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MCQMediumJEE 2023Derivatives of Functions

JEE Mathematics 2023 Question with Solution

If f(x)=x3−x2f′(1)+xf′′(2)−f′′′(3)f(x) = x^3 - x^2 f'(1) + x f''(2) - f'''(3), x∈Rx \in \mathbb{R}, then:

  • A

    3f(1)+f(2)=f(3)3f(1) + f(2) = f(3)

  • B

    f(3)−f(2)=f(1)f(3) - f(2) = f(1)

  • C

    2f(0)−f(1)+f(3)=f(2)2f(0) - f(1) + f(3) = f(2)

  • D

    f(1)+f(2)+f(3)=f(0)f(1) + f(2) + f(3) = f(0)

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

f(x)=x3−x2f′(1)+xf′′(2)−f′′′(3)f(x) = x^3 - x^2 f'(1) + x f''(2) - f'''(3)

Find: Which relation among the given options is true.

Let

f′(1)=a,f′′(2)=b,f′′′(3)=cf'(1) = a, \quad f''(2) = b, \quad f'''(3) = c

Then

f(x)=x3−ax2+bx−cf(x) = x^3 - ax^2 + bx - c

Differentiate:

f′(x)=3x2−2ax+bf'(x) = 3x^2 - 2ax + b f′′(x)=6x−2af''(x) = 6x - 2a f′′′(x)=6f'''(x) = 6

Using the given definitions,

c=f′′′(3)=6c = f'''(3) = 6 a=f′(1)=3−2a+ba = f'(1) = 3 - 2a + b b=f′′(2)=12−2ab = f''(2) = 12 - 2a

From

b=12−2ab = 12 - 2a

substitute into

a=3−2a+ba = 3 - 2a + b

so

a=3−2a+12−2aa = 3 - 2a + 12 - 2a 5a=155a = 15 a=3a = 3

Hence,

b=12−2(3)=6,c=6b = 12 - 2(3) = 6, \quad c = 6

Therefore,

f(x)=x3−3x2+6x−6f(x) = x^3 - 3x^2 + 6x - 6

Now evaluate:

f(0)=−6,f(1)=−2,f(2)=2,f(3)=12f(0) = -6, \quad f(1) = -2, \quad f(2) = 2, \quad f(3) = 12

Check option CC:

2f(0)−f(1)+f(3)=2(−6)−(−2)+12=22f(0) - f(1) + f(3) = 2(-6) - (-2) + 12 = 2

And

f(2)=2f(2) = 2

So,

2f(0)−f(1)+f(3)=f(2)2f(0) - f(1) + f(3) = f(2)

Therefore, the correct option is C.

Note: The solution labels the option as A, but the actual statement proved matches source option (3), which is label C.

Also, the second provided approach contains algebraic inconsistencies in intermediate steps, but its final verified relation still matches option C.

Direct verification from evaluated values

After obtaining

f(x)=x3−3x2+6x−6f(x) = x^3 - 3x^2 + 6x - 6

we compute the needed values:

f(0)=−6f(0) = -6 f(1)=1−3+6−6=−2f(1) = 1 - 3 + 6 - 6 = -2 f(2)=8−12+12−6=2f(2) = 8 - 12 + 12 - 6 = 2 f(3)=27−27+18−6=12f(3) = 27 - 27 + 18 - 6 = 12

Now test the relevant option:

2f(0)−f(1)+f(3)=2(−6)−(−2)+12=−12+2+12=2=f(2)2f(0) - f(1) + f(3) = 2(-6) - (-2) + 12 = -12 + 2 + 12 = 2 = f(2)

Hence the true statement is 2f(0)−f(1)+f(3)=f(2)2f(0) - f(1) + f(3) = f(2), so the correct option is C.

Common mistakes

  • Assuming the option label written in the solution is automatically correct. Here the working proves the statement of source option (3), so the correct mapped label is C, not the displayed label A.

  • Differentiating f(x)=x3−ax2+bx−cf(x) = x^3 - ax^2 + bx - c incorrectly. The correct derivatives are f′(x)=3x2−2ax+bf'(x) = 3x^2 - 2ax + b, f′′(x)=6x−2af''(x) = 6x - 2a, and f′′′(x)=6f'''(x) = 6.

  • Using f′(1)=af'(1)=a, f′′(2)=bf''(2)=b, and f′′′(3)=cf'''(3)=c as independent constants without substituting back into the differentiated expressions. These definitions create equations that must be solved for a,b,ca, b, c.

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