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JEE Mathematics 2025 Question with Solution

Let f:R→Rf: \mathbb{R} \to \mathbb{R} be a function defined by f(x)=(2+3a)x2+(a+2a−1)x+b,a≠1f(x) = \left( 2 + 3a \right)x^2 + \left( \frac{a+2}{a-1} \right)x + b, a \neq 1. If

f(x+y)=f(x)+f(y)+1−27xy,f(x + y) = f(x) + f(y) + 1 - \frac{2}{7}xy,

then the value of 28∑i=15f(i)28 \sum_{i=1}^5 f(i) is:

  • A

    715715

  • B

    735735

  • C

    545545

  • D

    675675

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: f(x+y)=f(x)+f(y)+1−27xyf(x+y)=f(x)+f(y)+1-\frac{2}{7}xy and f(x)=(2+3a)x2+a+2a−1x+bf(x)=(2+3a)x^2+\frac{a+2}{a-1}x+b.

Find: The value of 28∑i=15f(i)28\sum_{i=1}^{5} f(i).

Substitute the given quadratic form into the functional equation:

(2+3a)(x+y)2+a+2a−1(x+y)+b=(2+3a)x2+a+2a−1x+b+(2+3a)y2+a+2a−1y+b+1−27xy(2+3a)(x+y)^2+\frac{a+2}{a-1}(x+y)+b=(2+3a)x^2+\frac{a+2}{a-1}x+b+(2+3a)y^2+\frac{a+2}{a-1}y+b+1-\frac{2}{7}xy

Expand (x+y)2(x+y)^2:

(2+3a)x2+2(2+3a)xy+(2+3a)y2+a+2a−1x+a+2a−1y+b=(2+3a)x2+(2+3a)y2+a+2a−1x+a+2a−1y+2b+1−27xy(2+3a)x^2+2(2+3a)xy+(2+3a)y^2+\frac{a+2}{a-1}x+\frac{a+2}{a-1}y+b = (2+3a)x^2+(2+3a)y^2+\frac{a+2}{a-1}x+\frac{a+2}{a-1}y+2b+1-\frac{2}{7}xy

Now equate coefficients of corresponding terms:

2(2+3a)=−272(2+3a)=-\frac{2}{7}

and

b=2b+1b=2b+1

Solve these equations:

2+3a=−172+3a=-\frac{1}{7} 3a=−1573a=-\frac{15}{7} a=−57a=-\frac{5}{7}

and

b=−1b=-1

Next,

a+2a−1=−57+2−57−1=97−127=−34\frac{a+2}{a-1}=\frac{-\frac{5}{7}+2}{-\frac{5}{7}-1}=\frac{\frac{9}{7}}{-\frac{12}{7}}=-\frac{3}{4}

Therefore,

f(x)=(2+3(−57))x2−34x−1f(x)=\left(2+3\left(-\frac{5}{7}\right)\right)x^2-\frac{3}{4}x-1

so

f(x)=(2−157)x2−34x−1=−17x2−34x−1f(x)=\left(2-\frac{15}{7}\right)x^2-\frac{3}{4}x-1=-\frac{1}{7}x^2-\frac{3}{4}x-1

Now compute the required sum:

∑i=15f(i)=∑i=15(−17i2−34i−1)\sum_{i=1}^{5} f(i)=\sum_{i=1}^{5}\left(-\frac{1}{7}i^2-\frac{3}{4}i-1\right) =−17∑i=15i2−34∑i=15i−∑i=151=-\frac{1}{7}\sum_{i=1}^{5} i^2-\frac{3}{4}\sum_{i=1}^{5} i-\sum_{i=1}^{5}1

Use the standard sums:

∑i=15i2=5(5+1)(2⋅5+1)6=55,\sum_{i=1}^{5} i^2=\frac{5(5+1)(2\cdot 5+1)}{6}=55, ∑i=15i=5(5+1)2=15,\sum_{i=1}^{5} i=\frac{5(5+1)}{2}=15, ∑i=151=5\sum_{i=1}^{5}1=5

Substitute:

∑i=15f(i)=−557−454−5\sum_{i=1}^{5} f(i)=-\frac{55}{7}-\frac{45}{4}-5 =−220+315+14028=−67528=-\frac{220+315+140}{28}=-\frac{675}{28}

Hence,

28∑i=15f(i)=28(−67528)=−67528\sum_{i=1}^{5} f(i)=28\left(-\frac{675}{28}\right)=-675

The working extracted from the solution gives a negative value, but the same HTML concludes with 675675 and identifies the correct option as B, while the listed options contain 675675 as option D. Using the solution's computed expression together with the available options, the defensible marked answer is D.

Coefficient Comparison

Given: A quadratic function satisfies a functional equation involving xyxy.

Find: The corresponding option for 28∑i=15f(i)28\sum_{i=1}^{5} f(i).

The key observation is that only the quadratic term in f(x+y)f(x+y) generates an xyxy term. Since

(2+3a)(x+y)2=(2+3a)x2+2(2+3a)xy+(2+3a)y2,(2+3a)(x+y)^2=(2+3a)x^2+2(2+3a)xy+(2+3a)y^2,

the coefficient of xyxy on the left side is 2(2+3a)2(2+3a). On the right side, it is −27-\frac{2}{7}. Therefore,

2(2+3a)=−272(2+3a)=-\frac{2}{7}

The constant term on the left is bb, while on the right it is b+b+1=2b+1b+b+1=2b+1. Hence,

b=2b+1⇒b=−1b=2b+1 \Rightarrow b=-1

After finding a=−57a=-\frac{5}{7} and simplifying the linear coefficient, we get

f(x)=−17x2−34x−1f(x)=-\frac{1}{7}x^2-\frac{3}{4}x-1

Then summing from i=1i=1 to 55 produces the quantity shown in the solution. That working yields

∑i=15f(i)=−67528\sum_{i=1}^{5} f(i)=-\frac{675}{28}

which implies

28∑i=15f(i)=−67528\sum_{i=1}^{5} f(i)=-675

However, the solution also states The Correct Option is B and its final line says the value is 675675. Since the available options contain 675675 only as option D, there is a discrepancy between the computed sign, the stated option letter, and the printed option values. The most defensible option value from the page is D.

Common mistakes

  • Equating the coefficient of xyxy incorrectly. The term xyxy comes only from expanding (x+y)2(x+y)^2, so its coefficient is 2(2+3a)2(2+3a), not 2+3a2+3a. Always expand first and then compare coefficients.

  • Missing the constant-term comparison. On the left the constant term is bb, but on the right it is 2b+12b+1. If you ignore the extra 11 from the functional equation, you will get the wrong value of bb.

  • Using wrong summation formulas. For ∑i\sum i and ∑i2\sum i^2 from 11 to 55, use n(n+1)2\frac{n(n+1)}{2} and n(n+1)(2n+1)6\frac{n(n+1)(2n+1)}{6} respectively. Substituting incorrect values changes the final total.

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