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KC Sinha: Exercise 22.1- Mathematics Solution Class 11 Chapter 22 अतिपरवलय

KC Sinha: Exercise 22.1- Mathematics Solution Class 11 Chapter 22 अतिपरवलय

Question 1 अतिपरवलय का समीकरण निकालें जिसके लिए उत्कोन्द्रता 2, एक नाभि (2,2) तथा संगत नियता x+y=9 है । [Find the equation to the hyperbola for which eccentricity is 2, one focus is (2,2) and the corresponding directrix is x+y=9.] Sol : माना P(x,y) अतिपरवलय पर स्थित बिन्दु है। $\frac{P S}{P M}=e$ PS=e.PM $\sqrt{(x-2)^{2}+(y-2)^{2}}=2 \cdot\left|\frac{x+y-9}{\sqrt{1^{2}+1^{2}}}\right|$…

KC Sinha: Exercise 17.4 - Mathematics Solution Class 11 Chapter 17 बिंदुओं के निर्देशांक

KC Sinha: Exercise 17.1 – Mathematics Solution Class 11 Chapter 17 बिंदुओं के निर्देशांक

Question 1 निम्नलिखित बिन्दुओ के बीच की दूरी निकाले [Find the distance between the following pair of points] (i) (0,0),(-5,12) Sol : दूरी सुत्रः $\sqrt{\left(x_{2}+x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$ OP=$\sqrt{\left(-5-0\right)^{2}+\left(12-0\right)^{2}}$ =√25+144=√169=13 इकाई (ii) (4,5),(-3,2) (iii)…

KC Sinha: Exercise 16.5 - Mathematics Solution Class 11 Chapter 16 कुछ महत्वपूर्ण अनंत श्रेणी

KC Sinha: Exercise 16.5 – Mathematics Solution Class 11 Chapter 16 कुछ महत्वपूर्ण अनंत श्रेणी

निम्नलिखित कोे सिद्ध कीजिए  (Prove the following) Question 1 $\log _{e} 2-\frac{1}{2}=\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{3 \cdot 4 \cdot 5}+\ldots$ to $\infty$ Sol : $\log _{e} 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\cdots \infty$ $\log_e2=\frac{2-1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\dots \infty$…

KC Sinha: Exercise 16.5 - Mathematics Solution Class 11 Chapter 16 कुछ महत्वपूर्ण अनंत श्रेणी

KC Sinha: Exercise 16.3 – Mathematics Solution Class 11 Chapter 16 कुछ महत्वपूर्ण अनंत श्रेणी

Question 1 (i) $\frac{e^{2 x}+1}{2 e^{x}}$ Sol : $=\frac{1}{2}\left[\frac{e^{2 x}+1}{e^{x}}\right]$ $=\frac{1}{2}\left[\frac{e^{2 x}}{e^{x}}+\frac{1}{e^{2}}\right]$ $=\frac{1}{2}\left[e^{x}+e^{-x}\right]$ $=\frac{1}{2}\left[2 \cdot\left\{1+\frac{x^{2}}{2!}+\frac{x^{4}}{4 !}+\cdots \cdot \infty\right\}\right]$ $=1+\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}+\cdots \cdot \infty$ (ii) $\frac{e^{x}-e^{-x}}{2}$ Sol : $=\frac{e^{x}\left(e^{4 x}+1\right)}{e^{3 x}}-1$…

KC Sinha: Exercise 16.5 - Mathematics Solution Class 11 Chapter 16 कुछ महत्वपूर्ण अनंत श्रेणी

KC Sinha: Exercise 16.1 – Mathematics Solution Class 11 Chapter 16 कुछ महत्वपूर्ण अनंत श्रेणी

Question 2 यदि (if) |x|<1 तो साबित कीजिए कि (show that) $(1+x)^{-\frac{1}{5}}=1-\frac{x}{5}+\frac{3x^2}{25}-\frac{11x^3}{125}+…to~ \infty$ Sol : L.H.S $=(1+x)^{\frac{-1}{5}}+\frac{\frac{-1}{5} \cdot\left(-\frac{1}{5}-1\right)}{2 !} x^{2}+\frac{-1}{5} \frac{\left(-\frac{1}{5}-1\right)\left(-\frac{1}{5}-2\right)}{3 !} x^{3}+\dots \text{to}~ \infty$ $=1-\frac{x}{5}+\frac{\frac{-1}{5}\left(\frac{-6}{5}\right)}{2} x^{2}+\frac{\frac{-1}{5}\left(-\frac{6}{5}\right)\left(-\frac{11}{5}\right)}{6} x^{3}+\dots~\infty$ $=1-\frac{x}{5}+\frac{3}{25} x^{2}-\frac{11}{125}…

KC Sinha: Exercise 15.10 - Mathematics Solution Class 11 Chapter 15 अनुक्रम और श्रेणी

KC Sinha: Exercise 15.10 – Mathematics Solution Class 11 Chapter 15 अनुक्रम और श्रेणी

Question 1  किसी अनुक्रम के n पदो का जोड़ 2n2+4 है तो इसका n वाँ पद निकालिए । क्या यह अनुक्रम A.P मे है?Sol :Sn=2n2+4 Sn+1=2(n-1)2+4=2(n2-2n+12)2+4=2n2-4n+2+4 Sn+1=2n2-4n+6 an=Sn-Sn-1 =(2n2+4)-(2n2-4n+6) =2n2+4-2n2-4n+6…