Maths Q20 – JEE-MAIN 2026

Home Maths Differential EquationsQuestion #178

Let

f (x)

and

g (x)

be twice differentiable functions satisfying

f^{″} (x) = g^{″} (x)

for all

x \in R

f^{'} (1) = 2 g^{'} (1) = 4

and

g (2) = 3 f (2) = 9

. Then

f (25) - g (25)

is equal to :

Clue 1 of 3: Define a new function

Consider defining a new function $h (x) = f (x) - g (x)$ to simplify the given differential equation.

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● Video Walkthrough

Step 1: Define a new function and integrate✦ Active

Let $h (x) = f (x) - g (x)$ . Given $f^{″} (x) = g^{″} (x)$ , it implies $h^{″} (x) = f^{″} (x) - g^{″} (x) = 0$ . Integrating $h^{″} (x) = 0$ once gives $h^{'} (x) = C_{1}$ . Integrating $h^{'} (x) = C_{1}$ once more gives $h (x) = C_{1} x + C_{2}$ . Thus, we have $f (x) - g (x) = C_{1} x + C_{2}$ .

Step 2: Determine the integration constants using given conditions○ Expand

From $f^{'} (1) = 2 g^{'} (1) = 4$ , we have $f^{'} (1) = 4$ and $g^{'} (1) = 2$ . Since $h^{'} (x) = f^{'} (x) - g^{'} (x) = C_{1}$ , substituting $x = 1$ gives $C_{1} = f^{'} (1) - g^{'} (1) = 4 - 2 = 2$ . So, $f (x) - g (x) = 2 x + C_{2}$ . From $g (2) = 3 f (2) = 9$ , we have $g (2) = 9$ and $f (2) = 3$ . Substituting $x = 2$ into $f (x) - g (x) = 2 x + C_{2}$ : $f (2) - g (2) = 2 (2) + C_{2} 3 - 9 = 4 + C_{2} - 6 = 4 + C_{2} C_{2} = - 10$ . Thus, the expression for $f (x) - g (x)$ is $2 x - 10$ .

Step 3: Calculate the required value○ Expand

We need to find $f (25) - g (25)$ . Substitute $x = 25$ into the expression $f (x) - g (x) = 2 x - 10$ : $f (25) - g (25) = 2 (25) - 10 = 50 - 10 = 40$ .

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Maths Question 20 – JEE-MAIN 2026

Clue 1 of 3: Define a new function

🧠 Full Solution Path