Mathematics Test-29

As we know,

Area of \(\triangle \mathrm{OAB}=\frac{1}{2}|\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}|\)

\(=\frac{1}{2}|(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \times(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}})|\)

\(=\frac{1}{2}\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -3 & 2 \\ 2 & 3 & 1\end{array}\right|\)

\(=\frac{1}{2}|[\hat{\mathrm{i}}(-3-6)-\hat{\mathrm{j}}(2-4)+\hat{\mathrm{k}}(6+6)]|\)

\(=\frac{1}{2}|-9 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}|\)

\(\therefore\) Area of \(\triangle \mathrm{OAB}=\frac{1}{2} \sqrt{81+4+144}\)

\(=\frac{1}{2} \sqrt{229}\)

Given:

\(\left(1+x-2 x^{2}\right)^{6}=1+a_{1} x+a_{2} x^{2}+\ldots \ldots+a_{12} x^{12}\)

Put \(\mathrm{x}=1\) in the given equation

\(0=1+\mathrm{a}_{1}+\mathrm{a}_{2}+\mathrm{a}_{3}+\ldots .+\mathrm{a}_{12}\).....(1)

Put \(\mathrm{x}=-1\) in the given equation

\((-2)^{6}=1-\mathrm{a}_{1}+\mathrm{a}_{2}-\mathrm{a}_{3}+\ldots+\mathrm{a}_{12}\).....(2)

Adding (1) and (2), we get

\(0+(-2)^{6}=\left(1+\mathrm{a}_{1}+\mathrm{a}_{2}+\ldots+\mathrm{a}_{12}\right)+\left(1-\mathrm{a}_{1}+\mathrm{a}_{2}-\mathrm{a}_{3} \ldots \ldots+\mathrm{a}_{12}\right) \)

\(\Rightarrow 64=2\left(1+\mathrm{a}_{2}+\mathrm{a}_{4}+\ldots .+\mathrm{a}_{12}\right) \)

\(\Rightarrow 1+\mathrm{a}_{2}+\mathrm{a}_{4}+\ldots \ldots+\mathrm{a}_{12}=32 \)

\(\Rightarrow \mathrm{a}_{2}+\mathrm{a}_{4}+\ldots \ldots+\mathrm{a}_{12}=31\)

Starting from left to right, the person will be omitted standing next to the person shaking hand.

As the person shake hands with \(3\) so starting from left and shaking hands with first person we will omit the next one. Shaking hands with third person, the fourth person will be omitted.

So we have \(4\) persons left after omitting \(2\) and to find distinct possible combinations we will any \(3\) among them for handshakes.

Number of distinct possible handshakes \(=(\frac{4}{3})\)

\(=\frac{4 !}{3 ! \times 1 !}=4\)

Step \(1\): Write the given angles referring to given diagram.

Given \(\angle SRM =60^{\circ}, \angle SQ =50^{\circ}\).

Step \(2\): Use the geometric rules and property to find the value of \(\angle Q S R\).

\(\text { If, } \angle SRM =60^{\circ}\)

\(\Rightarrow \angle SRM +\angle SRO =90^{\circ}\)

\(\therefore \angle SRO =30^{\circ}\)

For \(\triangle OSR\)

\(OS = OR [\) radii of same circle \(]\)

\(\angle OSR =\angle SRO =30^{\circ}\) [angles opposite to equal sides are equal]

As, \(\angle SQL =50^{\circ}\)

\(\Rightarrow \angle SQL +\angle SQO=90^{\circ}\)

\(\therefore \angle SQO =40^{\circ}\)

For \(\triangle OSQ\),

\(O S=O Q[\) radii of same circle \(]\)

\(\angle SQO =\angle OSQ =40^{\circ}\) [angles opposite to equal sides are equal]

\(\therefore \angle QSR =\angle OSR +\angle OSQ\)

\(=30^{\circ}+40^{\circ}\)

\(=70^{\circ}\)

The value of \(\angle Q S R\) is \(70^{\circ}\).

