Exponential Function MCQ Quiz in मराठी - Objective Question with Answer for Exponential Function - मोफत PDF डाउनलोड करा

Formula used:

X(a + b) = X^c then, (a + b) = c

Calculation:

Let 2x = 3y = 6^+z = k 

2x = k then, 2 = k^1/x

3^y = k then, 3 = k1/y

6^+z = k then, 6 = k1/z 

Now, We know that

2 × 3 = 6

⇒ k^1/x × k^1/y = k^1/z

⇒ k^{(1/x + 1/y)} = k^1/z

⇒ 1/x + 1/y = 1/z

⇒ -(1/x + 1/y) + 1/z = 0 

Note:- The official question is wrong, the updated question and the solution is provided.

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 \(\displaystyle \lim _{x \rightarrow 0^{-}} h(x)\)

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^- :

\(lim_{{x → 0^-}} h(x) = \frac{0 - 1}{2} = -\frac{1}{2}\)

Step 2: Evaluate \(\lim_{{x → 0^+}} h(x)\)

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⁺ :

\(\lim_{{x → 0^+}} h(x) = -(0 + 1) = -1\)

Step 3: Sum of the limits

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

\(\lim_{{x → 0^-}} h(x) + \lim_{{x → 0^+}} h(x) = -\frac{1}{2} + (-1) = -\frac{1}{2} - 1 = -\frac{3}{2}\)

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:

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

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 \(0 ≤ x < 1 , \) [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,

⇒ \(\frac{d(log \ y)}{dx} = x \times \frac{d(log\ x)}{dx} + log \ x \times \frac{dx}{dx}\)

⇒ \(\frac{1}{y}\frac{dy}{dx} = x \times \frac{1}{x} \frac{dx}{dx}+ log \ x\)

⇒ \(\frac{1}{y}\frac{dy}{dx} = 1+ log \ x\)

 \(\frac{dy}{dx} = y[1+ log \ x]\) 

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,

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

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

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

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

Now, \(\frac{d^2y}{dx^2}\) at x = 1 = 2.

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

Given:

The given limiting function is as follows,

\(​\rm\displaystyle\lim _{x \rightarrow 0}\frac{e^x−1−x−x^2}{x^2}\) 

Concept:

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

\(\lim_{x \to a} \frac{f(x)}{g(x)}=\lim_{x \to a} \frac{f'(x)}{g'(x)}\)

Where,

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

Solution:

The given limit is as follows,

\(​\rm\displaystyle\lim _{x \rightarrow 0}\frac{e^x−1−x−x^2}{x^2}\) 

\(​\rm\displaystyle\lim _{x \rightarrow 0}\frac{e^x−1−x−x^2}{x^2}=\frac{0}{0}\)

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

\(​\rm\displaystyle\lim _{x \rightarrow 0}\frac{e^x−1−2x}{2x}=\frac{0}{0}\)

Still, it is of \(\frac{0}{0}\) form.

Derivating again we will get,

\(​\rm\displaystyle\lim _{x \rightarrow 0}\frac{e^x-2}{2}=\frac{1-2}{2}=\frac{-1}{2}\)

Hence, option 3 is correct.

Calculation:

Given that,

\(\rm\displaystyle\lim_{x \rightarrow 0} \frac{a^x−b^x}{x}\),   form \(\left(\frac{0}{0}\right)\)

Using L' Hospital rule then

\(\rm\displaystyle\lim_{x \rightarrow 0} \frac{a^x \cdot \log a−b^x \log b}{1}\)

Applying limit, we get -

\(\rm\displaystyle\lim_{x \rightarrow 0} \frac{a^x−b^x}{x}=\frac{a^{0}, \log a−b^{0} \cdot \log b}{1}\) = log a − log b

= log \(\rm\left(\frac{a}{b}\right)\)

The correct answer is option "1"

Given:

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

Calculation:

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

\(\displaystyle \lim _{x \rightarrow 0^+} h(x)=\displaystyle \lim _{x \rightarrow 0^+}\frac{([x]-1) (|[x]-1|)}{[x|x|]-1}\)

For x > 0, |x| = x,

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

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

\(⇒ \frac{(0-1)(|0-1|)}{-1}\)

As, |-1| = 1, 

\(⇒ \frac{(-1)(1)}{-1}\) = 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).