QBManager

Matrix and Determinant

79245 If \(f(x)=\left|\begin{array}{ccc}1+\sin x+\sin 2 x+\sin 3 x & \frac{3+\sin 2 x}{2} & \frac{-2+\sin 3 x}{3} \\ 3+4 \sin x & \frac{3}{2} & \frac{4}{3} \sin x \\ 1+\sin x & \frac{1}{2} \sin x & \frac{1}{3}\end{array}\right|\) then \(\int_0^{\pi / 2}\left(f(x)+f^{\prime}(x)\right) d x=\)

1 \(\frac{-1}{6}\)

2 \(\frac{-1}{9}\)

3 \(\frac{-2}{9}\)

4 \(\frac{1}{27}\)

Explanation:

(B) : Given that,
\(f(x)=\left|\begin{array}{ccc}1+\sin x+\sin 2 x+\sin 3 x & \frac{3+\sin 2 x}{2} & \frac{-2+\sin 3 x}{3} \\ 3+4 \sin x & \frac{3}{2} & \frac{4}{3} \sin x \\ 1+\sin x & \frac{1}{2} \sin x & \frac{1}{3}\end{array}\right|\)
Applying \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-2 \mathrm{C}_{2}-3 \mathrm{C}_{3}\)
\(\mathrm{f}(\mathrm{x})=\left|\begin{array}{ccc} \sin \mathrm{x} & \frac{3+\sin 2 \mathrm{x}}{2} & \frac{-2+\sin 3 \mathrm{x}}{3} \\ 0 & \frac{3}{2} & \frac{4}{3} \sin \mathrm{x} \\ 0 & \frac{1}{2} \sin \mathrm{x} & \frac{1}{3} \end{array}\right|\)
On expanding, we get
\(\mathrm{f}(\mathrm{x}) =\sin \mathrm{x}\left|\begin{array}{cc} \frac{3}{2} & \frac{4}{3} \sin \mathrm{x} \\ \frac{1}{2} \sin \mathrm{x} & \frac{1}{3} \end{array}\right|+0+0\)
\(=\sin \mathrm{x}\left(\frac{3}{2} \times \frac{1}{3}-\frac{4}{3} \sin \mathrm{x} \times \frac{1}{2} \sin \mathrm{x}\right)\)
\(=\sin \mathrm{x}\left(\frac{1}{2}-\frac{2}{3} \sin ^{2} \mathrm{x}\right)\)
\(\mathrm{f}(\mathrm{x}) =\frac{1}{2} \sin \mathrm{x}-\frac{2}{3} \sin ^{3} \mathrm{x}\)
\(\mathrm{f}(\mathrm{x}) =\frac{1}{6}\left(3 \sin \mathrm{x}-4 \sin ^{3} \mathrm{x}\right) \)
\(\mathrm{f}(\mathrm{x}) =\frac{\sin 3 \mathrm{x}}{6} \quad\left[\because \sin 3 \mathrm{x}=3 \sin \mathrm{x}-4 \sin ^{3} \mathrm{x}\right]\)
Differentiating on both sides w.r.t. ' \(x\) ', we get
\(\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\sin 3 \mathrm{x}}{6}\right)=\frac{1}{6} \cdot \cos 3 \mathrm{x} \cdot 3\)
\(\mathrm{f}^{\prime}(\mathrm{x})=\frac{\cos 3 \mathrm{x}}{2}\)
Let \(\quad I=\int_{0}^{\pi / 2}\left(f(x)+f^{\prime}(x)\right) d x\)
\(I=\int_{0}^{\pi / 2}\left(\frac{\sin 3 x}{6}+\frac{\cos 3 x}{2}\right) d x \tag{i}\)
\(I=\int_{0}^{\pi / 2}\left[\frac{\sin 3\left(\frac{\pi}{2}-x\right)}{6}+\frac{\cos 3\left(\frac{\pi}{2}-x\right)}{2}\right] d x\)
\(I=\int_{0}^{\pi / 2}\left[\frac{\sin \left(\frac{3 \pi}{2}-3 x\right)}{6}+\frac{\cos \left(\frac{3 \pi}{2}-3 x\right)}{2}\right] d x\)
\(I=\int_{0}^{\pi / 2}\left(-\frac{\cos 3 x}{6}-\frac{\sin 3 x}{2}\right) d x \tag{ii}\)
Adding equation (i) and (ii), we get
\(2 \mathrm{I} =\int_{0}^{\pi / 2}\left(\frac{\cos 3 \mathrm{x}}{2}-\frac{\cos 3 \mathrm{x}}{6}+\frac{\sin 3 \mathrm{x}}{6}-\frac{\sin 3 \mathrm{x}}{2}\right) \mathrm{dx}\)
\(2 \mathrm{I} =\frac{1}{3} \int_{0}^{\pi / 2}(\cos 3 \mathrm{x}-\sin 3 \mathrm{x}) \mathrm{dx}\)
\(\text { or } \quad 2 \mathrm{I} =\frac{1}{3}\left[\frac{\sin 3 \mathrm{x}}{3}+\frac{\cos 3 \mathrm{x}}{3}\right]_{0}^{\pi / 2}\)
\(=\frac{1}{9}\left[\sin 3 \times \frac{\pi}{2}+\cos 3 \times \frac{\pi}{2}-\sin 3 \times 0-\cos 3 \times 0\right]\)
\(\mathrm{I} =\frac{1}{9}[-1+0-0-1]=-\frac{2}{9}\)
Or \(\quad \mathrm{I}=-\frac{1}{9}\)

