NEET Physics Solutions For Class 11 Chapter 10 Mathematical Tools – Algebra

NEET Physics Class 11 Chapter 10 Mathematical Tools – Algebra

Quadratic Equation And Its Solution:

An algebraic equation of second order (the highest power of the variable is equal to 2) is called a quadratic equation. Equation ax² + bx + c = 0 is the general quadratic equation.

The general solution of the above quadratic equation or value of variable x

⇒ \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x_1=\frac{-b+\sqrt{b^2-4 a c}}{2 a}\)

and \(x_2=\frac{-b-\sqrt{b^2-4 a c}}{2 a}\)

Sum of roots = x₁ + x₂ = – \(\frac{\mathrm{b}}{\mathrm{a}}\) and product of roots = x₁ + x₂ = \(\frac{\mathrm{c}}{\mathrm{a}}\)

For real roots b²– 4ac ≥ 0 and for imaginary roots b² – 4ac < 0.

Question 1. find roots of f (x) = x² – 4x + 3, f (x) = – x² + 3x – 5 Every quadratic equation has 2 roots (x₁ x₂) such that f (x₁) and f (x₂) is zero.

Graph of quadratic Equation: ax² + bx + c

Graph of a quadratic equation is of parabolic nature.

Points where the graph cuts the x-axis are roots of a quadratic equation.

Binomial Expression: An algebraic expression containing two terms is called a binomial expression.

For example (a + b), (a + b)³, (2x – 3y)^-1, \(\left(x+\frac{1}{y}\right)\),etc are binomial theorem

Binomial Theorem

⇒ \((a+b)^n=a^n+n a^{n-1} b^1+\frac{n(n-1)}{2 \times 1} a^{n-2} b^2+\ldots \ldots \ldots \ldots,(1+x)^n\)

⇒ \(1+n x+\frac{n(n-1)}{2 \times 1} x^2+\ldots \ldots \ldots\)

Binomial Approximation

If x is very small, then terms containing higher powers of x can be neglected so (1 + x)n = 1 + nx

Logarithm Definition: Every positive real number N can be expressed in exponential form as

N = a^x…… (1)for example.,  49 = 7²

where ‘a’ is also a positive real different than unity and is called the base and ‘x’ is called the exponent. We can write the relation (1) in logarithmic form as

loga N = x ……..  (2)

Hence the two relations and

\(\left.\log _a \begin{array}{c}
a^x=N \\
N=x
\end{array}\right]\)

 

are identical where N > 0, a > 0, a ≠ 1

Hence logarithm of a number to some base is the exponent by which the base must be raised to get that number.

The logarithm of zero does not exist and the logarithm of (–) ve reals is not defined in the system of real numbers. a is raised what power to get N

Question 2. Find value of

1. \(\log _{81} 27\)
2. \(\log _{10} 100\)
3. \(\log 9 \sqrt{3}\)

Answer:

1. \(\log _{81} 27\)

⇒ \(3^3=3^{4 x}\)

gives x = 3/4

2. \(\log _{10} 100\)

⇒ \(10^2=10^x\)

gives x = 2

3. \(\log 9 \sqrt{3}\)

⇒ \(9 \sqrt{3}\left(\frac{1}{3}\right)^x\)

⇒ \(3^{5 / 2}=3^{-x}\)

gives x = -5/2

Note:

Unity has been excluded from the base of the logarithm as in this case, log1N will not be possible, and if N = 1 then log11 will have infinitely many solutions and will not be unique which is necessary in the functional notation.

a N log N a = is an identify for all N > 0 and a > 0, a ≠ 1 for example.,  \(2^{\log _2 5}\) = 5

The number N in (2) is called the antilog of ‘x’ to the base ‘a’. Hence If log₂512 is 9 then antilog₂9 is equal to 2² = 512

Using the basic definition of log we have 3 important deductions :

log_NN = 1 i.e. logarithm of the number to the same base is 1.

⇒ \(\log _{\frac{1}{N}}\) =–1 i.e. logarithm of a number to its reciprocal is –1.

log_a 1 = 0 i.e. logarithm of unity to any base is zero. (basic constraints on number and base must be observed.)

⇒ \(\log ^{\log _a n}\)= n is an identify for all N > 0 and a > 0; a ≠ 1 for example., \(\log ^{\log _a n}\)

Whenever the number and base are on the same side of unity then the logarithm of that number to the base is (+ve), however, if the number and base are located on different sides of unity then the logarithm of that number to the base is (–ve) for example.,

log₁₀ 100 = 2

log_1/10 100 =– 2

For a non negative number ‘a’ and \(n \geq 2, n \in N \sqrt[n]{a}=a^{1 / n}\)

Componend And Dividend Rule

If \(\frac{p}{q}=\frac{a}{b}\) then \(\frac{p+q}{p-q}=\frac{a+b}{a-b}\)

Arithmetic Progression (AP)

General from: a, a + d, a + 2d, ……………a + (n – 1)d

Here a = first term, d = common difference

Sum of n terms \(S_n=\frac{n}{2}[a+a+(n-1) d]=\frac{n}{2}\left[I^{\text {st }} \text { term }+n^{\text {th }} \text { term }\right]\)