Consider a polynomial y=P(x) of the least degree passing through A(-1,1) and whose graph has two

Name: Problem 87, Chapter 7: Differential Calculus for JEE Main and Advanced
Uploaded: 2021-11-25T10:30:17-08:00
Duration: 42 s
Channel: Hast Aggarwal
Description: Consider a polynomial y=P(x) of the least degree passing through A(-1,1) and whose graph has two points of inflexion B(1,2) and C with abscissa 0 at which the curve is inclined to the positive axis of abscissas at an angle of sec^-1 √(2). If ∫-2^2 P(x) d x=(k)/(5), k ∈N then k equals (A) 17 (B) 24 (C) 32 (D) 42

step 1 So let's solve this question. So in this question we have integration minus 2 to 2 x 4 by 2 minu

Consider a polynomial y=P(x) of the least degree passing...

Consider a polynomial $\mathrm{y}=\mathrm{P}(\mathrm{x})$ of the least degree passing through $\mathrm{A}(-1,1)$ and whose graph has two points of inflexion $\mathrm{B}(1,2)$ and $\mathrm{C}$ with abscissa $0 \mathrm{at}$ which the curve is inclined to the positive axis of abscissas at an angle of $\sec ^{-1} \sqrt{2}$. If $\int_{-2}^{2} \mathrm{P}(\mathrm{x}) \mathrm{d} \mathrm{x}=\frac{\mathrm{k}}{5}, \mathrm{k} \in \mathrm{N}$ then $\mathrm{k}$ equals (A) 17 (B) 24 (C) 32 (D) 42

Consider a polynomial $\mathrm{y}=\mathrm{P}(\mathrm{x})$ of the least degree passing through $\mathrm{A}(-1,1)$ and whose graph has two points of inflexion $\mathrm{B}(1,2)$ and $\mathrm{C}$ with abscissa $0 \mathrm{at}$ which the curve is inclined to the positive axis of abscissas at an angle of $\sec ^{-1} \sqrt{2}$.
If $\int_{-2}^{2} \mathrm{P}(\mathrm{x}) \mathrm{d} \mathrm{x}=\frac{\mathrm{k}}{5}, \mathrm{k} \in \mathrm{N}$ then $\mathrm{k}$ equals
(A) 17
(B) 24
(C) 32
(D) 42

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Definite Integration of Polynomial Functions

Definite integration computes the net area under the curve between two limits. For polynomials, this involves finding an antiderivative and applying the Fundamental Theorem of Calculus, which is critical in problems where the integrated value provides additional conditions to solve for unknown parameters.

Derivative as Slope of Tangent

The derivative at a point represents the slope of the tangent line to the curve. When an angle of inclination of the tangent is given, the derivative can be computed using the tangent of that angle, linking geometric interpretations to analytical calculations in determining function behavior.

Inflection Points

Inflection points are where the curvature of a function changes sign, identified by setting the second derivative equal to zero. In polynomial analysis, these points provide key constraints on the function’s shape and assist in determining the unknown coefficients of the polynomial.

Least Degree Polynomial

This concept involves finding the simplest polynomial, with the smallest possible degree, that satisfies given conditions such as passing through specified points and meeting derivative constraints. It ensures uniqueness in the solution by using the minimal number of coefficients required to match the required values and behaviors.

Transcript

Please give Ace some feedback