If $A equals open square brackets table row cell fraction numerator 1 over denominator 2 end fraction open parentheses e to the power of i x end exponent plus e to the power of negative i x end exponent close parentheses end cell cell fraction numerator 1 over denominator 2 end fraction open parentheses e to the power of i x end exponent minus e to the power of negative i x end exponent close parentheses end cell row cell fraction numerator 1 over denominator 2 end fraction open parentheses e to the power of i x end exponent minus e to the power of negative i x end exponent close parentheses end cell cell fraction numerator 1 over denominator 2 end fraction open parentheses e to the power of i x end exponent plus e to the power of negative i x end exponent close parentheses end cell end table close square brackets$ then $A to the power of negative 1 end exponent$ exists

If $A equals open square brackets table row alpha 0 row 1 1 end table close square brackets$ and $B equals open square brackets table row 1 0 row 3 1 end table close square brackets comma$ then value of $alpha$ for which $A to the power of 2 end exponent equals B$ is

For such questions, we should know how to multiply to matrices. When there is equal sign between two matrices, the elements of both the matrix should be equal.

If $A equals open square brackets table row alpha 0 row 1 1 end table close square brackets$ and $B equals open square brackets table row 1 0 row 3 1 end table close square brackets comma$ then value of $alpha$ for which $A to the power of 2 end exponent equals B$ is

For such questions, we should know how to multiply to matrices. When there is equal sign between two matrices, the elements of both the matrix should be equal.

For the primitive integral equation $y d x plus y to the power of 2 end exponent d y equals x d y comma x element of R comma y greater than 0 comma y equals y open parentheses x close parentheses comma y open parentheses 1 close parentheses equals 1 comma$ then $y open parentheses negative 3 close parentheses$ is

For such questions, we should know different method of differentiation and integration.

For the primitive integral equation $y d x plus y to the power of 2 end exponent d y equals x d y comma x element of R comma y greater than 0 comma y equals y open parentheses x close parentheses comma y open parentheses 1 close parentheses equals 1 comma$ then $y open parentheses negative 3 close parentheses$ is

For such questions, we should know different method of differentiation and integration.

The differential equation of all circles which pass through the origin and whose centre lies on y-axis is

For such questions, we should know the equation of cricle with its centre at a point other than origin.

The differential equation of all circles which pass through the origin and whose centre lies on y-axis is

For such questions, we should know the equation of cricle with its centre at a point other than origin.

The differential equation of all parabolas whose axis are parallel to y-axis is

For such questions, we should know how to write the equation of a parabola. The axis of parabola can be parallel to x or y axis. So, we have to write the equation accordingly.

The differential equation of all parabolas whose axis are parallel to y-axis is

For such questions, we should know how to write the equation of a parabola. The axis of parabola can be parallel to x or y axis. So, we have to write the equation accordingly.

Solution of the differential equation $open parentheses fraction numerator d y over denominator d x end fraction close parentheses to the power of 2 end exponent minus fraction numerator d y over denominator d x end fraction open parentheses e to the power of x end exponent plus e to the power of negative x end exponent close parentheses plus 1 equals 0$ is given by

When we get multipliction of two brackets equal to zero, any of the bracket can be zero or both of the brackets can be zero. It is decided based on the conditions in the equation.

Solution of the differential equation $open parentheses fraction numerator d y over denominator d x end fraction close parentheses to the power of 2 end exponent minus fraction numerator d y over denominator d x end fraction open parentheses e to the power of x end exponent plus e to the power of negative x end exponent close parentheses plus 1 equals 0$ is given by

When we get multipliction of two brackets equal to zero, any of the bracket can be zero or both of the brackets can be zero. It is decided based on the conditions in the equation.

The differential equation of all ellipse centred at the origin and major and minor axes along coordinate axes is

Area of the region bounded by $y equals tan invisible function application x comma$ tangent drawn to the curve at $x equals fraction numerator pi over denominator 4 end fraction$ and the $x minus$ axis is

Maximum value of $x left parenthesis 1 minus x right parenthesis to the power of 2 end exponent$ , when $0 less or equal than x less or equal than 2$ , is

If x = $c o s to the power of negative 1 end exponent$ t, y = $square root of 1 minus t to the power of 2 end exponent end root$ , then $y subscript 2 end subscript equals$

Two capacitors $C subscript 1 end subscript equals 2 mu F$ and $C subscript 2 end subscript equals 6 mu F$ in series, are connected in parallel to a third capacitor $C subscript 3 end subscript equals 4 mu F$ . This arrangement is then connected to a battery of e.m.f. = 2V, as shown in the figure. How much energy is lost by the battery in charging the capacitors

The combination of capacitors with $C subscript 1 end subscript equals 3 mu F comma C subscript 2 end subscript equals 4 mu F$ and $C subscript 3 end subscript equals 2 mu F$ is charged by connecting AB to a battery. Consider the following statements
I. Energy stored in $C subscript 1 end subscript$ = Energy stored in $C subscript 2 end subscript$ + Energy stored in $C subscript 3 end subscript$
II. Charge on C₁ = Charge on C₂ + Charge on C₃
III. Potential drop across C₁ = Potential drop across C₂ = Potential drop across C₃
Which of these is/are correct

Consider a parallel plate capacitor of $10 mu rightwards arrow over short leftwards arrow F$ (micro-farad) with air filled in the gap between the plates. Now one half of the space between the plates is filled with a dielectric of dielectric constant 4, as shown in the figure. The capacity of the capacitor changes to

In the figure a capacitor is filled with dielectrics. The resultant capacitance is

Equivalent capacitance between A and B is