If \(n-1C_r= (k^2-8)^nC_r+1\) Find \(k\).
- \(k\in[-3,-2\sqrt2) \cup(2\sqrt2,3)\)
- \(k\in [-4,-2\sqrt3\cup(2\sqrt3,4]\)
- \(k\in [2\sqrt3,4]\)
- \(k\in[3,2\sqrt3]\)
The Correct Option is A
Solution and Explanation
The correct option is (A): \(k\in[-3,-2\sqrt2) \cup(2\sqrt2,3)\)
Top Questions on Derivatives
- Let \(( f(x) =\) \(\int_0^{x^2 \frac{t^2 - 8t + 15}{e^t} dt}\), \(x \in \mathbb{R}\). Then the numbers of local maximum and local minimum points of \( f \), respectively, are:
- JEE Main - 2025
- Mathematics
- Derivatives
- If $y = 5 \cos x - 3 \sin x$, prove that $\frac{d^2y}{dx^2} + y = 0$.
- Differentiate $2\cos^2 x$ w.r.t. $\cos^2 x$.
Let the length of the side perpendicular to the partition be \( x \) metres and the side parallel to the partition be \( y \) metres.
Based on this information, answer the following questions:
(i) Write the equation for the total boundary material used in the boundary and parallel to the partition in terms of \( x \) and \( y \).
(ii) Write the area of the solar panel as a function of \( x \).
(iii) Find the critical points of the area function. Use the second derivative test to determine critical points at the maximum area. Also, find the maximum area.
OR
(iii) Using the first derivative test, calculate the maximum area the company can enclose with the 300 metres of boundary material, considering the parallel partition.- If \( x = a \left( \cos \theta + \log \tan \frac{\theta}{2} \right) \) and \( y = \sin \theta \), then find \( \frac{d^2y}{dx^2} \) at \( \theta = \frac{\pi}{4} \).
Questions Asked in JEE Main exam
A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is _____.
- JEE Main - 2025
- Magnetic Field
- Concentrated nitric acid is labelled as 75%by mass. The volume in mL of the solution which contains 30 g of nitric acid is: Given: Density of nitric acid solution is 1.25 g/mL.
- JEE Main - 2025
- Solutions
- A physical quantity \( Q \) is related to four observables \( a \), \( b \), \( c \), and \( d \) as follows: \[ Q = \frac{a b^4}{c d^2} \] Where:
- \( a = (60 \pm 3) \, \text{Pa} \),
- \( b = (20 \pm 0.1) \, \text{m} \),
- \( c = (40 \pm 0.2) \, \text{N·s/m}^2 \),
- \( d = (50 \pm 0.1) \, \text{m} \).
Then the percentage error in \( Q \) is:- JEE Main - 2025
- Error analysis
- Let circle $C$ be the image of
$$ x^2 + y^2 - 2x + 4y - 4 = 0 $$
in the line
$$ 2x - 3y + 5 = 0 $$
and $A$ be the point on $C$ such that $OA$ is parallel to the x-axis and $A$ lies on the right-hand side of the centre $O$ of $C$.
If $B(\alpha, \beta)$, with $\beta < 4$, lies on $C$ such that the length of the arc $AB$ is $\frac{1}{6}$ of the perimeter of $C$, then $\beta - \sqrt{3}\alpha$ is equal to: - Assume a living cell with 0.9%(\(w/w\)) of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data up to first decimal place only) The cell will:
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- osmosis
Concepts Used:
Derivatives
Derivatives are defined as a function's changing rate of change with relation to an independent variable. When there is a changing quantity and the rate of change is not constant, the derivative is utilised. The derivative is used to calculate the sensitivity of one variable (the dependent variable) to another one (independent variable). Derivatives relate to the instant rate of change of one quantity with relation to another. It is beneficial to explore the nature of a quantity on a moment-to-moment basis.
Few formulae for calculating derivatives of some basic functions are as follows:
