The correct option is B A−iv,B−iii,C−ii,D−i
A) A=∣∣
∣∣x+2x+3x+5x+4x+6x+9x+8x+11x+15∣∣
∣∣
Applying R1→R1−R2,R2→R2−R3,R3→R3−R1
A=∣∣
∣∣−1−2−3−2−3−5−3−4−7∣∣
∣∣=−∣∣
∣∣123235347∣∣
∣∣=−(1(21−20)−2(14−15)+3(8−9))=−2
B) B=∣∣
∣
∣∣76x2−132x2−132x2−1337∣∣
∣
∣∣
Applying R1→R1+R2+R3
B=∣∣
∣
∣∣x2−4x2−4x2−42x2−132x2−1337∣∣
∣
∣∣=(x2−4)∣∣
∣
∣∣1112x2−132x2−1337∣∣
∣
∣∣
Now applying C2→C2−C1,C3→C3−C1
B=(x2−4)∣∣
∣
∣∣1002x2−90x2−1316−x220−x2∣∣
∣
∣∣=(x2−4)(x2−9)(20−x2)
As x+2 is one of the roots the equation become
B=(x2−9)(20−x2)(x−2)
Hence sum off roots is 0
C) C=∣∣
∣
∣∣√62i3+√6√12√3+√8i3√2+√6i√18√2+√12i√27+2i∣∣
∣
∣∣=√6∣∣
∣
∣∣12i3+√6√2√3+√8i3√2+√6i√3√2+√12i√27+2i∣∣
∣
∣∣
Applying R2→R2−√2R1,R3→R3−√3R1, we get
C=∣∣
∣
∣∣12i3+√60√3√6i−2√30√22i−3√2∣∣
∣
∣∣=√6(√3(2i−3√2)−√2(√6i−2√3))=−6
D) f(θ)=∣∣
∣
∣∣cos2θcosθsinθ−sinθcosθsinθsin2θcosθsinθ−cosθ0∣∣
∣
∣∣=1sinθ∣∣
∣
∣∣cos2θsinθcosθsinθ−sinθcosθsin2θsin2θcosθsin2θ−cosθ0∣∣
∣
∣∣
Applying C1→C1−cosθC2, we get
f(θ)=1sinθ∣∣
∣
∣∣0cosθsinθ−sinθ0sin2θcosθ1−cosθ0∣∣
∣
∣∣=1sinθ(cos2θsinθ+sin3θ)=cos2θ+sin2θ=1⇒f(π3)=1