If the area of the bounded region R = \ (x,y):\max \ 0,\log _ex\ \le y \le 2^x,1 \over 2 \le x \le 2 \ is , \alpha (\log _e2)^ - 1 + \beta (\log _e2) + \gamma , then the value of (\alpha + \beta - 2\lambda )^2 is equal to :

Area Under the Curves PYQ — JEE Main 2021 · 27 Jul · Shift 1 · Q27 with Solution

View written solutionFree

Correct answer: B

Interpret the region

The region is

R=\{(x,y): \max\{0,\ln x\}\le y\le 2^x,\ \tfrac12\le x\le 2\}.

So the area is

A=\int_{1/2}^{2}\left(2^x-\max\{0,\ln x\}\right)dx.

Since

for $\tfrac12\le x\le 1$ , we have $\ln x\le 0$ , so $\max\{0,\ln x\}=0$ ,
for $1\le x\le 2$ , we have $\ln x\ge 0$ , so $\max\{0,\ln x\}=\ln x$ ,

we split the integral:

A=\int_{1/2}^{1}2^x\,dx+\int_{1}^{2}(2^x-\ln x)\,dx.

Thus,

A=\int_{1/2}^{2}2^x\,dx-\int_{1}^{2}\ln x\,dx.

Evaluate $\int_{1/2}^{2}2^x\,dx$

We use

\int 2^x\,dx=\frac{2^x}{\ln 2}.

Hence,

\int_{1/2}^{2}2^x\,dx=\left[\frac{2^x}{\ln 2}\right]_{1/2}^{2} =\frac{2^2-2^{1/2}}{\ln 2} =\frac{4-\sqrt2}{\ln 2}.

Evaluate $\int_{1}^{2}\ln x\,dx$

We use

\int \ln x\,dx=x\ln x-x.

So,

\int_{1}^{2}\ln x\,dx= \left[x\ln x-x\right]_{1}^{2} =(2\ln 2-2)-(0-1)=2\ln 2-1.

Compute the total area

Therefore,

A=\frac{4-\sqrt2}{\ln 2}-(2\ln 2-1) =\frac{4-\sqrt2}{\ln 2}-2\ln 2+1.

Compare this with the form

\alpha(\ln 2)^{-1}+\beta(\ln 2)+\gamma.

So we identify

\alpha=4-\sqrt2,\qquad \beta=-2,\qquad \gamma=1.

The expression asked is evidently intended as

(\alpha+\beta-2\gamma)^2

(the symbol in the question appears as $\lambda$ , but only $\gamma$ is defined in the area expression).

Thus,

\alpha+\beta-2\gamma=(4-\sqrt2)+(-2)-2(1)=-\sqrt2.

Hence,

(\alpha+\beta-2\gamma)^2=(-\sqrt2)^2=2.

Check with options

The correct option is

\boxed{2}

which is Option B.

Comparison with stored answer

Stored correct answer: B.

My derived answer is also B.

More from Area Under the Curves

View written solutionFree

Correct answer: B

Interpret the region

The region is

R=\{(x,y): \max\{0,\ln x\}\le y\le 2^x,\ \tfrac12\le x\le 2\}.

So the area is

A=\int_{1/2}^{2}\left(2^x-\max\{0,\ln x\}\right)dx.

Since

for $\tfrac12\le x\le 1$ , we have $\ln x\le 0$ , so $\max\{0,\ln x\}=0$ ,
for $1\le x\le 2$ , we have $\ln x\ge 0$ , so $\max\{0,\ln x\}=\ln x$ ,

we split the integral:

A=\int_{1/2}^{1}2^x\,dx+\int_{1}^{2}(2^x-\ln x)\,dx.

Thus,

A=\int_{1/2}^{2}2^x\,dx-\int_{1}^{2}\ln x\,dx.

Evaluate $\int_{1/2}^{2}2^x\,dx$

We use

\int 2^x\,dx=\frac{2^x}{\ln 2}.

Hence,

\int_{1/2}^{2}2^x\,dx=\left[\frac{2^x}{\ln 2}\right]_{1/2}^{2} =\frac{2^2-2^{1/2}}{\ln 2} =\frac{4-\sqrt2}{\ln 2}.

Evaluate $\int_{1}^{2}\ln x\,dx$

We use

\int \ln x\,dx=x\ln x-x.

So,

\int_{1}^{2}\ln x\,dx= \left[x\ln x-x\right]_{1}^{2} =(2\ln 2-2)-(0-1)=2\ln 2-1.

Compute the total area

Therefore,

A=\frac{4-\sqrt2}{\ln 2}-(2\ln 2-1) =\frac{4-\sqrt2}{\ln 2}-2\ln 2+1.

Compare this with the form

\alpha(\ln 2)^{-1}+\beta(\ln 2)+\gamma.

So we identify

\alpha=4-\sqrt2,\qquad \beta=-2,\qquad \gamma=1.

The expression asked is evidently intended as

(\alpha+\beta-2\gamma)^2

(the symbol in the question appears as $\lambda$ , but only $\gamma$ is defined in the area expression).

Thus,

\alpha+\beta-2\gamma=(4-\sqrt2)+(-2)-2(1)=-\sqrt2.

Hence,

(\alpha+\beta-2\gamma)^2=(-\sqrt2)^2=2.

Check with options

The correct option is

\boxed{2}

which is Option B.

Comparison with stored answer

Stored correct answer: B.

My derived answer is also B.

Area Under the Curves question

Area Under the Curves question

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