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NVAMediumJEE 2026Linear Differential Equations

JEE Mathematics 2026 Question with Solution

Let f:R→Rf: R \to R be a twice differentiable function such that the quadratic equation f(x)m2−2f′(x)m+f′′(x)=0f(x)m^2-2f'(x) m+f''(x) = 0 in mm, has two equal roots for every x∈Rx \in R. If f(0)=1,f′(0)=2f(0)=1, f'(0) = 2, and (α,β)(\alpha, \beta) is the largest interval in which the function g(x)=f(log⁡ex−x)g(x) = f(\log_e x - x) is increasing, then α+β\alpha+\beta is equal to:

Answer

Correct answer:1

Step-by-step solution

Standard Method

Given: The quadratic equation

f(x)m2−2f′(x)m+f′′(x)=0f(x)m^2 - 2f'(x)m + f''(x) = 0

in mm has equal roots for every x∈Rx \in R. Also, f(0)=1f(0)=1 and f′(0)=2f'(0)=2.

Find: The value of α+β\alpha+\beta, where (α,β)(\alpha,\beta) is the largest interval in which

g(x)=f(log⁡ex−x)g(x)=f(\log_e x - x)

is increasing.

For a quadratic to have equal roots, its discriminant must be zero:

(2f′(x))2−4f(x)f′′(x)=0(2f'(x))^2 - 4f(x)f''(x) = 0

So,

4(f′(x))2=4f(x)f′′(x)4(f'(x))^2 = 4f(x)f''(x)

which gives

(f′(x))2=f(x)f′′(x)(f'(x))^2 = f(x)f''(x)

This implies

(f′(x)f(x))′=0\left(\frac{f'(x)}{f(x)}\right)' = 0

Hence,

f′(x)f(x)=c\frac{f'(x)}{f(x)} = c

for some constant cc. Therefore,

f′(x)=cf(x)f'(x)=cf(x)

and so

f(x)=Kecxf(x)=Ke^{cx}

Using f(0)=1f(0)=1, we get

K=1K=1

Thus,

f(x)=ecxf(x)=e^{cx}

Now,

f′(x)=cecxf'(x)=ce^{cx}

Using f′(0)=2f'(0)=2, we get

c=2c=2

Therefore,

f(x)=e2xf(x)=e^{2x}

Now evaluate g(x)g(x):

g(x)=f(log⁡ex−x)=e2(log⁡ex−x)=x2e−2xg(x)=f(\log_e x - x)=e^{2(\log_e x - x)}=x^2e^{-2x}

Differentiate:

g′(x)=ddx(x2e−2x)g'(x)=\frac{d}{dx}(x^2e^{-2x})

Using the product rule,

g′(x)=2xe−2x+x2(−2e−2x)g'(x)=2xe^{-2x}+x^2(-2e^{-2x})

So,

g′(x)=2xe−2x(1−x)g'(x)=2xe^{-2x}(1-x)

Since e−2x>0e^{-2x}>0 for all xx in the domain and log⁡ex\log_e x requires x>0x>0, we need

2x(1−x)>02x(1-x)>0

For x>0x>0, this gives

1−x>01-x>0

So,

00

Common mistakes

  • Ignoring the domain of log⁡ex\log_e x. Since g(x)=f(log⁡ex−x)g(x)=f(\log_e x-x), we must have x>0x>0. Taking intervals that include x≤0x\le 0 is invalid. Always apply the domain restriction before testing where g′(x)>0g'(x)>0.

  • Using the equal-roots condition incorrectly. For the quadratic in mm, equal roots mean the discriminant is zero, so (2f′(x))2−4f(x)f′′(x)=0\big(2f'(x)\big)^2-4f(x)f''(x)=0. Do not equate coefficients or roots directly without using the discriminant.

  • Differentiating x2e−2xx^2e^{-2x} incorrectly. The function is a product, so the product rule is required. Writing g′(x)=2xe−2xg'(x)=2xe^{-2x} or differentiating only one factor gives the wrong sign pattern and hence the wrong interval of increase.

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