1. ⇒ ( MHT CET 2023 12th May Evening Shift)

For $x > 1$ , if $(2 x)^{2 y} = 4 e^{2 x - 2 y}$ , then ${(1 + \log_{e} 2 x)}^{2} \frac{d y}{d x}$ is equal to

A. $\frac{x \log_{e} 2 x + \log_{e} 2}{x}$

B. $\frac{x \log_{e} 2 x - \log_{e} 2}{x}$

C. $x \log_{e} 2 x + \frac{\log_{e} 2}{x}$

D. $x \log_{e} 2 x - \frac{\log_{e} 2}{2}$

Correct Option is (B)

$(2 x)^{2 y} = 4 e^{2 x - 2 y}$

Taking ' $\log_{e}$ ' on both sides, we get

$\begin{aligned} 2 y \log_{e} (2 x) = \log_{e} 4 + (2 x - 2 y) \log_{e} e \\ y \log_{e} (2 x) = \log_{e} 2 + x - y ..... (i) \end{aligned}$

Differentiating w.r.t. $x$ , we get

$\begin{array}{ll} [\log_{e} 2 x] \frac{d y}{d x} + \frac{2 y}{2 x} = 0 + 1 - \frac{d y}{d x} \\ [1 + \log_{e} 2 x] \frac{d y}{d x} = 1 - \frac{y}{x} .... (ii) \\ Now, (i) \Rightarrow y = \frac{\log_{e} 2 + x}{1 + \log_{e} 2 x} \\ ∴ & (ii) \Rightarrow (1 + \log_{e} 2 x) \frac{d y}{d x} = \frac{x \log_{e} 2 x - \log_{e} 2}{x (1 + \log_{e} 2 x)} \\ ∴ & {(1 + \log_{e} 2 x)}^{2} \frac{d y}{d x} = \frac{x \log_{e} 2 x - \log_{e} 2}{x} \end{array}$

2. ⇒ (MHT CET 2023 12th May Morning Shift )

$y = (1 + x) (1 + x^{2}) (1 + x^{4}) \dots \dots \dots (1 + x^{2 n})$ , then the value of $\frac{d y}{d x}$ at $x = 0$ is

A. 0

B. $-$ 1

C. 1

D. 2

Correct Option is (C)

$y = (1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2 n})$ .... (i)

Taking ' $\log$ ' on both sides, we get

$\begin{aligned} \log y = \log (1 + x) + \log (1 + x^{2}) & + \log (1 + x^{4}) \\ + \dots + \log (1 + x^{2 n}) \end{aligned}$

Differentiating w.r.t. $x$ , we get

$\frac{1}{y} \frac{d y}{d x} = \frac{1}{1 + x} + \frac{2 x}{1 + x^{2}} + \frac{4 x^{3}}{1 + x^{4}} + \dots + \frac{2 n \times x^{2 n - 1}}{1 + x^{2 n}}$ ..... (ii)

At $x = 0$ , (i) $\Rightarrow y = 1$

$∴$ (ii) ${\Rightarrow \frac{d y}{d x} |}_{x = 0} = 1 + 0 + 0 + \dots + 0 = 1$

3. ⇒ (MHT CET 2023 11th May Evening Shift )

If $y = [(x + 1) (2 x + 1) (3 x + 1) \dots (n x + 1)]^{\frac{3}{2}}$ , then $\frac{d y}{d x}$ at $x = 0$ is

A. $\frac{3 n (n + 1)}{4}$

B. $\frac{n (n + 1)}{2}$

C. $\frac{3 n (n + 1)}{2}$

D. $\frac{n (n + 1)}{4}$

Correct Option is (A)

$y = [(x + 1) (2 x + 1) (3 x + 1) \dots (n x + 1)]^{\frac{3}{2}}$

Taking ' $\log$ ' on both sides, we get

$\begin{array}{r} \log y = \frac{3}{2} [\log (x + 1) + \log (2 x + 1) + \log (3 x + 1) \\ + \dots + \log (n x + 1)] \end{array}$

Differentiating w.r.t. $x$ , we get

$\begin{aligned} \frac{1}{y} \frac{d y}{d x} = \frac{3}{2} [\frac{1}{x + 1} + \frac{2}{2 x + 1} + \frac{3}{3 x + 1} + \dots + \frac{n}{n x + 1}] \\ \frac{d y}{d x} = \frac{3 y}{2} [\frac{1}{x + 1} + \frac{2}{2 x + 1} + \frac{3}{3 x + 1} + \dots + \frac{n}{n x + 1}] \end{aligned}$

Now at $x = 0, y = [\underset{n times}{\underset{⏟}{(1) (1) (1) \dots (1)}}]^{\frac{3}{2}} = 1$

$\begin{aligned} {∴ \frac{d y}{d x} |}_{x = 0} & = \frac{3 (1)}{2} [\frac{1}{0 + 1} + \frac{2}{0 + 1} + \frac{3}{0 + 1} + \dots + \frac{n}{0 + 1}] \\ = \frac{3}{2} (1 + 2 + 3 + \dots + n) \\ = \frac{3}{2} \times \frac{n (n + 1)}{2} \\ = \frac{3 n (n + 1)}{4} \end{aligned}$

⇒Derivative