Exponential Function MCQ Quiz in தமிழ் - Objective Question with Answer for Exponential Function - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Greatest Integer Function:

Greatest Integer Function [x] indicates an integral part of the real number x which is the nearest and smaller integer to x. It is also known as the floor of x

• In general, If n≤ x ≤ n+1 Then [x] = n     (n ∈ Integer)
• This means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.

Calculation:

Let

f(x) = |x| and g(x) = [x] - 1, where [.] is the greatest integer function.

$h(x)=\frac{f(g(x))}{g(f(x))}$.

$h(x)=\frac{|[x]-1|}{[|x|]-1}$

$\Rightarrow {\displaystyle \underset{x\to 0-}{lim}\frac{|[x]-1|}{[|x|]-1}}$

For a number lesser than 0, Let us say x = - 0.1, so the greatest integer [x] = -1

$\Rightarrow {\displaystyle \underset{x\to 0-}{lim}\frac{|-1-1|}{[-x]-1}}$

$\Rightarrow {\displaystyle \underset{x\to 0-}{lim}\frac{|-2|}{[-x]-1}}$

|-2| = 2, For a number lesser than 0, Let us say x = - 0.1, so the greatest integer [-x] = 0

$\Rightarrow \frac{2}{0-1}$

$\therefore {\displaystyle \underset{x\to 0-}{lim}\frac{|-2|}{[-x]-1}=-2}$

Explanation:

It is given that $p(n)=x^{3^n}+y^{3^n}$ is divisible by x + y.

Let us put n = 0, we get

$p(0)=x^{3^0}+y^{3^0}$

⇒ p(0) = x + y      (∵ a⁰ = 1)

⇒ p(0) is divisible by x + y

Now, let's put n = 1, we get

p(1) = x³ + y³ 

⇒ p(1) = (x + y)(x² + y² - xy)      [∵ x3 + y3 = (x + y)(x2 + y2 - xy)]

⇒ p(1) is divisible by x + y

Now, let's put n = 2, 

$p(2)=x^{3^2}+y^{3^2}$

⇒ p(2) = x⁹ + y⁹ 

⇒ p(2) = (x³ + y³)(x⁶ + y⁶ - x³y³)

⇒ p(2) is divisible by x + y

Thus, we can say that p(n) is divisible by x + y for all values of n ≥ 0, where n is an integer. 

Given -

Let f(x) = |x| + 1 and g(x) = [x] - 1, $h(x)=\frac{f(x)}{g(x)}$

where [.] is the greatest integer function.

Let $h(x)=\frac{f(x)}{g(x)}$

Explanation -

Step 1: Evaluate ${\displaystyle \underset{x\to 0^{-}}{lim}h(x)}$

For x → 0^- (approaching 0 from the left):

f(x) = |x| + 1 = -x + 1 (since x < 0 )

g(x) = [x] - 1 = -1 - 1 = -2 (since [x] = -1 when -1 < x < 0 )

Thus, 

h(x) = $\frac{f(x)}{g(x)}=\frac{-x+1}{-2}=\frac{x-1}{2}$

As x → 0^- :

$li{m}_{x\to 0^{-}}h(x)=\frac{0-1}{2}=-\frac{1}{2}$

Step 2: Evaluate $\underset{x\to 0^{+}}{lim}h(x)$

For x → 0⁺ (approaching 0 from the right):

f(x) = |x| + 1 = x + 1 (since x > 0 )

g(x) = [x] - 1 = -1 (since [x] = 0 when 0 ≤ x < 1 )

Thus,

h(x) = $\frac{f(x)}{g(x)}=\frac{x+1}{-1}=-(x+1)$

As x → 0⁺ :

$\underset{x\to 0^{+}}{lim}h(x)=-(0+1)=-1$

Step 3: Sum of the limits

Now, sum the left-hand limit and the right-hand limit:

$\underset{x\to 0^{-}}{lim}h(x)+\underset{x\to 0^{+}}{lim}h(x)=-\frac{1}{2}+(-1)=-\frac{1}{2}-1=-\frac{3}{2}$

So, the answer is -3/2.

Given -

Let f(x) = |x| + 1 and g(x) = [x] - 1, $h(x)=\frac{f(x)}{g(x)}$

where [.] is the greatest integer function.

Let $h(x)=\frac{f(x)}{g(x)}$

Explanation -

Statement 1: f(x) is differentiable for all x < 0 .

The function f(x) = |x| + 1 can be written as:

$f(x)=-x+1\text{for}x<0$

The function -x + 1 is differentiable for all x < 0 because it is a linear function.

Therefore, Statement 1 is correct.

Statement 2: g(x) is continuous at x = 0.0001 .

The function g(x) = [x] - 1 :

For $0\le x<1,$ [x] = 0

Therefore, g(x) =[x] - 1 = 0 - 1 = -1

At x = 0.0001 , which is in the interval 0 ≤ x < 1 :

⇒ g(0.0001) = -1

Since the greatest integer function is not continuous at integer points, but 0.0001 is not an integer, g(x) does not have any discontinuity at x = 0.0001 .

Therefore, Statement 2 is correct.

Statement 3: The derivative of g(x) at x = 2.5 is 1.

The function g(x) =[x] - 1 is a piecewise constant function, which means its derivative is zero almost everywhere.

For 2 ≤ x < 3 , [x] = 2

Therefore, g(x) = 2 - 1 = 1

Since g(x) is constant in the interval 2 ≤ x < 3 , its derivative is 0 in this interval.

Thus, Statement 3 is incorrect.

Hence Option (1) is Correct.

The correct answer is 6⁴. 

Key Points

• As the above Question:  
  • 152 = 15*15 = 225
  • 6⁴ = 6*6*6*6 = 1296
  • 2⁷ = 2*2*2*2*2*2*2 = 128
  • 3⁵ = 3*3*3*3*3 = 243

So, 6⁴ has the highest value. 

