Special Functions MCQ Quiz in मल्याळम - Objective Question with Answer for Special Functions - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

|x| ≥ 0 for every x ∈ R

Calculation:

Let f(x) = |x - 4| + 2

As we know that |x| ≥ 0 for every x ∈ R

∴ |x - 4| ≥ 0

The minimum value of function is attained when |x - 4| = 0

Hence, Minimum value of f(x) = 0 + 2 = 2 

Alternate Method

f(x) =  |x - 4| + 2

There is one critical point i.e. x = 4

f(4) = |4 - 4| + 2

= 0 + 2

= 2

Hence 2 is the minimum value of f(x).

Concept:

Logarithmic formula:

• ${\log }_{\mathrm{a}}\mathrm{b}=\frac{1}{{\log }_{\mathrm{b}}\mathrm{a}}$
• log_a M + log_a N = log_a (MN)

Where a ≠ 1, a > 0 and b ≠ 1, b > 0 and M, N are arbitrary positive numbers and p is any real number.

Calculation:

Given: ${\log }_{4}2=a$

To find: ${\log }_{2}2$

Using property, ${\log }_{\mathrm{a}}\mathrm{b}=\frac{1}{{\log }_{\mathrm{b}}\mathrm{a}}$

$\Rightarrow {\log }_{4}2=\frac{1}{{\log }_{2}4}=\frac{1}{{\log }_2(2\times 2)}$

Using property, log_a M + log_a N = log_a (MN)

$\Rightarrow \mathrm{a}=\frac{1}{({\log }_{2}2+\mathrm{l}\mathrm{o}{\mathrm{g}}_{2}2)}$

$\Rightarrow \mathrm{a}=\frac{1}{2{\log }_{2}2}$

$\Rightarrow 2{\log }_{2}2=\frac{1}{\mathrm{a}}$

$\therefore {\log }_{2}2=\frac{1}{2\mathrm{a}}$

Concept:

If x = log_a b then x = ${\displaystyle \frac{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{b}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{a}}}$

log (mn) = log m + log n

Calculations:

Given, x = logc (ab), y = loga (bc), z = logb (ca),

Consider,x = logc (ab) 

By logarithm property, we have

$\mathrm{x}={\displaystyle \frac{\mathrm{l}\mathrm{o}\mathrm{g}(\mathrm{a}\mathrm{b})}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{c}}}$, $\mathrm{y}={\displaystyle \frac{\mathrm{l}\mathrm{o}\mathrm{g}(\mathrm{b}\mathrm{c})}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{a}}}$and $\mathrm{z}={\displaystyle \frac{\mathrm{l}\mathrm{o}\mathrm{g}(\mathrm{c}\mathrm{a})}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{b}}}$

⇒ from this $\mathrm{x}\mathrm{y}\mathrm{z}\ne 1$ and $\mathrm{x}+\mathrm{y}+\mathrm{z}\ne 1$

Consider,

⇒1 + x = 1 + ${\displaystyle \frac{\mathrm{l}\mathrm{o}\mathrm{g}(\mathrm{a}\mathrm{b})}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{c}}}$

By usiing logarithm property  log (mn) = log m + log n, we have

⇒1 + x =  ${\displaystyle \frac{\log \mathrm{c}+\mathrm{l}\mathrm{o}\mathrm{g}(\mathrm{a}\mathrm{b})}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{c}}}$

⇒ 1+ x = ${\displaystyle \frac{\log \mathrm{a}\mathrm{b}\mathrm{c}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{c}}}$

Similarly, 1 + y = ${\displaystyle \frac{\log \mathrm{a}\mathrm{b}\mathrm{c}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{a}}}$ and 1 + z = ${\displaystyle \frac{\log \mathrm{a}\mathrm{b}\mathrm{c}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{b}}}$

Consider, ${\displaystyle \frac{1}{1+\mathrm{x}}}+{\displaystyle \frac{1}{1+\mathrm{y}}}+{\displaystyle \frac{1}{1+\mathrm{z}}}={\displaystyle \frac{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{c}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{a}\mathrm{b}\mathrm{c}}}+{\displaystyle \frac{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{b}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{a}\mathrm{b}\mathrm{c}}}+{\displaystyle \frac{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{a}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{a}\mathrm{b}\mathrm{c}}}$                                                     

= ${\displaystyle \frac{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{c}+\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{a}+\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{b}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{a}\mathrm{b}\mathrm{c}}}$

= ${\displaystyle \frac{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{a}\mathrm{b}\mathrm{c}}{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{a}\mathrm{b}\mathrm{c}}}$

= 1

Concept:

Formulas and Properties of Logarithms:

1. ${\log }_{a}a=1$
2. ${\log }_a\left(x.y\right)={\log }_{a}x+{\log }_{a}y$
3. ${\log }_a\left(\frac{x}{y}\right)={\log }_{a}x-{\log }_{a}y$
4. ${\log }_a\left(\frac{1}{x}\right)=-{\log }_{a}x$
5. $\mathrm{l}\mathrm{o}{\mathrm{g}}_a{x}^p=p\mathrm{l}\mathrm{o}{\mathrm{g}}_{a}x$
6. $lo{g}_a\left(x\right)=\frac{lo{g}_b\left(x\right)}{lo{g}_b\left(a\right)}$

Calculation:

Given: log (x2 - 2x) = log (5x - 12)

By applying antilog on both the sides we get,

10^{log (x2 - 2x)} = 10^{log (5x - 12)}

⇒ x² - 2x = 5x - 12

⇒ x2 - 7x + 12 = 0

⇒ (x - 3) (x - 4) = 0

⇒ x = 3 and 4

Concept:

Formula of Logarithm:

$a^b=x\iff lo{g}_{a}x=b$, here a ≠ 1 and a > 0 and x be any number.

Properties of Logarithms:

1. ${\log }_{a}a=1$
2. ${\log }_a\left(x.y\right)={\log }_{a}x+{\log }_{a}y$
3. ${\log }_a\left(\frac{x}{y}\right)={\log }_{a}x-{\log }_{a}y$
4. ${\log }_a\left(\frac{1}{x}\right)=-{\log }_{a}x$
5. $\mathrm{l}\mathrm{o}{\mathrm{g}}_a{x}^p=p\mathrm{l}\mathrm{o}{\mathrm{g}}_{a}x$
6. $lo{g}_a\left(x\right)=\frac{lo{g}_b\left(x\right)}{lo{g}_b\left(a\right)}$

Calculation:

Given: $\frac{log\left(324\right)}{log\left(18\right)}=log\left(x\right)$

As we know that, $\frac{log\left(324\right)}{log\left(18\right)}=\frac{lo{g}_{10}\left(324\right)}{lo{g}_{10}\left(18\right)}$ and $log\left(x\right)=lo{g}_{10}\left(x\right)$

Now, $\frac{lo{g}_{10}\left(324\right)}{lo{g}_{10}\left(18\right)}=lo{g}_{10}\left(x\right)$

Using the rule, $a^b=x\iff lo{g}_{a}x=b$ we have,

${10}^{\frac{lo{g}_{10}\left(324\right)}{lo{g}_{10}\left(18\right)}}=x$

${10}^{\frac{lo{g}_{10}\left({18}^2\right)}{lo{g}_{10}\left(18\right)}}=x$

By the power rule we have,

${10}^{\frac{2lo{g}_{10}\left(18\right)}{lo{g}_{10}\left(18\right)}}=x$

⇒ x = 10²

⇒ x = 100

Concept:

Logarithm properties

1. Product rule: The log of a product equals the sum of two logs.

${\log }_a\left(mn\right)={\log }_{a}m+{\log }_{a}n$

2. Quotient rule: The log of a quotient equals the difference of two logs.

${\log }_a\frac{m}{n}={\log }_{a}m-{\log }_{a}n$

3. Power rule: In the log of a power the exponent becomes a coefficient.

${\log }_a{m}^n=n{\log }_{a}m$

4. Change of base rule

${\log }_{m}n=\frac{{\log }_{a}n}{{\log }_{a}m}$

If m = n;
⇒ ${\log }_{m}m=\frac{{\log }_{a}m}{{\log }_{a}m}=1$ (Example, log₁₀ 10 = 1)

5. ${\log }_{m}n=\frac{1}{{\log }_{n}m}$

Calculation:

Given:

(0.2)^x = 2

Taking log both sides,

log₁₀ (0.2)^x = log₁₀ 2

⇒ x log₁₀ (2/10) = 0.3010

⇒ x (log₁₀ 2 - log₁₀ 10) =0.3010

⇒ x (0.3010 - 1) = 0.3010

⇒ x ( - 0.6990) = 0.3010

⇒ x = 0.3010/ ( - 0.6990)

∴ x = - 0.43

So, option 3 is correct.

Concept:

Logarithm properties:

Product rule: The log of a product equals the sum of two logs.

${\log }_{\mathrm{a}}\left(\mathrm{m}\mathrm{n}\right)={\log }_{\mathrm{a}}\mathrm{m}+{\log }_{\mathrm{a}}\mathrm{n}$

Quotient rule: The log of a quotient equals the difference of two logs.

${\log }_{\mathrm{a}}\frac{\mathrm{m}}{\mathrm{n}}={\log }_{\mathrm{a}}\mathrm{m}-{\log }_{\mathrm{a}}\mathrm{n}$

Power rule: In the log of power the exponent becomes a coefficient.

${\log }_{\mathrm{a}}{\mathrm{m}}^{\mathrm{n}}=\mathrm{n}{\log }_{\mathrm{a}}\mathrm{m}$

Calculation:

Given: log (x + 4) + log (x - 4) = 2 log 4

⇒ log [(x + 4) (x - 4)] = 2 log 4        (∵ log m + log n = log mn)

⇒ log (x2 - 42) =  log 4²                   (∵ n log m = log mⁿ)

⇒ (x2 - 42) =  42

⇒ x² = 32

∴ x = $\sqrt{32}=4\sqrt{2}$

Important Points

Logarithm properties:

Change of base rule:

${\log }_{\mathrm{m}}\mathrm{n}=\frac{{\log }_{\mathrm{a}}\mathrm{n}}{{\log }_{\mathrm{a}}\mathrm{m}}$  and ${\log }_{\mathrm{m}}\mathrm{n}=\frac{1}{{\log }_{\mathrm{n}}\mathrm{m}}$

If m = n;

${\log }_{\mathrm{m}}\mathrm{m}=\frac{{\log }_{\mathrm{a}}\mathrm{m}}{{\log }_{\mathrm{a}}\mathrm{m}}=1$

Concepts:

• ${\log }_{e}y=x\iff y=e^x$

Calculation:

Given: log₁₀ a is negative.

⇒ log₁₀ a < 0

Let log₁₀ a = - b (Here b is positive)

⇒ a = 10^-b

Given a lies between 0 to 1

⇒ 0 < a < 1

⇒ 0 < 10^-b < 1

Hence it is negative power of 10 is between 0 and 1

GIVEN:

2log₁₀(x + 1) = log₁₀(7x + 1)

CONCEPT:

Algebric and logarithmic concept

FORMULA USED:

alog_b(x) = log_bx^a

CALCULATION:

Considering the given equation

2log₁₀(x + 1) = log₁₀(7x + 1)

⇒ log₁₀(x + 1)² = log₁₀(7x + 1)

After comparing:

⇒ (x + 1)² = (7x + 1)

⇒ x² + 1 + 2x = 7x + 1

⇒ x² – 5x = 0

⇒ x(x – 5) = 0

x = 0 and 5

Formulae Used:

log₁₀ x^m = m × log₁₀ x 

log₁₀ (ab) = log10 a + log10 b 

log10 (a/b) = log10 a – log10 b

Calculation:

5x - 3 = 8

5x - 3 = 2³

Taking log₁₀ on both sides,

(x – 3)log₁₀ 5 = 3log₁₀ 2

⇒ x – 3 =  3log10 2/log10 5

⇒ x = (3log10 2/log10 5) + 3

⇒ x = 3[(log10 2/log10 5) + 1]

⇒ x = 3[(log10 2 + log10 5)/log10 5]

⇒ x = 3{[log10 (2 × 5)]/log10 5}

⇒ x = 3(1/log10 5)

⇒ x = 3/log10 5

⇒ x = 3/[log10 (10/2)]

⇒ x = 3/(log10 10 – log₁₀ 2)

∴ x = 3/(1 – log10 2)