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 = \(\frac{32 \pi}{3}\)

\(\frac{4}{3}\)πr^{3} = \(\frac{32 \pi}{3}\)

4r^{3} = 32

r^{3} = \(\frac{32}{4}\) = 8

r^{3} = 2^{3}

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: