If \(x=\frac{a}{1+t}\) and \(y=\frac{a}{(1+t)^2}\) where \(a>0\), then \(\frac{d^2 y}{d x^2}\) at \(t=1\) is:

An objective function \(z=a x+b y\) is maximum at points \((15,15)\) and \((0,20)\). If \(a, b \geq 0\) and \(a b=27\), then the maximum value of the objective function is

For the function \(f(x)=e^x+e^{-x}\)
(A) \(f^{\prime}(x)=e^x-e^{-x}\)
(B) The critical point is \(x=0\)
(C) The minimum value is 1
(D) \(x=0\) is the point of local minimum.
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Match List-l with List-ll
Let \(A\) be any invertible square matrix. Then

List-l	List-ll
(A) \(A-A^T\)	(I) \(|A| A^{-1}\)
(B) \(A A^T\)	(II) Skew-symmetric
(C) \(\operatorname{det}\left(A^{-1}\right)\)	(III) Symmetric
(D) \(\operatorname{adj} A\)	(IV) \([\operatorname{det}(A)]^{-1}\)

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Let \(\Delta = \begin{vmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{vmatrix}\), then \(\Delta\) lies in the interval

The area of the region bounded by the curves \(y=x^2+2, y=x, x=0\) and \(x=2\) is

A line through the points \(A(3,4,1)\) and \(B(5,1,6)\) is drawn.
The coordinates of the points where the line through the points \(A \backslash \& B\) crosses the \(X Y\) plane is:

The feasible region of the constraints \(x \geq 0, x+y \leq 1\) and \(x-y \leq 1\) is situated in:

Read the Passage carefully and answer the Questions
Kinetic studies not only help us to determine the rate of a chemical reaction but also describe the conditions by which the reaction rates can be altered. The factors such as concentration, temperature, pressure and catalyst affect the rate of a reaction.
The speed of a reaction or the rate of a reaction can be defined as the change in concentration of a reactant or product in unit time.
Rate \(=k[A]^{\text {a }}[B]^{\text {y }}\)
Here \(k\) is the rate constant
\(x+y\) gives the overall order of a reaction, whereas \(x\) and \(y\) repreeent the order with respect to the reactants \(A\) and \(B\), respectively.
Table: Initial rate of formation of NO 2

S No.	Initial [NO]/mol L-1	Initial [O2]/mol L-1	Initial rate of formation of NO2/mol L-1s-1
1.	0.30	0.3	0.096
2.	0.60	0.3	0.384
3.	0.30	0.6	0.192
4.	0.60	0.6	0.768

Match List-I with List-II

List-I (Differential equation)	List-II (Order and Degree)
(A) \(\frac{d^3 y}{d x^3}+y^2+e^{d y / d x}=0\)	(I) order \(=3\), degree \(=1\)
(B) \(\left(\frac{d^2 y}{d x^2}\right)^3+\left(\frac{d y}{d x}\right)^2+\frac{d y}{d x}+1=0\)	(II) order \(=3\), degree not defined
(C) \(2 x^2 \frac{d^2 y}{d x^2}-3\left(\frac{d y}{d x}\right)^2+y=0\)	(III) order \(=2\), degree \(=3\)
(D) \(\frac{d^3 y}{d x^3}+2\left(\frac{d y}{d x}\right)^2+\frac{d y}{d x}=0\)	(IV) order \(=2\), degree \(=1\)

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The second trophic level in an aquatic ecosystem is:

Read the passage carefully and answer the question
The speed at which a chemical reaction takes place is called the rate of reaction. The rate of reaction depends on various factors like concentration of the reactants, temperature, etc. The relation between the rate of reaction and the concentration of reacting species is represented by the equation r = k[A]^x[B]^y, where x and y are the order of the reaction with respect to the reactants A and B, respectively. The overall order of the reaction is x + y. The rate of reaction can also be increased by the use of a catalyst which provides an alternate pathway of lower activation energy. It increases the rate of forward and backward reaction to an equal extent. It does not alter the Gibbs energy of the reaction.
Calculate the order of a reaction whose Rate = k[A]^1/2 [B]^3/2.

Diodes are most commonly used for rectification of alternating voltages. This is because, in diode
(A) current flows in one direction only
(B) forward bias resistance is low as compared to reverse bias resistance
(C) reverse current is independent of the voltage applied
(D) reverse current is linearly dependent on the voltage applied
Choose the correct answer from the options given below: