Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.11 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.11

Question 1.
Simplify
(a) (125)^2/3
(b) 16^-3/4
(c) (- 1000)^-2/3
(d) (3^-6)^1/3
(e) \(\frac{27^{-\frac{2}{3}}}{27^{-\frac{1}{3}}}\)
Answer:
(a) (125)^2/3

(b) 16^-3/4

(c) (- 1000)^-2/3

(d) (3^-6)^1/3

(e) \(\frac{27^{-\frac{2}{3}}}{27^{-\frac{1}{3}}}\)

Question 2.
Evaluate
Answer:

Question 3.
If \(\left(x^{\frac{1}{2}}+x^{-\frac{1}{2}}\right)^{2}=\frac{9}{2}\) then find the value of \(\left(x^{\frac{1}{2}}-x^{-\frac{1}{2}}\right)\) for x > 1
Answer:

Question 4.
Simplify and hence find the value of n:
\(\frac{3^{2 n} \cdot 9^{2} \cdot 3^{-n}}{3^{3 n}}\) = 27
Answer:
Given

4 – 2n = 3
2n = 4 – 3
2n = 1
n = \(\frac { 1 }{ 2 }\)

Question 5.
Find the radius of the spherical tank whose volume is \(\frac{32 \pi}{3}\) units.
Answer:
Let r be the radius of the spherical tank.
Given volume of the spherical tank = \(\frac{32 \pi}{3}\)
\(\frac{4}{3}\)πr³ = \(\frac{32 \pi}{3}\)
4r³ = 32
r³ = \(\frac{32}{4}\) = 8
r³ = 2³
r = 2
∴ Radius of the spherical tank r = 2 units.

Question 6.
Simplify by rationalizing the denominator \(\frac{7+\sqrt{6}}{3-\sqrt{2}}\)
Answer:
\(\frac{7+\sqrt{6}}{3-\sqrt{2}}\)
Multiply the numerator and denominator by 3 + √2

Question 7.
Simplify:

Answer:

Question 8.
If x = √2 + √3 find \(\frac{x^{2}+1}{x^{2}-2}\)
Answer: