
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
232.8k+ 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
Geometry of Complex Numbers Explained

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

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

Electricity and Magnetism Explained: Key Concepts & Applications

Algebra Made Easy: Step-by-Step Guide for Students

Trending doubts
JEE Main 2026: Admit Card Out, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

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 12 Limits and Derivatives (2025-26)

NCERT Solutions For Class 11 Maths Chapter 10 Conic Sections (2025-26)

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