TS EAMCET-2020-11.09.2020

Matrix and Determinant

79239 If \(f(x)=\left|\begin{array}{lll}\cos x & x & 1 \\ 2 \sin x & x^{2} & 2 x \\ \tan x & x & 1\end{array}\right|\), then the value of \(f^{\prime}(x)\) at \(x=0\) is equal to

1 -1

2 1

3 2

4 0

AP EAMCET-2020-22.09.2020

Matrix and Determinant

79240 If \(f(x)=\left|\begin{array}{ccc}\cos x & 1 & 0 \\ 1 & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x\end{array}\right|\), then \(\int_{0}^{\pi / 2} f(x) d x\) is equal to

1 \(\frac{1}{4}\)

2 \(-\frac{1}{3}\)

3 \(\frac{1}{2}\)

4 1

BITSAT-2005

Matrix and Determinant

79245 If \(f(x)=\left|\begin{array}{ccc}1+\sin x+\sin 2 x+\sin 3 x & \frac{3+\sin 2 x}{2} & \frac{-2+\sin 3 x}{3} \\ 3+4 \sin x & \frac{3}{2} & \frac{4}{3} \sin x \\ 1+\sin x & \frac{1}{2} \sin x & \frac{1}{3}\end{array}\right|\) then \(\int_0^{\pi / 2}\left(f(x)+f^{\prime}(x)\right) d x=\)

1 \(\frac{-1}{6}\)

2 \(\frac{-1}{9}\)

3 \(\frac{-2}{9}\)

4 \(\frac{1}{27}\)

Explanation:

(B) : Given that,
\(f(x)=\left|\begin{array}{ccc}1+\sin x+\sin 2 x+\sin 3 x & \frac{3+\sin 2 x}{2} & \frac{-2+\sin 3 x}{3} \\ 3+4 \sin x & \frac{3}{2} & \frac{4}{3} \sin x \\ 1+\sin x & \frac{1}{2} \sin x & \frac{1}{3}\end{array}\right|\)
Applying \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-2 \mathrm{C}_{2}-3 \mathrm{C}_{3}\)
\(\mathrm{f}(\mathrm{x})=\left|\begin{array}{ccc} \sin \mathrm{x} & \frac{3+\sin 2 \mathrm{x}}{2} & \frac{-2+\sin 3 \mathrm{x}}{3} \\ 0 & \frac{3}{2} & \frac{4}{3} \sin \mathrm{x} \\ 0 & \frac{1}{2} \sin \mathrm{x} & \frac{1}{3} \end{array}\right|\)
On expanding, we get
\(\mathrm{f}(\mathrm{x}) =\sin \mathrm{x}\left|\begin{array}{cc} \frac{3}{2} & \frac{4}{3} \sin \mathrm{x} \\ \frac{1}{2} \sin \mathrm{x} & \frac{1}{3} \end{array}\right|+0+0\)
\(=\sin \mathrm{x}\left(\frac{3}{2} \times \frac{1}{3}-\frac{4}{3} \sin \mathrm{x} \times \frac{1}{2} \sin \mathrm{x}\right)\)
\(=\sin \mathrm{x}\left(\frac{1}{2}-\frac{2}{3} \sin ^{2} \mathrm{x}\right)\)
\(\mathrm{f}(\mathrm{x}) =\frac{1}{2} \sin \mathrm{x}-\frac{2}{3} \sin ^{3} \mathrm{x}\)
\(\mathrm{f}(\mathrm{x}) =\frac{1}{6}\left(3 \sin \mathrm{x}-4 \sin ^{3} \mathrm{x}\right) \)
\(\mathrm{f}(\mathrm{x}) =\frac{\sin 3 \mathrm{x}}{6} \quad\left[\because \sin 3 \mathrm{x}=3 \sin \mathrm{x}-4 \sin ^{3} \mathrm{x}\right]\)
Differentiating on both sides w.r.t. ' \(x\) ', we get
\(\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\sin 3 \mathrm{x}}{6}\right)=\frac{1}{6} \cdot \cos 3 \mathrm{x} \cdot 3\)
\(\mathrm{f}^{\prime}(\mathrm{x})=\frac{\cos 3 \mathrm{x}}{2}\)
Let \(\quad I=\int_{0}^{\pi / 2}\left(f(x)+f^{\prime}(x)\right) d x\)
\(I=\int_{0}^{\pi / 2}\left(\frac{\sin 3 x}{6}+\frac{\cos 3 x}{2}\right) d x \tag{i}\)
\(I=\int_{0}^{\pi / 2}\left[\frac{\sin 3\left(\frac{\pi}{2}-x\right)}{6}+\frac{\cos 3\left(\frac{\pi}{2}-x\right)}{2}\right] d x\)
\(I=\int_{0}^{\pi / 2}\left[\frac{\sin \left(\frac{3 \pi}{2}-3 x\right)}{6}+\frac{\cos \left(\frac{3 \pi}{2}-3 x\right)}{2}\right] d x\)
\(I=\int_{0}^{\pi / 2}\left(-\frac{\cos 3 x}{6}-\frac{\sin 3 x}{2}\right) d x \tag{ii}\)
Adding equation (i) and (ii), we get
\(2 \mathrm{I} =\int_{0}^{\pi / 2}\left(\frac{\cos 3 \mathrm{x}}{2}-\frac{\cos 3 \mathrm{x}}{6}+\frac{\sin 3 \mathrm{x}}{6}-\frac{\sin 3 \mathrm{x}}{2}\right) \mathrm{dx}\)
\(2 \mathrm{I} =\frac{1}{3} \int_{0}^{\pi / 2}(\cos 3 \mathrm{x}-\sin 3 \mathrm{x}) \mathrm{dx}\)
\(\text { or } \quad 2 \mathrm{I} =\frac{1}{3}\left[\frac{\sin 3 \mathrm{x}}{3}+\frac{\cos 3 \mathrm{x}}{3}\right]_{0}^{\pi / 2}\)
\(=\frac{1}{9}\left[\sin 3 \times \frac{\pi}{2}+\cos 3 \times \frac{\pi}{2}-\sin 3 \times 0-\cos 3 \times 0\right]\)
\(\mathrm{I} =\frac{1}{9}[-1+0-0-1]=-\frac{2}{9}\)
Or \(\quad \mathrm{I}=-\frac{1}{9}\)

TS EAMCET-2020-11.09.2020

Matrix and Determinant

79239 If \(f(x)=\left|\begin{array}{lll}\cos x & x & 1 \\ 2 \sin x & x^{2} & 2 x \\ \tan x & x & 1\end{array}\right|\), then the value of \(f^{\prime}(x)\) at \(x=0\) is equal to

1 -1

2 1

3 2

4 0

AP EAMCET-2020-22.09.2020

Matrix and Determinant

79240 If \(f(x)=\left|\begin{array}{ccc}\cos x & 1 & 0 \\ 1 & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x\end{array}\right|\), then \(\int_{0}^{\pi / 2} f(x) d x\) is equal to

1 \(\frac{1}{4}\)

2 \(-\frac{1}{3}\)

3 \(\frac{1}{2}\)

4 1

BITSAT-2005

Matrix and Determinant

79245 If \(f(x)=\left|\begin{array}{ccc}1+\sin x+\sin 2 x+\sin 3 x & \frac{3+\sin 2 x}{2} & \frac{-2+\sin 3 x}{3} \\ 3+4 \sin x & \frac{3}{2} & \frac{4}{3} \sin x \\ 1+\sin x & \frac{1}{2} \sin x & \frac{1}{3}\end{array}\right|\) then \(\int_0^{\pi / 2}\left(f(x)+f^{\prime}(x)\right) d x=\)

1 \(\frac{-1}{6}\)

2 \(\frac{-1}{9}\)

3 \(\frac{-2}{9}\)

4 \(\frac{1}{27}\)

Explanation:

(B) : Given that,
\(f(x)=\left|\begin{array}{ccc}1+\sin x+\sin 2 x+\sin 3 x & \frac{3+\sin 2 x}{2} & \frac{-2+\sin 3 x}{3} \\ 3+4 \sin x & \frac{3}{2} & \frac{4}{3} \sin x \\ 1+\sin x & \frac{1}{2} \sin x & \frac{1}{3}\end{array}\right|\)
Applying \(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-2 \mathrm{C}_{2}-3 \mathrm{C}_{3}\)
\(\mathrm{f}(\mathrm{x})=\left|\begin{array}{ccc} \sin \mathrm{x} & \frac{3+\sin 2 \mathrm{x}}{2} & \frac{-2+\sin 3 \mathrm{x}}{3} \\ 0 & \frac{3}{2} & \frac{4}{3} \sin \mathrm{x} \\ 0 & \frac{1}{2} \sin \mathrm{x} & \frac{1}{3} \end{array}\right|\)
On expanding, we get
\(\mathrm{f}(\mathrm{x}) =\sin \mathrm{x}\left|\begin{array}{cc} \frac{3}{2} & \frac{4}{3} \sin \mathrm{x} \\ \frac{1}{2} \sin \mathrm{x} & \frac{1}{3} \end{array}\right|+0+0\)
\(=\sin \mathrm{x}\left(\frac{3}{2} \times \frac{1}{3}-\frac{4}{3} \sin \mathrm{x} \times \frac{1}{2} \sin \mathrm{x}\right)\)
\(=\sin \mathrm{x}\left(\frac{1}{2}-\frac{2}{3} \sin ^{2} \mathrm{x}\right)\)
\(\mathrm{f}(\mathrm{x}) =\frac{1}{2} \sin \mathrm{x}-\frac{2}{3} \sin ^{3} \mathrm{x}\)
\(\mathrm{f}(\mathrm{x}) =\frac{1}{6}\left(3 \sin \mathrm{x}-4 \sin ^{3} \mathrm{x}\right) \)
\(\mathrm{f}(\mathrm{x}) =\frac{\sin 3 \mathrm{x}}{6} \quad\left[\because \sin 3 \mathrm{x}=3 \sin \mathrm{x}-4 \sin ^{3} \mathrm{x}\right]\)
Differentiating on both sides w.r.t. ' \(x\) ', we get
\(\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\sin 3 \mathrm{x}}{6}\right)=\frac{1}{6} \cdot \cos 3 \mathrm{x} \cdot 3\)
\(\mathrm{f}^{\prime}(\mathrm{x})=\frac{\cos 3 \mathrm{x}}{2}\)
Let \(\quad I=\int_{0}^{\pi / 2}\left(f(x)+f^{\prime}(x)\right) d x\)
\(I=\int_{0}^{\pi / 2}\left(\frac{\sin 3 x}{6}+\frac{\cos 3 x}{2}\right) d x \tag{i}\)
\(I=\int_{0}^{\pi / 2}\left[\frac{\sin 3\left(\frac{\pi}{2}-x\right)}{6}+\frac{\cos 3\left(\frac{\pi}{2}-x\right)}{2}\right] d x\)
\(I=\int_{0}^{\pi / 2}\left[\frac{\sin \left(\frac{3 \pi}{2}-3 x\right)}{6}+\frac{\cos \left(\frac{3 \pi}{2}-3 x\right)}{2}\right] d x\)
\(I=\int_{0}^{\pi / 2}\left(-\frac{\cos 3 x}{6}-\frac{\sin 3 x}{2}\right) d x \tag{ii}\)
Adding equation (i) and (ii), we get
\(2 \mathrm{I} =\int_{0}^{\pi / 2}\left(\frac{\cos 3 \mathrm{x}}{2}-\frac{\cos 3 \mathrm{x}}{6}+\frac{\sin 3 \mathrm{x}}{6}-\frac{\sin 3 \mathrm{x}}{2}\right) \mathrm{dx}\)
\(2 \mathrm{I} =\frac{1}{3} \int_{0}^{\pi / 2}(\cos 3 \mathrm{x}-\sin 3 \mathrm{x}) \mathrm{dx}\)
\(\text { or } \quad 2 \mathrm{I} =\frac{1}{3}\left[\frac{\sin 3 \mathrm{x}}{3}+\frac{\cos 3 \mathrm{x}}{3}\right]_{0}^{\pi / 2}\)
\(=\frac{1}{9}\left[\sin 3 \times \frac{\pi}{2}+\cos 3 \times \frac{\pi}{2}-\sin 3 \times 0-\cos 3 \times 0\right]\)
\(\mathrm{I} =\frac{1}{9}[-1+0-0-1]=-\frac{2}{9}\)
Or \(\quad \mathrm{I}=-\frac{1}{9}\)

TS EAMCET-2020-11.09.2020

Matrix and Determinant

79239 If \(f(x)=\left|\begin{array}{lll}\cos x & x & 1 \\ 2 \sin x & x^{2} & 2 x \\ \tan x & x & 1\end{array}\right|\), then the value of \(f^{\prime}(x)\) at \(x=0\) is equal to

1 -1

2 1

3 2

4 0

AP EAMCET-2020-22.09.2020

Matrix and Determinant

79240 If \(f(x)=\left|\begin{array}{ccc}\cos x & 1 & 0 \\ 1 & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x\end{array}\right|\), then \(\int_{0}^{\pi / 2} f(x) d x\) is equal to

1 \(\frac{1}{4}\)

2 \(-\frac{1}{3}\)

3 \(\frac{1}{2}\)

4 1

BITSAT-2005