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1

### JEE Main 2021 (Online) 27th August Morning Shift

Numerical
If $${y^{1/4}} + {y^{ - 1/4}} = 2x$$, and

$$({x^2} - 1){{{d^2}y} \over {d{x^2}}} + \alpha x{{dy} \over {dx}} + \beta y = 0$$, then | $$\alpha$$ $$-$$ $$\beta$$ | is equal to __________.

## Explanation

$${y^{{1 \over 4}}} + {1 \over {{y^{{1 \over 4}}}}} = 2x$$

$$\Rightarrow {\left( {{y^{{1 \over 4}}}} \right)^2} - 2x{y^{\left( {{1 \over 4}} \right)}} + 1 = 0$$

$$\Rightarrow {y^{{1 \over 4}}} = x + \sqrt {{x^2} - 1}$$ or $$x - \sqrt {{x^2} - 1}$$

So, $${1 \over 4}{1 \over {{y^{{3 \over 4}}}}}{{dy} \over {dx}} = 1 + {x \over {\sqrt {{x^2} - 1} }}$$

$$\Rightarrow {1 \over 4}{1 \over {{y^{{3 \over 4}}}}}{{dy} \over {dx}} = {{{y^{{1 \over 4}}}} \over {\sqrt {{x^2} - 1} }}$$

$$\Rightarrow {{dy} \over {dx}} = {{4y} \over {\sqrt {{x^2} - 1} }}$$ .... (1)

Hence, $${{{d^2}y} \over {d{x^2}}} = 4{{\left( {\sqrt {{x^2} - 1} } \right)y' - {{yx} \over {\sqrt {{x^2} - 1} }}} \over {{x^2} - 1}}$$

$$\Rightarrow ({x^2} - 1)y'' = 4{{({x^2} - 1)y' - xy} \over {\sqrt {{x^2} - 1} }}$$

$$\Rightarrow ({x^2} - 1)y'' = 4\left( {\sqrt {{x^2} - 1} y' - {{xy} \over {\sqrt {{x^2} - 1} }}} \right)$$

$$\Rightarrow ({x^2} - 1)y'' = 4\left( {4y - {{xy'} \over 4}} \right)$$ (from I)

$$\Rightarrow ({x^2} - 1)y'' + xy' - 16y = 0$$

So, | $$\alpha$$ $$-$$ $$\beta$$ | = 17
2

### JEE Main 2021 (Online) 27th July Morning Shift

Numerical
If $$y = y(x),y \in \left[ {0,{\pi \over 2}} \right)$$ is the solution of the differential equation $$\sec y{{dy} \over {dx}} - \sin (x + y) - \sin (x - y) = 0$$, with y(0) = 0, then $$5y'\left( {{\pi \over 2}} \right)$$ is equal to ______________.

## Explanation

$$\sec y{{dy} \over {dx}} = 2\sin x\cos y$$

$${\sec ^2}ydy = 2\sin xdx$$

$$\tan y = - 2\cos x + c$$

$$c = 2$$

$$\tan y = - 2\cos x + 2 \Rightarrow$$ at $$x = {\pi \over 2}$$

$$\tan y = 2$$

$${\sec ^2}y{{dy} \over {dx}} = 2\sin x$$

$$\therefore$$ $$5{{dy} \over {dx}} = 2$$
3

### JEE Main 2021 (Online) 27th July Morning Shift

Numerical
Let $$F:[3,5] \to R$$ be a twice differentiable function on (3, 5) such that $$F(x) = {e^{ - x}}\int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt}$$. If $$F'(4) = {{\alpha {e^\beta } - 224} \over {{{({e^\beta } - 4)}^2}}}$$, then $$\alpha$$ + $$\beta$$ is equal to _______________.

## Explanation

$$F(3) = 0$$

$${e^x}F(x) = \int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt}$$

$${e^x}F(x) + {e^x}F'(x) = 3{x^2} + 2x + 4F'(x)$$

$$({e^x} - 4){{dy} \over {dx}} + {e^x}y = (3{x^2} + 2x)$$

$${{dy} \over {dx}} + {{{e^x}} \over {({e^x} - 4)}}y = {{(3{x^2} + 2x)} \over {({e^x} - 4)}}$$

$$y{e^{\int {{{{e^x}} \over {({e^x} - 4)}}dx} }} = \int {{{(3{x^2} + 2x)} \over {({e^x} - 4)}}{e^{\int {{{{e^x}} \over {{e^x} - 4}}dx} }}dx}$$

$$y.({e^x} - 4) = \int {(3{x^2} + 2x)dx + c}$$

$$y({e^x} - 4) = {x^3} + {x^2} + c$$

Put x = 3 $$\Rightarrow$$ c = $$-$$36

$$F(x) = {{({x^3} + {x^2} - 36)} \over {({e^x} - 4)}}$$

$$F'(x) = {{(3{x^2} + 2x)({e^x} - 4) - ({x^3} + {x^2} - 36){e^x}} \over {{{({e^x} - 4)}^2}}}$$

Now, put value of x = 4 we will get $$\alpha$$ = 12 & $$\beta$$ = 4
4

### JEE Main 2021 (Online) 27th July Evening Shift

Numerical
Let y = y(x) be the solution of the differential equation dy = e$$\alpha$$x + y dx; $$\alpha$$ $$\in$$ N. If y(loge2) = loge2 and y(0) = loge$$\left( {{1 \over 2}} \right)$$, then the value of $$\alpha$$ is equal to _____________.

## Explanation

$$\int {{e^{ - y}}} dy = \int {{e^{\alpha x}}} dx$$

$$\Rightarrow {e^{ - y}} = {{{e^{\alpha x}}} \over \alpha } + c$$ ..... (i)

Put (x, y) = (ln2, ln2)

$${{ - 1} \over 2} = {{{2^\alpha }} \over \alpha } + C$$ ..... (ii)

Put (x, y) $$\equiv$$ (0, $$-$$ln2) in (i)

$$- 2 = {1 \over \alpha } + C$$ ..... (iii)

(ii) $$-$$ (iii)

$${{{2^\alpha } - 1} \over \alpha } = {3 \over 2}$$

$$\Rightarrow$$ $$\alpha$$ = 2 (as $$\alpha$$ $$\in$$ N)

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