
If \[f\left( x \right) = m{x^2} + nx + p\], then \[{f^{'}}\left( 1 \right) + {f^{'}}\left( 4 \right) - {f^{'}}\left( 5 \right)\] is equal to
A. \[m\]
B. \[ - m\]
C. \[n\]
D. \[ - n\]
Answer
216.3k+ views
Hint: In this question, we are asked to find the value of \[{f^{'}}\left( 1 \right) + {f^{'}}\left( 4 \right) - {f^{'}}\left( 5 \right)\] . For that, we first take the derivative of the given function and then substitute the value \[ x= 1,4,5 \] to get the desired result.
Formula used:
1. \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
Complete step-by-step solution:
We are given a function \[f\left( x \right) = m{x^2} + nx + p\] that is a second-degree polynomial Now we need to act on the first derivation, due to a small difference in the process of derivation, we call it a differentiation function. So, the differentiation formula is \[\dfrac{{dy}}{{dx}}\] .
\[{f^{'}}\left( x \right) = 2mx + n\]
Now substitute \[x = 1\], we obtain
\[{f^{'}}\left( 1 \right) = 2m + n\]
Now substitute \[x = 4\], we obtain
\[
{f^{'}}\left( 4 \right) = 2m \times \,4 + n \\
= 8m + n
\]
Now substitute \[x = 5\], we obtain
\[
{f^{'}}\left( 5 \right) = 2m \times \,5 + n \\
= 10\,m + n
\]
Thus, the required values of function \[{f^{'}}\left( 1 \right) + {f^{'}}\left( 4 \right) - {f^{'}}\left( 5 \right)\] are
\[
{f^{'}}\left( 1 \right) + {f^{'}}\left( 4 \right) - {f^{'}}\left( 5 \right) = 2m + n + 8m + n - 10m - n \\
= 2n - n \\
= n
\]
Hence, option (C) is correct
Note: Integration and differentiation are inverse processes. The derivative of x raised to the power is denoted by \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\] in differentiation. Inverse differentiation is the process of integration. The process of determining which functions have a given derivative is known as anti-differentiation. It can calculate the area, volume, and central points. The anti-differentiation value at the upper limit and lower limit with the same anti-differentiation.
Formula used:
1. \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
Complete step-by-step solution:
We are given a function \[f\left( x \right) = m{x^2} + nx + p\] that is a second-degree polynomial Now we need to act on the first derivation, due to a small difference in the process of derivation, we call it a differentiation function. So, the differentiation formula is \[\dfrac{{dy}}{{dx}}\] .
\[{f^{'}}\left( x \right) = 2mx + n\]
Now substitute \[x = 1\], we obtain
\[{f^{'}}\left( 1 \right) = 2m + n\]
Now substitute \[x = 4\], we obtain
\[
{f^{'}}\left( 4 \right) = 2m \times \,4 + n \\
= 8m + n
\]
Now substitute \[x = 5\], we obtain
\[
{f^{'}}\left( 5 \right) = 2m \times \,5 + n \\
= 10\,m + n
\]
Thus, the required values of function \[{f^{'}}\left( 1 \right) + {f^{'}}\left( 4 \right) - {f^{'}}\left( 5 \right)\] are
\[
{f^{'}}\left( 1 \right) + {f^{'}}\left( 4 \right) - {f^{'}}\left( 5 \right) = 2m + n + 8m + n - 10m - n \\
= 2n - n \\
= n
\]
Hence, option (C) is correct
Note: Integration and differentiation are inverse processes. The derivative of x raised to the power is denoted by \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\] in differentiation. Inverse differentiation is the process of integration. The process of determining which functions have a given derivative is known as anti-differentiation. It can calculate the area, volume, and central points. The anti-differentiation value at the upper limit and lower limit with the same anti-differentiation.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections

NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series

Understanding Atomic Structure for Beginners

NCERT Solutions For Class 11 Maths Chapter 12 Limits And Derivatives

