Definite IntegrationhardFree

d³y/dx³ at x=e for y=x²∫√(ln t)dt | JEE

JEE Maths question with a full step-by-step solution.

Question

y=x^{2}\displaystyle\int_{1}^{x}\sqrt{\ln t}\,dt

, then the value of

\dfrac{d^{3}y}{dx^{3}}

x=e

33

11

\dfrac{33}{4}

correct

DNone of these

Solution

Step 1: Write

G(x)=\displaystyle\int_{1}^{x}\sqrt{\ln t}\,dt

, so

y=x^{2}G

and, by the Fundamental Theorem of Calculus,

G'(x)=\sqrt{\ln x}

. Differentiate

y

once:

y'=2xG+x^{2}G'.

Step 2: Differentiate again:

y''=2G+2xG'+2xG'+x^{2}G''=2G+4xG'+x^{2}G''.

Differentiate a third time:

y'''=2G'+4G'+4xG''+2xG''+x^{2}G'''=6G'+6xG''+x^{2}G'''.

(The

G

term has disappeared, so

G(e)

is never needed.) Step 3: Compute the derivatives of

G

. With

G'(x)=(\ln x)^{1/2}

G''(x)=\dfrac{1}{2x\sqrt{\ln x}},\qquad G'''(x)=-\dfrac{1}{2x^{2}\sqrt{\ln x}}-\dfrac{1}{4x^{2}(\ln x)^{3/2}}.

Step 4: Evaluate at

x=e

, where

\ln x=1

G'(e)=1,\qquad G''(e)=\dfrac{1}{2e},\qquad G'''(e)=-\dfrac{1}{2e^{2}}-\dfrac{1}{4e^{2}}=-\dfrac{3}{4e^{2}}.

Hence

y'''(e)=6(1)+6e\cdot\dfrac{1}{2e}+e^{2}\!\left(-\dfrac{3}{4e^{2}}\right)=6+3-\dfrac34=\dfrac{33}{4}.

Correct answer: (3)

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