QBManager

Differential Equation

87617 Taking two sub-intervals and using Simpson's $\frac{1}{3}$ rd rule, the value of $\int_{0}^{1} \frac{dx}{1 + x}$ will be

1

\frac{17}{24}

2

\frac{17}{36}

3

\frac{25}{36}

4

\frac{17}{25}

Explanation:

(C) : $\int_{0}^{1} \frac{dx}{1 + x}$
Taking two sub-interval or range, $[0, 1]$ of integral,
$h = \frac{1 - 0}{2} = \frac{1}{2}$
| $x$ | $y = \frac{1}{1 + x}$ |
| :---: | :---: |
| $x_{0} = 0$ | $y_{0} = 1$ |
| $x_{0} + h = \frac{1}{2}$ | $y_{1} = \frac{2}{3}$ |
| $x_{0} + 2 h = 1$ | $y_{2} = \frac{1}{2}$ |
Using Simpson's $\frac{1}{3}$ rd rule.
$\int_{0}^{1} ydx = \frac{h}{3} [y_{0} + 4 y_{1} + y_{2}]$
$\int_{0}^{1} \frac{dx}{1 + x} = \frac{1}{3.2} [1 + 4 (\frac{2}{3}) + \frac{1}{2}]$
$= \frac{1}{6} [1 + \frac{8}{3} + \frac{1}{2}] = \frac{1}{6} \cdot \frac{25}{6} = \frac{25}{36}$

CG PET- 2018

Differential Equation

87626 The differential equation of all parabolas whose axes are parallel to $Y$ -axis is

1

\frac{d^{3} y}{d x^{3}} = 0

2

\frac{d^{2} y}{d x^{2}} = 0

3

\frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} = 0

4

y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = 0

Explanation:

(A) : Equation of parabola is
$(x - h)^{2} = 4 a (y - k)$
Differentiating w.r.t. $x$
$2 (x - h) = 4 {ay}_{1}$
$x - h = 2 {ay}_{1}$
$1 = 2 {ay}_{2}$
$y_{2} = \frac{1}{2 a}$
Differentiating w.r.t.x,
$y_{3} = 0$
$\frac{d^{3} y}{d x^{3}} = 0$

AP EAMCET-21.04.2019

Differential Equation

87619 Let $y = y_{1} (x)$ and $y = y_{2} (x)$ b two distinct solutions of the differential equation $\frac{d y}{d x} = x + y$ , with $y_{1} (0)$ and $y_{2} (0) = 1$ respectively. Then the number of points of intersection of $y = y_{1} (x)$ and $y_{2} (x)$ is

1 0

2 1

3 2

4 3

Explanation:

(A) : Given, differential equation
$\frac{dy}{dx} = x + y$
$\begin{array}{ll} or \frac{d y}{d x} - y = x, I.F = e^{- x} \\ ∴ y \cdot e^{- x} = \int [x^{- x} \cdot e^{- x}] d x + c \\ y e^{- x} = - x^{- x} - e^{- x} + c \end{array}$
Therefore, $y = - x - 1 + c$ . $e^{x}$
$y_{1} (0) = 0$
$0 = - 1 + c$ or $c = + 1$
$∴ y_{1} (x) = - x + 1 + e^{x}$
$y_{2} (0) = 1, 1 = - 1 + c$
or $2 = c, c = 2$
$∴ y_{2} (x) = - x - 1 + 2 e^{x}$
For point of intersection,
$- x - 1 + e^{x} = - x - 1 + 2 e^{x}$
$0 = 2 e^{x} - e^{x} = e^{x}$
$e^{x} = 0$ , which is not possible.
$∴$ Number of point of intersection of $y_{1}$ and $y_{2}$ is 0 .

