If P - x 1 1 x and Q - x 1 1 1 x 1 1 1 x- then dQdx -

The correct option is D 3P P=x^2 1 ...(i) #160;[Solving the determinant] Q=x(x^2 1) (x 1)+1 x #160; [Solving the determinant] #160; ...

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Byju's Answer

Standard XII

Mathematics

Reflection of a Point about a Line

If P = x 1...

Question

If $P = ∣ ∣ ∣ \begin{matrix} x & 1 1 & x \end{matrix} ∣ ∣ ∣$ and $Q = ∣ ∣ ∣ ∣ \begin{matrix} x & 1 & 1 1 & x & 1 1 & 1 & x \end{matrix} ∣ ∣ ∣ ∣$ , then $\frac{d Q}{d x} =$ _____

3 P + 1

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1 - 3 P

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- 3 P

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3 P

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Solution

The correct option is D $3 P$
$P = x^{2} - 1$ ...(i) [Solving the determinant]
$Q = x (x^{2} - 1) - (x - 1) + 1 - x$ [Solving the determinant]
$= x^{3} - x - x + 1 + 1 - x$
$= x^{3} - 3 x + 2$
Therefore,
$\frac{d Q}{d x} = 3 x^{2} - 3$
$= 3 (x^{2} - 1)$
$= 3 P$ .

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Similar questions

If $p = ∣ ∣ ∣ \begin{matrix} x & 1 1 & x \end{matrix} ∣ ∣ ∣$ and $Q = ∣ ∣ ∣ ∣ \begin{matrix} x & 1 & 1 1 & x & 1 1 & 1 & x \end{matrix} ∣ ∣ ∣ ∣$ then $\frac{d Q}{d x}$ =.......

Q. If

P = ∣ ∣ ∣ \begin{matrix} x & 1 1 & x \end{matrix} ∣ ∣ ∣

and

Q = ∣ ∣ ∣ ∣ \begin{matrix} x & 1 & 1 1 & x & 1 1 & 1 & x \end{matrix} ∣ ∣ ∣ ∣,

then

\frac{d Q}{d x} =

Q. If

(3 p + q, p - q) = (p - q, 3 p + q)

, then:

Q. If

(3 p - \frac{1}{3 p}) = 5

and

(3 p + \frac{1}{3 p}) = 8

, then find the value of

9 p^{2} - \frac{1}{9 p^{2}}

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If P=∣∣∣x11x∣∣∣ and Q=∣∣ ∣∣x111x111x∣∣ ∣∣, then dQdx= _____

The correct option is D 3PP=x2−1 ...(i) [Solving the determinant]Q=x(x2−1)−(x−1)+1−x [Solving the determinant] =x3−x−x+1+1−x =x3−3x+2Therefore,dQdx=3x2−3 =3(x2−1) =3P.

If $P = ∣ ∣ ∣ \begin{matrix} x & 1 1 & x \end{matrix} ∣ ∣ ∣$ and $Q = ∣ ∣ ∣ ∣ \begin{matrix} x & 1 & 1 1 & x & 1 1 & 1 & x \end{matrix} ∣ ∣ ∣ ∣$ , then $\frac{d Q}{d x} =$ _____

The correct option is D $3 P$
$P = x^{2} - 1$ ...(i) [Solving the determinant]
$Q = x (x^{2} - 1) - (x - 1) + 1 - x$ [Solving the determinant]
$= x^{3} - x - x + 1 + 1 - x$
$= x^{3} - 3 x + 2$
Therefore,
$\frac{d Q}{d x} = 3 x^{2} - 3$
$= 3 (x^{2} - 1)$
$= 3 P$ .