For all y- R- f y -1 y y-1 2y y y-1 y y-1 3y y-1 y y-1 y-2 y y2-1 - then - -2- -2f y2-2 dy equals to

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Arithmetic Progression

For all yϵ ...

Question

For all $y ϵ R$ , $f (y) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & y & y + 1 2 y & y (y - 1) & y (y + 1) 3 y (y - 1) & y (y - 1) (y - 2) & y (y^{2} - 1) \end{matrix} ∣ ∣ ∣ ∣ ∣$ , then $\int_{- π / 2}^{π / 2} f (y^{2} + 2) d y$ equals to

π

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- π

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0

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2 π

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Solution

The correct option is C $0$
Taking $y$ common from $R_{2}$ and $y (y + 1)$ from $R_{3}$ , we get

$f (y) = y^{2} (y - 1) ∣ ∣ ∣ ∣ \begin{matrix} 1 & y & y + 1 2 & y - 1 & y + 1 3 & y - 2 & y + 1 \end{matrix} ∣ ∣ ∣ ∣$

Applying $C_{3} \to C_{3} - C_{2}$ , we get

$f (y) = y^{2} (y - 1) ∣ ∣ ∣ ∣ \begin{matrix} 1 & y & 1 2 & y - 1 & 2 3 & y - 2 & 3 \end{matrix} ∣ ∣ ∣ ∣ = 0$

Therefore, $\int_{- π / 2}^{π / 2} f (y^{2} + 2) d y = 0$

Hence, option 'C' is correct.

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For all yϵR, f(y)=∣∣ ∣ ∣∣1yy+12yy(y−1)y(y+1)3y(y−1)y(y−1)(y−2)y(y2−1)∣∣ ∣ ∣∣, then ∫π/2−π/2f(y2+2)dy equals to

The correct option is C 0Taking y common from R2 and y(y+1) from R3, we getf(y)=y2(y−1)∣∣ ∣∣1yy+12y−1y+13y−2y+1∣∣ ∣∣Applying C3→C3−C2, we getf(y)=y2(y−1)∣∣ ∣∣1y12y−123y−23∣∣ ∣∣=0Therefore, ∫π/2−π/2f(y2+2)dy=0Hence, option 'C' is correct.

For all $y ϵ R$ , $f (y) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & y & y + 1 2 y & y (y - 1) & y (y + 1) 3 y (y - 1) & y (y - 1) (y - 2) & y (y^{2} - 1) \end{matrix} ∣ ∣ ∣ ∣ ∣$ , then $\int_{- π / 2}^{π / 2} f (y^{2} + 2) d y$ equals to