If x=4 t^3+3, y=4+3 t^4 and ((d^2 x)/(d y^2)) ((d x)/(d y))^n is a constant, then find the value of (4)/(5)+(4)/(5 n)+(4)/(5 n^2)+ . . upto infinity.

Correct answer: 1 aligned (d x)/(d y) & =((d x)/(d t))/((d y)/(d t))=(12 t^2)/(12 t^3)=(1)/(t) (d^2 x)/(d y^2) & =(d)/(d t)((1)/(t)) ×((d t)/(d y))=(-1)/(t^2) × (1)/(12 t^3)=(-1)/(12 t^5) (((d^2 x)/(d y^2)))/(((d x)/(d y))^n) & = constant & n=5 Sum & =((4)/(5))/(1-(1)/(n))=((4)/(5))/(1-(1)/(5))=1 aligned

JEE Advanced

Exam: JEE Advanced
Subject: Mathematics
Chapter: Differentiation

JEE Advanced · Mathematics · Differentiation

If x=4 t^3+3, y=4+3 t^4 and ((d^2 x)/(d y^2)) ((d x)/(d y))^n is a...

x=4 t^3+3, \quad y=4+3 t^4

and

\frac{\left(\frac{d^2 x}{d y^2}\right)}

{

\left(\frac{d x}{d y}\right)^n}

is a constant, then find the value of

\frac{4}{5}+\frac{4}{5 n}+\frac{4}{5 n^2}+\ldots .

. upto infinity.

Correct answer

Step-by-step solution

\begin{aligned} \frac{d x}{d y} & =\frac{\frac{d x}{d t}}{\frac{d y}{d t}}=\frac{12 t^2}{12 t^3}=\frac{1}{t} \\ \frac{d^2 x}{d y^2} & =\frac{d}{d t}\left(\frac{1}{t}\right) \times\left(\frac{d t}{d y}\right)=\frac{-1}{t^2} \times \frac{1}{12 t^3}=\frac{-1}{12 t^5} \\ \frac{\left(\frac{d^2 x}{d y^2}\right)}{\left(\frac{d x}{d y}\right)^n} & =\text { constant } \\ \therefore \quad & n=5 \\ \therefore \quad \text { Sum } & =\frac{\frac{4}{5}}{1-\frac{1}{n}}=\frac{\frac{4}{5}}{1-\frac{1}{5}}=1\end{aligned}

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