}}$
$\Rightarrow (x^2+ y^2)^2= \frac{a^{2}+b^{2}}{c^{2}+d^{2}}$

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Question 83 Marks

$\left(\frac{1}{1-4 i}-\frac{2}{1+i}\right)\left(\frac{3-4 i}{5+i}\right)$ को मानक रूप में परिवर्तित कीजिए।

Answer

माना z = $\left(\frac{1}{1-4 i}-\frac{2}{1+i}\right)\left(\frac{3-4 i}{5+i}\right)$
= $\left\{\frac{1+i-2(1-4 i)}{(1-4 i)(1+i)}\right\}\left(\frac{3-4 i}{5+i}\right)$
= $\frac{(-1+9 i)(3-4 i)}{\left(1-4 i+i-4 i^{2}\right)(5+i)}$
= $\frac{-3+4 i+27 i-36 i^{2}}{(1-3 i+4)(5+i)}$
= $\frac{-3+31 i+36}{(5-3 i)(5+i)}=\frac{33+31 i}{25+5 i-15 i-3 i^{2}}$
= $\frac{33+31 i}{28-10 i}=\frac{33+31 i}{28-10 i} \times\left(\frac{28+10 i}{28+10 i}\right)$
= $\frac{614+1198 i}{784+100}$
= $\frac{614+1198 i}{884}=\frac{307}{442}+\frac{599}{442}i$

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Question 93 Marks

यदि $\alpha$ और $\beta$ भिन्न सम्मिश्र संख्याएँ हैं जहाँ |$\beta$| = 1, तब $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|$ का मान ज्ञात कीजिए।

Answer

$\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|^{2} =\left(\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right)\left(\frac{\overline{\beta-\alpha}}{1-\bar{\alpha} \beta}\right)$ $\left(\because|z|^{2}=z \bar{z}\right)$
= $\left(\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right)\left(\frac{\bar{\beta}-\bar{\alpha}}{1-\alpha \bar{\beta}}\right)$ $\{\because(\bar{z})=z\}$
= $\frac{\beta \bar{\beta}-\beta \bar{\alpha}-\alpha \bar{\beta}+\alpha \bar{\alpha}}{1-\alpha \bar{\beta}-\bar{\alpha} \beta+\alpha \bar{\alpha} \beta \bar{\beta}}$
= $\frac{|\beta|^{2}-\beta \bar{\alpha}-\alpha \bar{\beta}+|\alpha|^{2}}{1-\alpha \bar{\beta}-\beta \bar{\alpha}+|\alpha|^{2}|\beta|^{2}}$
= $\frac{1-\beta \bar{\alpha}-\alpha \bar{\beta}+|\alpha|^{2}}{1-\alpha \bar{\beta}-\bar{\alpha}+|\alpha|^{2}}$ {$\because|\beta|$ = 1 दिया है।}
= 1
$\Rightarrow \left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|^{2}$ = 1
$\therefore \left|\frac{\beta-\alpha}{1-\bar{\alpha \beta}}\right|$ = 1

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Question 103 Marks

यदि $(x + iy)^3= u + iv,$ तो दशाईए कि $\frac{u}{x}+\frac{v}{y} = 4(x^2 - y^2)$

Answer

$(x + iy)^3= u + iv$
$\Rightarrow x^3+ 3x^2 iy + 3x(iy)^2+ (iy)^3= u + iv$
$\Rightarrow x^3+ 3x^2 yi - 3xy^2- y^3 i = u + iv$
$\Rightarrow (x^3- 3xy^2) + (3x^2y - y^3) i = u + iv$
वास्तविक और काल्पनिक भागों की तुलना करने पर,
$u = x^3- 3xy^2, v = 3x^2 y - y^3$
$\therefore LHS = \frac{u}{x}+\frac{v}{y}=\left(\frac{x^{3}-3 x y^{2}}{x}\right)+(\left.\frac{3 x^{2} y-y^{3}}{y}\right)$
$= (x^2- 3y^2) + (3x^2- y^2)$
$= 4x^2- 4y^2$
$= 4(x^2- y^2) =$ RHS

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Question 113 Marks

$\frac{1+i}{1-i}-\frac{1-i}{1+i}$ का मापांक ज्ञात कीजिए।

Answer

z = $\frac{1+i}{1-i}-\frac{1-i}{1+i}$
= $\frac{(1+i)^{2}-(1-i)^{2}}{(1-i)(1+i)}$
= $\frac{\left(1+i^{2}+2 i\right)-\left(1+i^{2}-2 i\right)}{1-i^{2}}$
= $\frac{4 i}{1+1}$ = 2i
$\therefore$ z = 0 + 2i
$\Rightarrow$ |z| = $\sqrt{0^{2}+2^{2}}=\sqrt{0+4}=\sqrt{4}$ = 2

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Question 123 Marks

यदि $(x - iy)(3 + 5i), -6 - 24i$ की संयुग्मी है तो वास्तविक संख्याएँ $x$ और $y$ ज्ञात कीजिए।

Answer

माना $z_1 = (x - iy)(3 + 5i)$
और $z_2 = -6 - 24i,$
$\therefore \bar{z}_{2} = -6 + 24i$
$\because{z_{1}} =\overline{z_{2}} ($दिया है।$)$
$\Rightarrow (x - iy)(3 + 5i) = -6 + 24i$
$\Rightarrow x - iy = \frac{-6+24 i}{3+5 i}$
$= \frac{-6+24 i}{3+5 i} \times \frac{3-5 i}{3-5 i}$
$\Rightarrow x - iy = \frac{-18+30 i+72 i-120 i^{2}}{9-25 i^{2}}$
$\Rightarrow x - iy = \frac{-18+102 i+120}{9+25} $
$\Rightarrow x - iy = \frac{102+102 i}{34}$
$\Rightarrow x - iy = 3 + 3i$
वास्तविक और काल्पनिक भागों की तुलना से,
$x = 3, y = -3$

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Question 133 Marks

सम्मिश्र संख्या $\frac{1+2 i}{1-3 i}$ का मापांक और कोणांक ज्ञात कीजिए।

Answer

माना $z = \frac{1+2 i}{1-3 i}$
$\Rightarrow z=\frac{1+2 i}{1-3 i} \times\left(\frac{1+3 i}{1+3 i}\right)$
$\Rightarrow z=\frac{1+2 i+3 i+6 i^{2}}{1-9 i^{2}}$
$\Rightarrow z=\frac{1+5 i-6}{1+9}$
$\Rightarrow z=\frac{-5+5 i}{10}=\frac{-1}{2}+\frac{1}{2} i$
$\therefore$ मापांक $|z| = \sqrt{\left(-\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{4}+\frac{1}{4}}=\sqrt{\frac{1}{2}}$
$= \frac{1}{\sqrt{2}}$
और कोणांक$ (z) = \tan^{-1}\left|\frac{\frac{1}{2}}{\frac{-1}{2}}\right| = \tan^{-1}(1) \{\because z$ द्वितीय चतुर्थांश में है।$\}$
$= \pi-\frac{\pi}{4}=\frac{3 \pi}{4}$

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Question 143 Marks

यदि $ a + ib = \frac{(x+i)^{2}}{2 x^{2}+1}$, सिद्ध कीजिए कि, $a^2+ b^2= \frac{\left(x^{2}+1\right)^{2}}{\left(2 x^{2}+1\right)^{2}}$

Answer

$a + ib = \frac{(x+i)^{2}}{2 x^{2}+1} ...(i)$
दोनों पक्षों की संयुग्मी सम्मिश्र संख्या लेने पर
$\overline{(a+i b)}=\left(\frac{\overline{(x+i)^{2}}}{2 x^{2}+1}\right)$
$\Rightarrow a-i b=\frac{(x-i)^{2}}{2 x^{2}+1} ...(ii)$
समीकरण (i) और (ii) का गुणा करने पर,
$(a + ib)(a - ib) = \frac{(x+i)^{2}}{2 x^{2}+1} \times \frac{(x-i)^{2}}{2 x^{2}+1}$
$\Rightarrow a^2- i^2b^2= \frac{\left(x^{2}-i^{2}\right)^{2}}{\left(2 x^{2}+1\right)^{2}}$
$\Rightarrow a^2 + b^2= \frac{\left(x^{2}+1\right)^{2}}{\left(2 x^{2}+1\right)^{2}}$

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Question 153 Marks

यदि $z_1= 2 - i, z_2= 1 + i, \left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+i}\right|$ का मान ज्ञात कीजिए।

Answer

$z_1= 2 - i, z_2= 1 + i$
$\therefore \frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+i} =\frac{2-i+1+i+1}{2-i-1-i+i} $
= $\frac{4}{1-i} $
= $\frac{4}{1-i} \times\left(\frac{1+i}{1+i}\right)$
= $\frac{4(1+i)}{1-i^{2}}=\frac{4(1+i)}{1+1} $
= $\frac{4(1+i)}{2}$ = 2(1 + i)
$\Rightarrow \frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+i} $ = 2 + 2i
$\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+i} =\sqrt{2^{2}+2^{2}}=\sqrt{4+4} $
= $\sqrt{8}=2 \sqrt{2} $

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