JEE Main-27.07.2022

Differential Equation

87620 If $\frac{d y}{d x} + 2 y \tan x = \sin x, 0 < x < \frac{π}{2}$ and $y (\frac{π}{3}) = 0$ , then the maximum value of $y (x)$ is :

1

\frac{1}{8}

2

\frac{3}{4}

3

\frac{1}{4}

4

\frac{3}{8}

Explanation:

(A) : Given,
Differential equation $\frac{d y}{d x} + 2 y \tan x = \sin x$ ,
The given differential equation is of the form
$\frac{d y}{d x} + p y = Q on comparing, we get$
$p = 2 y \tan x and Q = \sin x$
Now, We fill find the integrating factor-
$IF = e^{\int ddx} = e^{\int 2 \tan x d x}$
$= e^{2} \int \tan x d x = e^{[\log \sec x] = e^{\log (\sec^{2} x)}} = \sec^{2} x$
$∴$ Solution to the differential equation-
$y (IF) = \int (Q \times IF) dx + c$
$y \sec^{2} x = \int \sin x \sec^{2} x d x + c$
$y \sec^{2} x = \int \frac{\sin x}{\cos^{2} x} d x + c = \int \frac{\sin x}{\cos x} \times \frac{1}{\cos x} d x + c$
$y \sec^{2} x = \int \tan x \sec x d x + c$
$y \sec^{2} x = \sec x + c$
$y = \frac{1}{\sec x} + \frac{c}{\sec^{2} x} = \cos x + (\cos^{2} x) c$
When, $x = \frac{π}{3}, y = 0$
$y = \cos (\frac{π}{3}) + \cos^{2} (\frac{π}{3}) c$
$= \frac{1}{2} + \frac{1}{4} (c) = 0 = \frac{2 + c}{4} = 0 or 2 + c = 0$
$∴ c = - 2$
$\frac{y}{2} = - \cos^{2} x + \frac{1}{2} \cos x - 2 \cos^{2} x$
$\frac{y}{2} = - [(\cos^{2} x - \frac{1}{2} \cos x) = - [{(\cos x - \frac{1}{4})}^{2} + {(\frac{1}{4})}^{2}]$
${(\cos x - \frac{1}{4})}^{2} + {(\frac{1}{4})}^{2}] = - {(\cos x - \frac{1}{4})}^{2} + \frac{1}{16}$
${\cos x - \frac{1}{4})}^{2} will always be possible and therefore-$
${(\cos x - \frac{1}{4})}^{2} \leq 0$
$Hence, will be max. value of {(\frac{1}{4})}^{2} = \frac{1}{16}$
$So, The max. value of y = \frac{1}{16} \times 2 = \frac{1}{8}$

JEE Main-26.07.2022

Differential Equation

87617 Taking two sub-intervals and using Simpson's $\frac{1}{3}$ rd rule, the value of $\int_{0}^{1} \frac{dx}{1 + x}$ will be

1

\frac{17}{24}

2

\frac{17}{36}

3

\frac{25}{36}

4

\frac{17}{25}

Explanation:

(C) : $\int_{0}^{1} \frac{dx}{1 + x}$
Taking two sub-interval or range, $[0, 1]$ of integral,
$h = \frac{1 - 0}{2} = \frac{1}{2}$
| $x$ | $y = \frac{1}{1 + x}$ |
| :---: | :---: |
| $x_{0} = 0$ | $y_{0} = 1$ |
| $x_{0} + h = \frac{1}{2}$ | $y_{1} = \frac{2}{3}$ |
| $x_{0} + 2 h = 1$ | $y_{2} = \frac{1}{2}$ |
Using Simpson's $\frac{1}{3}$ rd rule.
$\int_{0}^{1} ydx = \frac{h}{3} [y_{0} + 4 y_{1} + y_{2}]$
$\int_{0}^{1} \frac{dx}{1 + x} = \frac{1}{3.2} [1 + 4 (\frac{2}{3}) + \frac{1}{2}]$
$= \frac{1}{6} [1 + \frac{8}{3} + \frac{1}{2}] = \frac{1}{6} \cdot \frac{25}{6} = \frac{25}{36}$

CG PET- 2018

Differential Equation

87626 The differential equation of all parabolas whose axes are parallel to $Y$ -axis is

1

\frac{d^{3} y}{d x^{3}} = 0

2

\frac{d^{2} y}{d x^{2}} = 0

3

\frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} = 0

4

y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = 0

Explanation:

(A) : Equation of parabola is
$(x - h)^{2} = 4 a (y - k)$
Differentiating w.r.t. $x$
$2 (x - h) = 4 {ay}_{1}$
$x - h = 2 {ay}_{1}$
$1 = 2 {ay}_{2}$
$y_{2} = \frac{1}{2 a}$
Differentiating w.r.t.x,
$y_{3} = 0$
$\frac{d^{3} y}{d x^{3}} = 0$

