JEE Main 2026 24 January Evening Shift
Exam JEE Main
Year 2026
Date 24 January
Shift Evening Shift
Subject Mathematics
Chapter Functions JEE Main 2026 · 24 January · Evening Shift · Mathematics · Functions
Let f be a function such that 3 f(x)+2 f((m)/(19 x))=5 x, x ≠ 0 ,...
Let
f f f be a function such that
3 f ( x ) + 2 f ( m 19 x ) = 5 x , x ≠ 0 3 f(x)+2 f\left(\frac{m}{19 x}\right)=5 x, x \neq 0 3 f ( x ) + 2 f ( 19 x m ) = 5 x , x = 0 , where
m = ∑ i = 1 9 ( i ) 2 m=\sum_{i=1}^{9}(i)^{2} m = ∑ i = 1 9 ( i ) 2 . Then
f ( 5 ) − f ( 2 ) f(5)-f(2) f ( 5 ) − f ( 2 ) is equal to
Step-by-step solution
m = ∑ i = 1 9 i 2 = 9 ⋅ 10 ⋅ 19 6 = 285 m = \sum_{i=1}^{9} i^2 = \frac{9 \cdot 10 \cdot 19}{6} = 285 m = ∑ i = 1 9 i 2 = 6 9 ⋅ 10 ⋅ 19 = 285 , so
m 19 = 15 \frac{m}{19} = 15 19 m = 15 .
The equation becomes
3 f ( x ) + 2 f ( 15 x ) = 5 x 3f(x) + 2f\left(\frac{15}{x}\right) = 5x 3 f ( x ) + 2 f ( x 15 ) = 5 x ...(i)
Replace
x x x by
15 x \frac{15}{x} x 15 :
3 f ( 15 x ) + 2 f ( x ) = 75 x 3f\left(\frac{15}{x}\right) + 2f(x) = \frac{75}{x} 3 f ( x 15 ) + 2 f ( x ) = x 75 ...(ii)
From
3 × ( i ) − 2 × ( i i ) 3 \times (i) - 2 \times (ii) 3 × ( i ) − 2 × ( ii ) :
5 f ( x ) = 15 x − 150 x 5f(x) = 15x - \frac{150}{x} 5 f ( x ) = 15 x − x 150 , so
f ( x ) = 3 x − 30 x f(x) = 3x - \frac{30}{x} f ( x ) = 3 x − x 30 .
f ( 5 ) = 15 − 6 = 9 f(5) = 15 - 6 = 9 f ( 5 ) = 15 − 6 = 9 ,
f ( 2 ) = 6 − 15 = − 9 f(2) = 6 - 15 = -9 f ( 2 ) = 6 − 15 = − 9 .
f ( 5 ) − f ( 2 ) = 9 − ( − 9 ) = 18 f(5) - f(2) = 9 - (-9) = 18 f ( 5 ) − f ( 2 ) = 9 − ( − 9 ) = 18 .
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