Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 கணங்கள், தொடர்புகள் மற்றும் naசார்புகள் Ex 11.12 Textbook Questions and Answers, Notes.
TN Board 11th Maths Solutions Chapter 11 கணங்கள், தொடர்புகள் மற்றும் சார்புகள் Ex 11.12
பின்வரும் சார்புகளின் தொகைக் காண்க.
Question 1.
(i) \(\sqrt{x^{2}+2 x+10}\)
(ii) \(\sqrt{x^{2}-2 x-3}\)
(iii) \(\sqrt{(6-x)(x-4)}\)
தீர்வு :
(i) \(\sqrt{x^{2}+2 x+10}\)
I = ∫ \(\sqrt{x^{2}+2 x+10}\) dx
= ∫ \(\sqrt{x^{2}+2 x+1-1+10}\) dx
= ∫ \(\sqrt{(x+1)^{2}+9}\) dx
= ∫ \(\sqrt{(x+1)^{2}+3^{2}}\) dx
∫ \(\sqrt{x^{2}+a^{2}}\) dx = ∫ \(\frac{x}{2}\) \(\sqrt{x^{2}+a^{2}}\) + \(\frac{a^{2}}{2}\) log |x + \(\sqrt{x^{2}+a^{2}}\)| + c
∴ I = \(\frac{x+1}{2} \sqrt{x^{2}+2 x+10}\) + \(\frac{9}{2}\) log |x + 1 + \(\sqrt{x^{2}+2 x+10}\)| + c
(ii) \(\sqrt{x^{2}-2 x-3}\)
I = ∫ \(\sqrt{x^{2}-2 x-3}\) dx என்க
= ∫ \(\sqrt{x^{2}-2 x+1-1-3}\) dx
= ∫ \(\sqrt{(x-1)^{2}-4}\) dx
= ∫ \(\sqrt{(x-1)^{2}-2^{2}}\) dx
ஆனால் ∫ \(\sqrt{x^{2}-a^{2}}\) dx = \(\frac{x}{2}\) \(\sqrt{x^{2}-a^{2}}\) – \(\frac{a^{2}}{2}\) log |x + \(\sqrt{x^{2}-a^{2}}\)| + c
∴ I = \(\frac{x-1}{2} \sqrt{x^{2}-2 x-3}\) – \(\frac{4}{2}\) log |x – 1 + \(\sqrt{x^{2}-2 x-3}\)| + c
= \(\frac{x-1}{2} \sqrt{x^{2}-2 x-3}\) – 2 log |x – 1 + \(\sqrt{x^{2}-2 x-3}\)| + c
(iii) \(\sqrt{(6-x)(x-4)}\)
I = ∫ \(\sqrt{(6-x)(x-4)}\) dx என்க
= ∫ \(\sqrt{6 x-24-x^{2}+4 x}\) dx
∴ I = ∫ \(\sqrt{-x^{2}+10 x-24}\) dx
= ∫ \(\sqrt{-\left(x^{2}-10 x+24\right)}\) dx
= ∫ \(\sqrt{-\left(x^{2}-10 x+25-25+24\right)}\) dx
= ∫ \(\sqrt{-\left[(x-5)^{2}-1^{2}\right]}\) dx
= ∫ \(\sqrt{1^{2}-(x-5)^{2}}\) dx
∫ \(\sqrt{a^{2}-x^{2}}\) dx = \(\frac{x}{2}\) \(\sqrt{a^{2}-x^{2}}\) + \(\sqrt{a^{2}-x^{2}}\) sin-1 (\(\frac{x}{a}\)) + c
∴ I = \(\frac{x-5}{2} \sqrt{(6-x)(x-4)}+\frac{1}{2} \sin ^{-1}\left(\frac{x-5}{1}\right)\) + c
= \(\frac{x-5}{2} \sqrt{(6-x)(x-4)}+\frac{1}{2} \sin ^{-1}(x-5)\) + c
Question 2.
(i) \(\sqrt{9-(2 x+5)^{2}}\)
தீர்வு :
I = ∫ \(\sqrt{9-(2 x+5)^{2}}\) dx
t = 2x + 5 ⇒ dt = 2 dx ⇒ \(\frac{d t}{2}\) = dx
∴ I = \(\sqrt{3^{2}-t^{2}}\) \(\frac{d t}{2}\)
= \(\frac{1}{2}\) ∫ \(\sqrt{3^{2}-t^{2}}\) dt
= \(\frac{1}{2}\) [\(\frac{t}{2}\) \(\sqrt{3^{2}-t^{2}}\) + \(\frac{9}{2}\) sin-1 (\(\frac{t}{3}\)) + c
= \(\frac{1}{2}\) [\(\frac{2 x+5}{2} \sqrt{9-(2 x+5)^{2}}\) + \(\frac{9}{2}\) sin-1 \(\left(\frac{2 x+5}{3}\right)\)] + c
= \(\frac{1}{4}\) [(2x + 5) \(\sqrt{9-(2 x+5)^{2}}\) + 9 sin-1 \(\left(\frac{2 x+5}{3}\right)\)] + c
(ii) \(\sqrt{81+(2 x+1)^{2}}\)
தீர்வு :
I = ∫ \(\sqrt{81+(2 x+1)^{2}}\) dx என்க
t = 2x + 1 ⇒ dt = 2 . dx ⇒ \(\frac{d t}{2}\) = dx
∴ I = ∫ \(\sqrt{9^{2}+t^{2}} \frac{d t}{2}\)
= \(\frac{1}{2} \int \sqrt{9^{2}+t^{2}}\) dt
\(\int \sqrt{a^{2}+x^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c\)I = \(\frac{t}{2}\) \(\sqrt{81+t^{2}}\) + \(\frac{81}{2}\) log |t + \(\sqrt{81+t^{2}}\)| + c
∴ I = \(\frac{1}{2}\) ∫ \(\frac{2 x+1}{2} \sqrt{81+(2 x+1)^{2}}\) + \(\frac{81}{2}\) log |2x + 1 + \(\sqrt{81+(2 x+1)^{2}}\)|] + c
= ∫ \(\frac{1}{4}\) [(2x + 1) \(\sqrt{81+(2 x+1)^{2}}\) + 81 log |2x + 1 + \(\sqrt{81+(2 x+1)^{2}}\)|] + c
(iii) \(\sqrt{(x+1)^{2}-4}\)
தீர்வு :
I = ∫ \(\sqrt{(x+1)^{2}-4}\) dx
= ∫ \(\sqrt{(x+1)^{2}-2^{2}}\) dx என்க
t = x + 1 ⇒ dt = dx
I = ∫ \(\sqrt{t^{2}-2^{2}}\) dt
∫ \(\sqrt{x^{2}-a^{2}}\) dx = \(\frac{x}{2}\) \(\sqrt{x^{2}-a^{2}}\) – \(\frac{a^{2}}{2}\) log |x + \(\sqrt{x^{2}-a^{2}}\)| + c
I = \(\frac{t}{2}\) \(\sqrt{t^{2}-4}\) – \(\frac{4}{2}\) log|t + \(\sqrt{t^{2}-4}\)| + c
I = \(\frac{x+1}{2} \sqrt{(x+1)^{2}-4}\) – 2 log |x + 1 + \(\sqrt{(x+1)^{2}-4}\)| + c