Concept:

$\frac{d(uv)}{dx}=u\times \frac{dv}{dx}+v\times \frac{du}{dx}$

Calculation:

Given: y = ex log x

By applying log both the sides we get

⇒ log y = log (e)x log x

⇒ log y = x log x log e             [log xa = a log x]

⇒ log y = x log x      ------(i)      [log e = 1]

Differentiating both sides with respect to x, using product rule $\frac{d(uv)}{dx}=u\times \frac{dv}{dx}+v\times \frac{du}{dx}$, we get,

⇒ $\frac{d(logy)}{dx}=x\times \frac{d(logx)}{dx}+logx\times \frac{dx}{dx}$

⇒ $\frac{1}{y}\frac{dy}{dx}=x\times \frac{1}{x}\frac{dx}{dx}+logx$

⇒ $\frac{1}{y}\frac{dy}{dx}=1+logx$

 $\frac{dy}{dx}=y[1+logx]$ 

Differentiating both sides with respect to x, using product rule $\frac{d(uv)}{dx}=u\times \frac{dv}{dx}+v\times \frac{du}{dx}$, we get,

$\frac{d^{2}y}{d{x}^2}=[y.\frac{1}{x}.\frac{dx}{dx}+(1+logx)\frac{dy}{dx}]$

⇒ $\frac{d^{2}y}{d{x}^2}=[\frac{y}{x}+(1+logx)\frac{dy}{dx}]$

⇒ $\frac{d^{2}y}{d{x}^2}=[\frac{x^x}{x}+(1+logx)x^x(1+logx)]$.....(here, $\frac{dy}{dx}=x^x(1+logx)$)

⇒ $\frac{d^{2}y}{d{x}^2}=x^x(1+logx)[\frac{1}{x(1+logx)}+(1+logx)]$

Now, $\frac{d^{2}y}{d{x}^2}$ at x = 1 = 2.

∴ $\frac{d^{2}y}{d{x}^2}$ at x = 1 equal to 2.

Given:

The given limiting function is as follows,

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\frac{{\mathrm{e}}^{\mathrm{x}}-1-\mathrm{x}-{\mathrm{x}}^2}{{\mathrm{x}}^2}}$ 

Concept:

Use the L'hospital rule for the limit $\underset{x\to a}{lim}\frac{f(x)}{g(x)}$ of the form $\frac{0}{0}$

$\underset{x\to a}{lim}\frac{f(x)}{g(x)}=\underset{x\to a}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Where,

f'(x) and g'(x) are the derivative of f(x) and g(x).

Solution:

The given limit is as follows,

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\frac{{\mathrm{e}}^{\mathrm{x}}-1-\mathrm{x}-{\mathrm{x}}^2}{{\mathrm{x}}^2}}$ 

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\frac{{\mathrm{e}}^{\mathrm{x}}-1-\mathrm{x}-{\mathrm{x}}^2}{{\mathrm{x}}^2}=\frac{0}{0}}$

Derivating f(x) and g(x) we will get,

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\frac{{\mathrm{e}}^{\mathrm{x}}-1-2\mathrm{x}}{2\mathrm{x}}=\frac{0}{0}}$

Still, it is of $\frac{0}{0}$ form.

Derivating again we will get,

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\frac{{\mathrm{e}}^{\mathrm{x}}-2}{2}=\frac{1-2}{2}=\frac{-1}{2}}$

Hence, option 3 is correct.

Calculation:

Given that,

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\frac{{\mathrm{a}}^{\mathrm{x}}-{\mathrm{b}}^{\mathrm{x}}}{\mathrm{x}}}$,   form $\left(\frac{0}{0}\right)$

Using L' Hospital rule then

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\frac{{\mathrm{a}}^{\mathrm{x}}\cdot \log \mathrm{a}-{\mathrm{b}}^{\mathrm{x}}\log \mathrm{b}}{1}}$

Applying limit, we get -

${\displaystyle \underset{\mathrm{x}\to 0}{lim}\frac{{\mathrm{a}}^{\mathrm{x}}-{\mathrm{b}}^{\mathrm{x}}}{\mathrm{x}}=\frac{{\mathrm{a}}^0,\log \mathrm{a}-{\mathrm{b}}^0\cdot \log \mathrm{b}}{1}}$ = log a − log b

= log $\left(\frac{\mathrm{a}}{\mathrm{b}}\right)$

The correct answer is option "1"

Given:

Let f(x) = [x] - 1 and g(x) = x|x|, where [.] is the greatest integer function.

Calculation:

$h(x)=\frac{g(f(x))}{f(g(x))}$.

${\displaystyle \underset{x\to 0^{+}}{lim}h(x)={\displaystyle \underset{x\to 0^{+}}{lim}\frac{([x]-1)(|[x]-1|)}{[x|x|]-1}}}$

For x > 0, |x| = x,

${\displaystyle \underset{x\to 0^{+}}{lim}h(x)={\displaystyle \underset{x\to 0^{+}}{lim}\frac{([x]-1)(|[x]-1|)}{[x^2]-1}}}$

For a number greater than 0, Let us say x = 0.1, so the greatest integer [x] = 0

For a number greater than 0, Let us say x² = 0.01, so the greatest integer [x] = 0

$\Rightarrow \frac{(0-1)(|0-1|)}{-1}$

As, |-1| = 1, 

$\Rightarrow \frac{(-1)(1)}{-1}$ = 1

∴ The correct answer is 1, i.e. option (3).

Solution- 

The constant r > 0 in the exponential growth function - p(t)= p0ert 

where , r is growth rate . 

Therefore, Correct answer is Option 1).