Matrices & DeterminantsmediumPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

System with Infinite Solutions: f Strictly Increasing | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
Consider the system of linear equations in x,y,zx,y,z: x+2y+tz=0, 6x+y+5tz=0, 3x+t2y+f(t)z=0x+2y+tz=0,\ 6x+y+5tz=0,\ 3x+t^2y+f(t)z=0, where f:R→Rf:\mathbb{R}\to\mathbb{R} is a differentiable function. If this system has infinitely many solutions for all t∈Rt\in\mathbb{R}, then ff
Ais a constant function
Bis strictly increasing on R\mathbb{R}correct
Cis strictly decreasing on R\mathbb{R}
Dhas two critical points
Solution
Step 1: Infinitely many solutions ⇒det⁡=0\Rightarrow\det=0 for all tt:
D=∣12t615t3t2f(t)∣=0.D=\begin{vmatrix}1&2&t\\6&1&5t\\3&t^2&f(t)\end{vmatrix}=0.
 Step 2: Expand along row 1:
1(f−5t3)−2(6f−15t)+t(6t2−3)=0.1\big(f-5t^3\big)-2\big(6f-15t\big)+t\big(6t^2-3\big)=0.
⇒f−5t3−12f+30t+6t3−3t=0⇒−11f+t3+27t=0.\Rightarrow f-5t^3-12f+30t+6t^3-3t=0\Rightarrow-11f+t^3+27t=0.
∴ f(t)=t3+27t11.\therefore\ f(t)=\frac{t^3+27t}{11}.
 Step 3: f′(t)=3t2+2711f'(t)=\dfrac{3t^2+27}{11}. Since 3t2≥0⇒3t2+27≥27>03t^2\ge0\Rightarrow 3t^2+27\ge27>0, so f′(t)>0f'(t)>0 for all tt. Step 4: f′(t)>0f'(t)>0 on R⇒f\mathbb{R}\Rightarrow f is strictly increasing. Correct answer: (2)
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