Suppose a, b, c, d are four real numbers andΔ(x)=x+ax+bx+a−cx+bx+cx−1x+cx+dx−b+d,Statement-1: If a, b, c, d are in A.P. and ∫02 Δ(x)dx=−4 then common difference of the A.P. is ±1Statement-2: If a, b, c, d are in A.P., then ∆(x) is independent of x.

Suppose  are four real numbers and$\mathrm{\Delta }\left(x\right)=\left|\begin{array}{ccc}x+a& x+b& x+a-c\\ x+b& x+c& x-1\\ x+c& x+d& x-b+d\end{array}\right|,$Statement$-1$: If  are in A.P. and ${\int }_{0}^{2} \mathrm{\Delta }\left(x\right)dx=-4$ then common difference of the A.P. is $±1$Statement$-2$: If  are in A.P., then $∆\left(x\right)$ is independent of $x.$

1. A

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

2. B

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

3. C

STATEMENT-1 is True, STATEMENT-2 is False

4. D

STATEMENT-1 is False, STATEMENT-2 is True

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Solution:

Let $D$ be the common difference of the A.P. |
Using and we get
$\mathrm{\Delta }\left(x\right)=\left|\begin{array}{ccc}x+a& x+b& x+a-c\\ D& D& 2D-1\\ D& D& 2D+1\end{array}\right|$

Again using we get
$\mathrm{\Delta }\left(x\right)=\left|\begin{array}{ccc}-D& x+b& x+a-d\\ 0& D& 2D-1\\ 0& D& 2D+1\end{array}\right|=-2{D}^{2}$
Now, ${\int }_{0}^{2} \left(-2{D}^{2}\right)dx=-4⇒4{D}^{2}=4⇒D=±1$
Also, $∆\left(x\right)$ is independent of $x.$