Skip to content
dronacharya.ai
← JEE Main 2026 · 6 Apr · Shift 2
MathematicsSingle correct+4 / -1Question 17 of 75

JEE Main 2026 · 6 Apr · Shift 2, Mathematics question 17

Let A=[13−121α01−1]A=\begin{bmatrix}1&3&-1\\2&1&\alpha\\0&1&-1\end{bmatrix} be a singular matrix. Let f(x)=∫0x(t2+2t+3)dtf(x)=\displaystyle\int_0^x\left(t^2+2t+3\right)dt, x∈[1,α]x\in[1,\alpha]. If M and m are respectively the maximum and the minimum values of ff in [1,α][1,\alpha], then 3(M−m)3(M-m) is equal to :

Choose one option
Answer and worked solution

Official answer: B

Singular. det⁡A=−1−α+6−2=3−α=0\det A=-1-\alpha+6-2=3-\alpha=0, so α=3\alpha=3.

Extremes. t2+2t+3>0t^2+2t+3>0, so f is increasing on [1,3][1,3]:
M−m=∫13(t2+2t+3) dt=27−133=683.M-m=\int_1^3(t^2+2t+3)\,dt=27-\frac{13}3=\frac{68}3.

So 3(M−m)=683(M-m)=68.

(B)

Topics:Fundamental theorem of calculusDeterminants and propertiesMaxima and minima
← Previous questionNext question →