Search Previous Year Questions

Calculation:

$\left( \frac{d^2 y}{dx^2} \right)^{3/2} = \left( \frac{dy}{dx} \right)^{5/2}$

Squaring both sides:

$\left( \frac{d^2 y}{dx^2} \right)^3 = \left( \frac{dy}{dx} \right)^5$

Degree = 3

Hence, the correct answer is Option 1.

Calculation:

$\int_n^{n+1} (x - [x]) , dx = \int_n^{n+1} (x - n) , dx$

$= \left[ \frac{x^2}{2} - nx \right]_n^{n+1}$

$= \frac{(n+1)^2}{2} - n(n+1) - \frac{n^2}{2} + n^2$

$= \frac{1 - n^2}{2} + \frac{n^2}{2} = \frac{1}{2}$

Hence, the correct answer is Option 3.

Calculation:

Given:

$y = xe^{2x} \Rightarrow \frac{dy}{dx} = e^{2x} + 2xe^{2x}$

$\frac{dy}{dx} = \frac{y}{x} + 2y = y\left(2 + \frac{1}{x}\right)$

So, statement I is correct.

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

$\Rightarrow \frac{dy}{dx} - \frac{1}{x} \cdot y = 1$

Integrating factor:

$\text{I.F.} = e^{\int -\frac{1}{x} dx} = e^{-\log x} = \frac{1}{x}$

$\therefore y \cdot \frac{1}{x} = \int \frac{1}{x} dx$

$\frac{y}{x} = \ln|x| + c \Rightarrow y = x \ln|x| + cx$

So, statement II is correct.

Hence, the correct answer is option 3.

Calculation:

$\frac{dy}{dx} = \frac{k^2}{(x+y)^2}$

Let $x + y = t \Rightarrow \frac{dt}{dx} = 1 + \frac{dy}{dx}$

⇒ $\frac{dt}{dx} - 1 = \frac{k^2}{t^2} \Rightarrow \frac{dt}{dx} = \frac{k^2 + t^2}{t^2}$

⇒ $\int \frac{t^2}{k^2 + t^2} , dt = \int dx$

⇒ $\int \left(1 - \frac{k^2}{t^2 + k^2} \right) dt = x$

⇒ $t - k^2 \cdot \frac{1}{k} \tan^{-1} \left( \frac{t}{k} \right) = x + c$

$\Rightarrow x + y - k \tan^{-1} \left( \frac{x + y}{k} \right) = x + c$

$\Rightarrow \frac{y - c}{k} = \tan^{-1} \left( \frac{x + y}{k} \right) \Rightarrow \frac{x + y}{k} = \tan \left( \frac{y - c}{k} \right) \Rightarrow x + y = k \tan \left( \frac{y - c}{k} \right)$

Hence, the correct answer is Option 2.

Calculation:

$\int \frac{dx}{10^x + 10^{-x}} = \int \frac{10^x}{(10^x)^2 + 1} , dx$

Let $10^x = t \Rightarrow 10^x \cdot \log_{10} e , dx = dt$

$= \int \frac{1}{1 + t^2} \cdot \log_{10} e , dt$

$= \log_{10} e \cdot \tan^{-1}(t) + c = \log_{10} e \cdot \tan^{-1}(10^x) + c$

$= \frac{1}{\log_e 10} \cdot \tan^{-1}(10^x) + c$

Hence, the correct answer is Option 3.

Calculation:

Given:

Perimeter of rectangle:

$2(x + y) = 20 \Rightarrow x + y = 10 \Rightarrow y = 10 - x$

Area:

$A = xy = x(10 - x) = 10x - x^2$

First derivative:

$\frac{dA}{dx} = 10 - 2x$

Setting derivative to zero for critical point:

$10 - 2x = 0 \Rightarrow x = 5$

Second derivative:

$\frac{d^2A}{dx^2} = -2 < 0$

∴ Area of maximum at =5

y = 10 - 5 = 5 = x

Statement I is correct.

Now testing if area 27 cm² is possible:

$A = 10x - x^2 = 27 \Rightarrow x^2 - 10x + 27 = 0$

Discriminant:

$D = (-10)^2 - 4 \cdot 27 = 100 - 108 = -8 < 0$

∴ Area 27 cm² is not possible So, statement II is not correct.

Hence, the correct answer is Option 1.

Calculation:

Given:

$I_1 = \int_e^{e^2} \frac{dx}{\ln x}$

Let $\ln x = t \Rightarrow x = e^t$, then $dx = e^t dt$

Limits change:

When $x = e \Rightarrow t = 1$

When $x = e^2 \Rightarrow t = 2$

Substituting:

$I_1 = \int_1^2 \frac{e^t}{t} dt = I_2$

$\therefore I_1 - I_2 = 0$

Hence, the correct answer is Option 1.

Calculation:

$|x|\le 2k ;\Rightarrow; -2k \le x \le 2k$

$|y|\le k ;\Rightarrow; -k \le y \le k$

$AB = CD = 2k - (-2k) = 4k$

$AD = BC = k - (-k) = 2k$

Area of rectangle:

$\text{Area} = 4k \cdot 2k = 8k^2$

Hence, the correct answer is Option 4.

Concept:

Monotonicity and Concavity of Function:

• The first derivative f′(x) determines whether a function is increasing or decreasing.
• If f′(x) < 0, the function is decreasing.
• If f′(x) > 0, the function is increasing.
• The second derivative f″(x) determines the concavity of the function.
• If f″(x) > 0, the function is concave upward.
• If f″(x) < 0, the function is concave downward.
• For rational functions like f(x) = 1/(x − a), domain excludes x = a.

Calculation:

Given, f(x) = 1/(x − 5)

First derivative:

⇒ f′(x) = −1/(x − 5)²

⇒ (x − 5)² > 0 for x ≠ 5

⇒ f′(x) < 0 for all x ≠ 5

⇒ Function decreasing for x < 5 and x > 5

Second derivative:

⇒ f″(x) = 2/(x − 5)³

⇒ For x > 5, (x − 5)³ > 0 ⇒ f″(x) > 0

⇒ For x < 5, (x − 5)³ < 0 ⇒ f″(x) < 0

⇒ f″(x) not positive for all x ≠ 5

∴ Only Statement I is correct.

Calculation:

Given function:

$f(x) = \frac{x^3 + 128}{x}$

⇒ $f'(x) = \frac{3x^2 \cdot x - (x^3 + 128)}{x^2} = \frac{2x^3 - 128}{x^2}$

Setting $f'(x) = 0$:

⇒  $2x^3 - 128 = 0 \Rightarrow x = 4$

⇒ $f''(x) = \frac{6x^2 \cdot x^2 - (2x^3 - 128)(2x)}{x^4}$

At $x = 4$:

⇒ $f''(4) = \frac{6 \cdot 16 \cdot 16 - 0}{16 \cdot 16} = 6 > 0$

Minimum occurs at $x = 4$

Minimum value: $f(4) = \frac{64 + 128}{4} = 48$

So, statement I is correct.

Behavior of function:

⇒ $f(x)$ is decreasing for $x < 4$

⇒ $f(x)$ is increasing for $x > 4$

Sign of $f'(x)$ changes from negative to positive as $x$ increases through 4.

Hence, the correct answer is Option 3.