\(\frac{d y}{d x}=\sec \left(\frac{y}{x}\right)+\frac{y}{x}\)

Let \(\frac{ y }{ x }= t\)

\(\Rightarrow y = xt\)

Differentiating with respect to \(x\), we get

\(\Rightarrow \frac{ dy }{ dx }= x \frac{ dt }{ dx }+ t\)

Now,

\( x \frac{ dt }{ dx }+ t =\sec t + t\)

\( \Rightarrow x \frac{ dt }{ dx }=\sec t \)

\( \Rightarrow \frac{ dt }{\sec t }=\frac{ dx }{ x }\)

Integrating both sides, we get

\( \Rightarrow \int \frac{d t}{\sec t}=\int \frac{d x}{x} \)

\(\Rightarrow \int \cos t d t=\int \frac{d x}{x} \)

\(\Rightarrow \sin t=\log x+\log c \)

\( \Rightarrow \sin t=\log (c x) \quad(\because \log m+\log n=\log (m n))\)

\(\therefore \sin \left(\frac{y}{x}\right)=\log (c x)\)

Given: \(f(x)=2 x-x^{2}\)

Now,

\(f(x+2)=2(x+2)-(x+2)^{2}\)

\(f(x-2)=2(x-2)-(x-2)^{2}\)

Adding above equations,

\(f(x+2)+f(x-2)=2(x+2)-(x+2)^{2}+2(x-2)-(x-2)^{2}\)

Put \(x=0\)

\(f(2)+f(-2)=2(0+2)-(0+2)^{2}+2(0-2)-(0-2)^{2}\)

\(=4-4-4-4\)

\(=-8\)

\(\mathrm{A}+\mathrm{B}=90^{\circ}\)

\(\mathrm{A}=90^{\circ}-\mathrm{B}\)

\(\frac{\tan \mathrm{A} \tan \mathrm{B}+\tan \mathrm{A} \cot \mathrm{B}}{\sin \mathrm{A} \sec \mathrm{B}}-\frac{\sin ^2 \mathrm{~B}}{\cos ^2 \mathrm{~A}}\)

\(=\frac{\tan \left(90^{\circ}-\mathrm{B}\right) \tan \mathrm{B}+\tan \left(90^{\circ}-\mathrm{B}\right) \cot \mathrm{B}}{\sin \left(90^{\circ}-\mathrm{B}\right) \sec \mathrm{B}}-\frac{\sin^2B}{\cos^{2}(90^{\circ}-B)}\)

\(=\frac{\cot \mathrm{B} \tan \mathrm{B}+\cot \mathrm{B} \cot \mathrm{B}}{\cos \mathrm{B} \sec \mathrm{B}}-\frac{\sin ^2 \mathrm{~B}}{\sin ^2 \mathrm{~B}}\)

\(=\frac{1+\cot ^2\mathrm{B}}{1}-1=1+\cot ^2 \mathrm{B}-1\)\(=\cot^{2} \mathrm{B}\)

Three non-zero terms a, b, c are in GP if and only if b2 = ac

For a quadratic equation ax² + bx + c = 0,

If α is a root then aα² + bα + c = 0

Given: \(\Delta=\left|\begin{array}{ccc}a & b & a \alpha+b \\ b & c & b \alpha+c \\ a \alpha+b & b \alpha+c & 0\end{array}\right|=0\)

\(C_3 \rightarrow C_3-C_1 a-C_2\)

\(\Rightarrow \Delta=\left|\begin{array}{ccc}a & b & 0 \\ b & c & 0 \\ a \alpha+b & b \alpha+c & -\left(a \alpha^2+b \alpha+b \alpha+c\right)\end{array}\right|=0\)

\(\Rightarrow\left(a a^2+2 b a+c\right)\left(a c-b^2\right)=0 \)

\(\Rightarrow a c-b^2=0 \text { or } a a^2+2 b a+c=0\)

\(a, b, c\) are in GP and \((x-\alpha)\) is the factor \(a x^2+2 b x+c\)

The maximum value of \(\lambda\) is \(2\).

So, \(|\lambda \vec{a}|=|\lambda| \cdot|\vec{a}|\)

\(\Rightarrow|\lambda \vec{a}|=2.4=8\)

The minimum value of \(\lambda\) is \(-3\).

So, \(|\lambda \vec{a}|=|\lambda| \cdot|\vec{a}|\)

\(\Rightarrow|\lambda \vec{a}|=|-3| \cdot 4 \)

\(\Rightarrow|\lambda \vec{a}|=3.4=12\)

So, there are no value of \(\lambda\) which is negative.

For, \(\lambda=0\), we get,

\(|\lambda \vec{ a }|=|\lambda| \cdot|\vec{ a }| \)

\(\Rightarrow|\lambda \vec{ a }|=0.4=0\)

Therefore, the smallest value of \(|\lambda \vec{a}|\) is \(0\).

Therefore, \(|\lambda \vec{a}|\) lies in \([0,12]\).