📐
JEE Main · MCQ · Differentiation · Functions
MCQ · Mathematics · Differentiation
Q. Let $f(x) = x^3 + x^2 f'(1) + 2x f”(2) + f”'(3),\ x \in \mathbb{R}$. Then the value of $f'(5)$ is :
A$\dfrac{657}{5}$
B$\dfrac{117}{5}$ ✓
C$\dfrac{2}{5}$
D$\dfrac{62}{5}$
✅ Correct Answer: (B) $\dfrac{117}{5}$
Step-by-Step Solution
1
Let $f'(1) = a$, $f”(2) = b$, $f”'(3) = c$ (constants)
Then the function becomes:
$f(x) = x^3 + ax^2 + 2bx + c$
2
Differentiate $f(x)$ successively
$f'(x) = 3x^2 + 2ax + 2b$
$f”(x) = 6x + 2a$
$f”'(x) = 6$
$f”(x) = 6x + 2a$
$f”'(x) = 6$
3
Find $c$ from $f”'(3) = c$
$c = f”'(3) = 6$
4
Find $b$ from $f”(2) = b$
$b = f”(2) = 12 + 2a \quad \cdots (1)$
5
Find $a$ from $f'(1) = a$
$a = f'(1) = 3 + 2a + 2b$
$\Rightarrow a – 2a – 2b = 3$
$\Rightarrow -a – 2b = 3 \quad \cdots (2)$
Substitute (1) into (2):
$\Rightarrow a – 2a – 2b = 3$
$\Rightarrow -a – 2b = 3 \quad \cdots (2)$
$-a – 2(12 + 2a) = 3$
$-a – 24 – 4a = 3$
$-5a = 27 \Rightarrow a = -\dfrac{27}{5}$
Then: $b = 12 + 2\!\left(-\dfrac{27}{5}\right) = 12 – \dfrac{54}{5} = \dfrac{60-54}{5} = \dfrac{6}{5}$
$-a – 24 – 4a = 3$
$-5a = 27 \Rightarrow a = -\dfrac{27}{5}$
6
Compute $f'(5)$
$f'(5) = 3(25) + 2a(5) + 2b$
$= 75 + 10\!\left(-\dfrac{27}{5}\right) + 2\!\left(\dfrac{6}{5}\right)$
$= 75 – \dfrac{270}{5} + \dfrac{12}{5}$
$= 75 – \dfrac{258}{5}$
$= \dfrac{375 – 258}{5} = \dfrac{117}{5}$
$= 75 + 10\!\left(-\dfrac{27}{5}\right) + 2\!\left(\dfrac{6}{5}\right)$
$= 75 – \dfrac{270}{5} + \dfrac{12}{5}$
$= 75 – \dfrac{258}{5}$
$= \dfrac{375 – 258}{5} = \dfrac{117}{5}$
$f'(5) = \dfrac{117}{5}$ → Option (B) ✓
Related Theory
📌 Self-Referential Function Technique
When $f(x)$ is defined in terms of its own derivatives at specific points, treat those derivative values as unknown constants ($a, b, c$). Write $f(x)$ as a polynomial, differentiate, then substitute the specific points to get a linear system. Solve for $a, b, c$.
f′(1)=a, f″(2)=b, f‴(3)=c → solve simultaneously
f′(1)=a, f″(2)=b, f‴(3)=c → solve simultaneously
📌 Key Derivatives of x³+ax²+2bx+c
f′ = 3x²+2ax+2b
f″ = 6x+2a
f‴ = 6 (constant)
📌 Common Mistakes
❌ Mistake 1: Treating a, b, c as functions of x — they are constants (specific values).
❌ Mistake 2: Arithmetic error in solving the linear system — solve step by step carefully.
❌ Mistake 2: Arithmetic error in solving the linear system — solve step by step carefully.
Frequently Asked Questions
1. What substitution simplifies this problem?
Let f′(1)=a, f″(2)=b, f‴(3)=c, treating them as unknown constants.
2. Why is f‴(x) = 6?
f(x)=x³+ax²+2bx+c → f‴(x)=6, a constant for all x.
3. What is the value of c?
c = f‴(3) = 6.
4. What is a?
a = −27/5.
5. What is b?
b = 6/5.
6. What is f′(5)?
75 − 258/5 = 117/5.
7. Is this MCQ or NAT?
MCQ — Option (B) 117/5 is correct.
8. What topic does this belong to?
Differentiation — self-referential polynomial functions. Common in JEE Main.
9. How many equations are formed?
Three: one each from f‴(3)=c, f″(2)=b, f′(1)=a.
10. What is the formula for f′(x)?
f′(x) = 3x² + 2ax + 2b.
Related JEE Main Questions
$6\displaystyle\int_0^{\pi}|(\sin 3x + \sin 2x + \sin x)|\,dx$ is equal to
Numerical Answer Type
✅ Correct Answer: 17
Let $S = \{(m,n) : m,n \in \{1,2,\ldots,50\}\}$. If the number of elements $(m,n)$ such that $6^m+9^n$ is a multiple of 5 is $p$ and those with $m+n$ a square of prime is $q$, then $p+q$ is equal to
Numerical Answer Type
✅ Correct Answer: 1333
Let $a_1=1$ and for $n\geq1$, $a_{n+1}=\frac{1}{2}a_n+\frac{n^2-2n-1}{n^2(n+1)^2}$. Then $\left|\sum_{n=1}^{\infty}(a_n-\frac{2}{n^2})\right|$ is equal to
Numerical Answer Type
✅ Correct Answer: 2
Let $f:\mathbf{R}\to\mathbf{R}$ be twice differentiable such that $f(x)m^2-2f'(x)m+f”(x)=0$ has two equal roots for every $x\in\mathbf{R}$. If $f(0)=1$, $f'(0)=2$, and $(\alpha,\beta)$ is the largest interval where $f(\log_e x-x)$ is increasing, then $\alpha+\beta$ is
Numerical Answer Type
✅ Correct Answer: 1
Let $y=y(x)$ be the solution of $(1+x^2)dy+(y-\tan^{-1}x)dx=0$, $y(0)=1$. Then $y(1)$ is :
(A) $\frac{4}{e^{\pi/4}}-\frac{\pi}{2}-1$
(B) $\frac{2}{e^{\pi/4}}+\frac{\pi}{4}-1$
✅ Correct Answer: (B)