JEE Mathematics 2026 Question with Solution

Given:

• $f(t)=\frac{|t+1|}{t^2}$ for $t<0$
• The largest interval of strict decrease is $(2\alpha,\alpha)$
• $g(x)=2\log_e(x-2)+\alpha x^2+4x-\alpha$ for $x>2$

Find: The local maximum value of $g(x)$.

For $t<0$, split $f(t)$ using the modulus.

$$|t+1|=-(t+1), \quad t<-1$$ $$|t+1|=t+1, \quad -1$$

Given: The interval $(2\alpha,\alpha)$ is the largest decreasing interval for $f(t)=\frac{|t+1|}{t^2}$, and then this value of $\alpha$ is used in $g(x)$.

Find: The local maximum value of $g(x)$.

The key idea is to first identify $\alpha$ from the decreasing interval of $f(t)$.

From the solution working, the decreasing interval is obtained as

$$(-2,-1)$$

Comparing with

$$(2\alpha,\alpha)$$

we get

$$2\alpha=-2, \quad \alpha=-1$$

Now write

$$g(x)=2\log_e(x-2)+\alpha x^2+4x-\alpha$$

Substituting $\alpha=-1$,

$$g(x)=2\log(x-2)-x^2+4x+1$$

Differentiate and equate to zero:

$$g'(x)=\frac{2}{x-2}-2x+4=0$$

So,

$$\frac{2}{x-2}=2x-4$$

which gives the critical point

$$x=3$$

Now compute the function value there:

$$g(3)=2\log 1-9+12+1$$

Since

$$\log 1=0$$

we get

$$g(3)=2$$

Therefore, the local maximum value is $2$.