AP EAMCET-21.04.2019

Differential Equation

87619 Let $y = y_{1} (x)$ and $y = y_{2} (x)$ b two distinct solutions of the differential equation $\frac{d y}{d x} = x + y$ , with $y_{1} (0)$ and $y_{2} (0) = 1$ respectively. Then the number of points of intersection of $y = y_{1} (x)$ and $y_{2} (x)$ is

1 0

2 1

3 2

4 3

Explanation:

(A) : Given, differential equation
$\frac{dy}{dx} = x + y$
$\begin{array}{ll} or \frac{d y}{d x} - y = x, I.F = e^{- x} \\ ∴ y \cdot e^{- x} = \int [x^{- x} \cdot e^{- x}] d x + c \\ y e^{- x} = - x^{- x} - e^{- x} + c \end{array}$
Therefore, $y = - x - 1 + c$ . $e^{x}$
$y_{1} (0) = 0$
$0 = - 1 + c$ or $c = + 1$
$∴ y_{1} (x) = - x + 1 + e^{x}$
$y_{2} (0) = 1, 1 = - 1 + c$
or $2 = c, c = 2$
$∴ y_{2} (x) = - x - 1 + 2 e^{x}$
For point of intersection,
$- x - 1 + e^{x} = - x - 1 + 2 e^{x}$
$0 = 2 e^{x} - e^{x} = e^{x}$
$e^{x} = 0$ , which is not possible.
$∴$ Number of point of intersection of $y_{1}$ and $y_{2}$ is 0 .

JEE Main-27.07.2022

Differential Equation

87620 If $\frac{d y}{d x} + 2 y \tan x = \sin x, 0 < x < \frac{π}{2}$ and $y (\frac{π}{3}) = 0$ , then the maximum value of $y (x)$ is :

1

\frac{1}{8}

2

\frac{3}{4}

3

\frac{1}{4}

4

\frac{3}{8}

Explanation:

(A) : Given,
Differential equation $\frac{d y}{d x} + 2 y \tan x = \sin x$ ,
The given differential equation is of the form
$\frac{d y}{d x} + p y = Q on comparing, we get$
$p = 2 y \tan x and Q = \sin x$
Now, We fill find the integrating factor-
$IF = e^{\int ddx} = e^{\int 2 \tan x d x}$
$= e^{2} \int \tan x d x = e^{[\log \sec x] = e^{\log (\sec^{2} x)}} = \sec^{2} x$
$∴$ Solution to the differential equation-
$y (IF) = \int (Q \times IF) dx + c$
$y \sec^{2} x = \int \sin x \sec^{2} x d x + c$
$y \sec^{2} x = \int \frac{\sin x}{\cos^{2} x} d x + c = \int \frac{\sin x}{\cos x} \times \frac{1}{\cos x} d x + c$
$y \sec^{2} x = \int \tan x \sec x d x + c$
$y \sec^{2} x = \sec x + c$
$y = \frac{1}{\sec x} + \frac{c}{\sec^{2} x} = \cos x + (\cos^{2} x) c$
When, $x = \frac{π}{3}, y = 0$
$y = \cos (\frac{π}{3}) + \cos^{2} (\frac{π}{3}) c$
$= \frac{1}{2} + \frac{1}{4} (c) = 0 = \frac{2 + c}{4} = 0 or 2 + c = 0$
$∴ c = - 2$
$\frac{y}{2} = - \cos^{2} x + \frac{1}{2} \cos x - 2 \cos^{2} x$
$\frac{y}{2} = - [(\cos^{2} x - \frac{1}{2} \cos x) = - [{(\cos x - \frac{1}{4})}^{2} + {(\frac{1}{4})}^{2}]$
${(\cos x - \frac{1}{4})}^{2} + {(\frac{1}{4})}^{2}] = - {(\cos x - \frac{1}{4})}^{2} + \frac{1}{16}$
${\cos x - \frac{1}{4})}^{2} will always be possible and therefore-$
${(\cos x - \frac{1}{4})}^{2} \leq 0$
$Hence, will be max. value of {(\frac{1}{4})}^{2} = \frac{1}{16}$
$So, The max. value of y = \frac{1}{16} \times 2 = \frac{1}{8}$

JEE Main-26.07.2022

Differential Equation

87617 Taking two sub-intervals and using Simpson's $\frac{1}{3}$ rd rule, the value of $\int_{0}^{1} \frac{dx}{1 + x}$ will be

1

\frac{17}{24}

2

\frac{17}{36}

3

\frac{25}{36}

4

\frac{17}{25}

Explanation:

(C) : $\int_{0}^{1} \frac{dx}{1 + x}$
Taking two sub-interval or range, $[0, 1]$ of integral,
$h = \frac{1 - 0}{2} = \frac{1}{2}$
| $x$ | $y = \frac{1}{1 + x}$ |
| :---: | :---: |
| $x_{0} = 0$ | $y_{0} = 1$ |
| $x_{0} + h = \frac{1}{2}$ | $y_{1} = \frac{2}{3}$ |
| $x_{0} + 2 h = 1$ | $y_{2} = \frac{1}{2}$ |
Using Simpson's $\frac{1}{3}$ rd rule.
$\int_{0}^{1} ydx = \frac{h}{3} [y_{0} + 4 y_{1} + y_{2}]$
$\int_{0}^{1} \frac{dx}{1 + x} = \frac{1}{3.2} [1 + 4 (\frac{2}{3}) + \frac{1}{2}]$
$= \frac{1}{6} [1 + \frac{8}{3} + \frac{1}{2}] = \frac{1}{6} \cdot \frac{25}{6} = \frac{25}{36}$

CG PET- 2018

Differential Equation

87626 The differential equation of all parabolas whose axes are parallel to $Y$ -axis is

1

\frac{d^{3} y}{d x^{3}} = 0

2

\frac{d^{2} y}{d x^{2}} = 0

3

\frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} = 0

4

y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = 0

Explanation:

(A) : Equation of parabola is
$(x - h)^{2} = 4 a (y - k)$
Differentiating w.r.t. $x$
$2 (x - h) = 4 {ay}_{1}$
$x - h = 2 {ay}_{1}$
$1 = 2 {ay}_{2}$
$y_{2} = \frac{1}{2 a}$
Differentiating w.r.t.x,
$y_{3} = 0$
$\frac{d^{3} y}{d x^{3}} = 0$

AP EAMCET-21.04.2019

Differential Equation

87619 Let $y = y_{1} (x)$ and $y = y_{2} (x)$ b two distinct solutions of the differential equation $\frac{d y}{d x} = x + y$ , with $y_{1} (0)$ and $y_{2} (0) = 1$ respectively. Then the number of points of intersection of $y = y_{1} (x)$ and $y_{2} (x)$ is

1 0

2 1

3 2

4 3

Explanation:

(A) : Given, differential equation
$\frac{dy}{dx} = x + y$
$\begin{array}{ll} or \frac{d y}{d x} - y = x, I.F = e^{- x} \\ ∴ y \cdot e^{- x} = \int [x^{- x} \cdot e^{- x}] d x + c \\ y e^{- x} = - x^{- x} - e^{- x} + c \end{array}$
Therefore, $y = - x - 1 + c$ . $e^{x}$
$y_{1} (0) = 0$
$0 = - 1 + c$ or $c = + 1$
$∴ y_{1} (x) = - x + 1 + e^{x}$
$y_{2} (0) = 1, 1 = - 1 + c$
or $2 = c, c = 2$
$∴ y_{2} (x) = - x - 1 + 2 e^{x}$
For point of intersection,
$- x - 1 + e^{x} = - x - 1 + 2 e^{x}$
$0 = 2 e^{x} - e^{x} = e^{x}$
$e^{x} = 0$ , which is not possible.
$∴$ Number of point of intersection of $y_{1}$ and $y_{2}$ is 0 .

JEE Main-27.07.2022

Differential Equation

87620 If $\frac{d y}{d x} + 2 y \tan x = \sin x, 0 < x < \frac{π}{2}$ and $y (\frac{π}{3}) = 0$ , then the maximum value of $y (x)$ is :

1

\frac{1}{8}

2

\frac{3}{4}

3

\frac{1}{4}

4

\frac{3}{8}

Explanation:

(A) : Given,
Differential equation $\frac{d y}{d x} + 2 y \tan x = \sin x$ ,
The given differential equation is of the form
$\frac{d y}{d x} + p y = Q on comparing, we get$
$p = 2 y \tan x and Q = \sin x$
Now, We fill find the integrating factor-
$IF = e^{\int ddx} = e^{\int 2 \tan x d x}$
$= e^{2} \int \tan x d x = e^{[\log \sec x] = e^{\log (\sec^{2} x)}} = \sec^{2} x$
$∴$ Solution to the differential equation-
$y (IF) = \int (Q \times IF) dx + c$
$y \sec^{2} x = \int \sin x \sec^{2} x d x + c$
$y \sec^{2} x = \int \frac{\sin x}{\cos^{2} x} d x + c = \int \frac{\sin x}{\cos x} \times \frac{1}{\cos x} d x + c$
$y \sec^{2} x = \int \tan x \sec x d x + c$
$y \sec^{2} x = \sec x + c$
$y = \frac{1}{\sec x} + \frac{c}{\sec^{2} x} = \cos x + (\cos^{2} x) c$
When, $x = \frac{π}{3}, y = 0$
$y = \cos (\frac{π}{3}) + \cos^{2} (\frac{π}{3}) c$
$= \frac{1}{2} + \frac{1}{4} (c) = 0 = \frac{2 + c}{4} = 0 or 2 + c = 0$
$∴ c = - 2$
$\frac{y}{2} = - \cos^{2} x + \frac{1}{2} \cos x - 2 \cos^{2} x$
$\frac{y}{2} = - [(\cos^{2} x - \frac{1}{2} \cos x) = - [{(\cos x - \frac{1}{4})}^{2} + {(\frac{1}{4})}^{2}]$
${(\cos x - \frac{1}{4})}^{2} + {(\frac{1}{4})}^{2}] = - {(\cos x - \frac{1}{4})}^{2} + \frac{1}{16}$
${\cos x - \frac{1}{4})}^{2} will always be possible and therefore-$
${(\cos x - \frac{1}{4})}^{2} \leq 0$
$Hence, will be max. value of {(\frac{1}{4})}^{2} = \frac{1}{16}$
$So, The max. value of y = \frac{1}{16} \times 2 = \frac{1}{8}$

JEE Main-26.07.2022

Differential Equation

87617 Taking two sub-intervals and using Simpson's $\frac{1}{3}$ rd rule, the value of $\int_{0}^{1} \frac{dx}{1 + x}$ will be

1

\frac{17}{24}

2

\frac{17}{36}

3

\frac{25}{36}

4

\frac{17}{25}

Explanation:

(C) : $\int_{0}^{1} \frac{dx}{1 + x}$
Taking two sub-interval or range, $[0, 1]$ of integral,
$h = \frac{1 - 0}{2} = \frac{1}{2}$
| $x$ | $y = \frac{1}{1 + x}$ |
| :---: | :---: |
| $x_{0} = 0$ | $y_{0} = 1$ |
| $x_{0} + h = \frac{1}{2}$ | $y_{1} = \frac{2}{3}$ |
| $x_{0} + 2 h = 1$ | $y_{2} = \frac{1}{2}$ |
Using Simpson's $\frac{1}{3}$ rd rule.
$\int_{0}^{1} ydx = \frac{h}{3} [y_{0} + 4 y_{1} + y_{2}]$
$\int_{0}^{1} \frac{dx}{1 + x} = \frac{1}{3.2} [1 + 4 (\frac{2}{3}) + \frac{1}{2}]$
$= \frac{1}{6} [1 + \frac{8}{3} + \frac{1}{2}] = \frac{1}{6} \cdot \frac{25}{6} = \frac{25}{36}$

CG PET- 2018

Differential Equation

87626 The differential equation of all parabolas whose axes are parallel to $Y$ -axis is

1

\frac{d^{3} y}{d x^{3}} = 0

2

\frac{d^{2} y}{d x^{2}} = 0

3

\frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} = 0

4

y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = 0

Explanation:

(A) : Equation of parabola is
$(x - h)^{2} = 4 a (y - k)$
Differentiating w.r.t. $x$
$2 (x - h) = 4 {ay}_{1}$
$x - h = 2 {ay}_{1}$
$1 = 2 {ay}_{2}$
$y_{2} = \frac{1}{2 a}$
Differentiating w.r.t.x,
$y_{3} = 0$
$\frac{d^{3} y}{d x^{3}} = 0$

AP EAMCET-21.04.2019

Differential Equation

87619 Let $y = y_{1} (x)$ and $y = y_{2} (x)$ b two distinct solutions of the differential equation $\frac{d y}{d x} = x + y$ , with $y_{1} (0)$ and $y_{2} (0) = 1$ respectively. Then the number of points of intersection of $y = y_{1} (x)$ and $y_{2} (x)$ is

1 0

2 1

3 2

4 3

Explanation:

(A) : Given, differential equation
$\frac{dy}{dx} = x + y$
$\begin{array}{ll} or \frac{d y}{d x} - y = x, I.F = e^{- x} \\ ∴ y \cdot e^{- x} = \int [x^{- x} \cdot e^{- x}] d x + c \\ y e^{- x} = - x^{- x} - e^{- x} + c \end{array}$
Therefore, $y = - x - 1 + c$ . $e^{x}$
$y_{1} (0) = 0$
$0 = - 1 + c$ or $c = + 1$
$∴ y_{1} (x) = - x + 1 + e^{x}$
$y_{2} (0) = 1, 1 = - 1 + c$
or $2 = c, c = 2$
$∴ y_{2} (x) = - x - 1 + 2 e^{x}$
For point of intersection,
$- x - 1 + e^{x} = - x - 1 + 2 e^{x}$
$0 = 2 e^{x} - e^{x} = e^{x}$
$e^{x} = 0$ , which is not possible.
$∴$ Number of point of intersection of $y_{1}$ and $y_{2}$ is 0 .

JEE Main-27.07.2022

Differential Equation

87620 If $\frac{d y}{d x} + 2 y \tan x = \sin x, 0 < x < \frac{π}{2}$ and $y (\frac{π}{3}) = 0$ , then the maximum value of $y (x)$ is :

1

\frac{1}{8}

2

\frac{3}{4}

3

\frac{1}{4}

4

\frac{3}{8}

Explanation:

(A) : Given,
Differential equation $\frac{d y}{d x} + 2 y \tan x = \sin x$ ,
The given differential equation is of the form
$\frac{d y}{d x} + p y = Q on comparing, we get$
$p = 2 y \tan x and Q = \sin x$
Now, We fill find the integrating factor-
$IF = e^{\int ddx} = e^{\int 2 \tan x d x}$
$= e^{2} \int \tan x d x = e^{[\log \sec x] = e^{\log (\sec^{2} x)}} = \sec^{2} x$
$∴$ Solution to the differential equation-
$y (IF) = \int (Q \times IF) dx + c$
$y \sec^{2} x = \int \sin x \sec^{2} x d x + c$
$y \sec^{2} x = \int \frac{\sin x}{\cos^{2} x} d x + c = \int \frac{\sin x}{\cos x} \times \frac{1}{\cos x} d x + c$
$y \sec^{2} x = \int \tan x \sec x d x + c$
$y \sec^{2} x = \sec x + c$
$y = \frac{1}{\sec x} + \frac{c}{\sec^{2} x} = \cos x + (\cos^{2} x) c$
When, $x = \frac{π}{3}, y = 0$
$y = \cos (\frac{π}{3}) + \cos^{2} (\frac{π}{3}) c$
$= \frac{1}{2} + \frac{1}{4} (c) = 0 = \frac{2 + c}{4} = 0 or 2 + c = 0$
$∴ c = - 2$
$\frac{y}{2} = - \cos^{2} x + \frac{1}{2} \cos x - 2 \cos^{2} x$
$\frac{y}{2} = - [(\cos^{2} x - \frac{1}{2} \cos x) = - [{(\cos x - \frac{1}{4})}^{2} + {(\frac{1}{4})}^{2}]$
${(\cos x - \frac{1}{4})}^{2} + {(\frac{1}{4})}^{2}] = - {(\cos x - \frac{1}{4})}^{2} + \frac{1}{16}$
${\cos x - \frac{1}{4})}^{2} will always be possible and therefore-$
${(\cos x - \frac{1}{4})}^{2} \leq 0$
$Hence, will be max. value of {(\frac{1}{4})}^{2} = \frac{1}{16}$
$So, The max. value of y = \frac{1}{16} \times 2 = \frac{1}{8}$

JEE Main-26.07.2022