Mathematics Test - 22

Given:

25 cosec² x + 36 sec² x

As we know that, minimum value of\(m \sec ^{2} \theta+n \operatorname{cosec}^{2} \theta=(\sqrt{m}+\sqrt{n})^{2}\)

Here, m = 36 and n = 25.

∴ The minimum value of 25 cosec² x + 36 sec² x :\((\sqrt{36}+\sqrt{25})^{2}=121\)

Given:

\(P(n): 1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+n}=\frac{2 n}{n+1}\)

For \(\boldsymbol{n}=\mathbf{1}\)

L.H.S \(=1\) and R. H. S. \(=\frac{2 \times 1}{1+1}=\frac{2}{2}=1\)

Thus, \(P(1)\) is true.

Let \(P(n)\) be true for some \(n=k\)

\(1+\frac{1}{1+2}+\ldots+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+k}\) \(=\frac{2 k}{k+1}\)

Now, we have to prove that \(P(n)\) is true \(n=k+1\)

i.e., \(1+\frac{1}{1+2}+\ldots+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+(k+1)}\)

\(\frac{=2(k+1)}{k+2}\)

Adding \(\frac{1}{1+2+3+\ldots+(k+1)}\) on both sides of \((1)\), we get

\(1+\frac{1}{1+2}+\ldots+\frac{1}{1+2+3}+\ldots \frac{1}{1+2+3+\ldots+k}+\frac{1}{1+2+3+\ldots+(k+1)} \)

\(=\frac{2 k}{k+1}+\frac{1}{1+2+3+\ldots+(k+1)} \)

\(=\frac{2 k}{k+1)}+\frac{1}{\frac{(k+1)(k+2)}{2}} \)

\({\left[\text{Using}~ 1+2+3+\ldots+n=\frac{n(n+1)}{2}\right]} \)

\(=\frac{2 k}{(k+1)}+\frac{2}{(k+1)(k+2)} \)

\(=\frac{2}{k+1}\left(\frac{k(k+2)+1}{k+2}\right) \)

\(=\frac{2(k+1)^{2}}{(k+1)(k+2)} \)

\(=\frac{2(k+1)}{(k+2)}\)

Thus, \(P(k+1)\) is true whenever \(P(k)\) is true.

By the principle of mathematical induction, statement \(\mathrm{P}(\mathrm{n})\) is true for all natural numbers.

\(f(-x)=\left[\begin{array}{ccc} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right] \)

\(f(x) \cdot f(-x)=\left[\begin{array}{llll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=I\)

Hence statement-I is correct

Now, checking statement II

\(f(y)=\left[\begin{array}{ccc} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right] \)

\(f(x) \cdot f(y)=\left[\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \cos (x+y) & 0 \\ 0 & 0 & 1 \end{array}\right]\)

\(\Rightarrow \mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y})=\mathrm{f}(\mathrm{x}+\mathrm{y})\)

Hence statement-II is also correct.

Given,

\( 2 l+2 m-n=0  \)...........(i)

\( m n+n l+I m=0\)...........(ii)

From Eq. (i), we get

\(n=2 l+2 m\)...........(iii)

Substituting, \(n=2 l+2 \mathrm{~m}\) in Eq. (ii), we have

\(\mathrm{m}(2 \mid+2 \mathrm{~m})+\mid(2 \mid+2 \mathrm{~m})+\mathrm{Im}=0 \)

\(\Rightarrow 2 \mathrm{Im}+2 \mathrm{~m}^2+2 \mathrm{I}^2+2 \operatorname{Im}+\operatorname{Im}=0 \)

\(\Rightarrow 2 \mathrm{I}^2+4 \mathrm{Im}+\operatorname{Im}+2 \mathrm{~m}^2=0 \)

\(\Rightarrow 2 \mid(\mid+2 \mathrm{~m})+\mathrm{m}(\mid+2 \mathrm{~m})=0 \)

\(\Rightarrow(2 \mid+m)(\mid+2 \mathrm{~m})=0\)

When

\(21=-m\)

From Eq (iii),

\(\mathrm{n}=\mathrm{m} \)

\(\Rightarrow \frac{2 \mathrm{I}}{-1}=\frac{\mathrm{m}}{1}=\frac{\mathrm{n}}{1} \)

\(\Rightarrow \frac{\mathrm{I}}{-\frac{1}{2}}=\frac{\mathrm{m}}{1}=\frac{\mathrm{n}}{1}\)

\( \text { or } \frac{\mathrm{I}}{1}=\frac{\mathrm{m}}{-2}=\frac{\mathrm{n}}{-2} \)

\( \Rightarrow(1, \mathrm{~m}, \mathrm{n})=(1,-2,-2)\)

When

\(I=-2 m\)

From Eq. (iii),

\(n=-2 m \)

\(\Rightarrow I=-2 m=n \)

\(\Rightarrow \frac{I}{-2}=\frac{m}{1}=\frac{n}{-2} \)

\(\Rightarrow(I, m, n)=(-2,1,-2)\)

\(\therefore\) Angles between straight lines

\(\cos \theta=\frac{(\hat{i}-2 \hat{j}-2 \hat{k})(-2 \hat{i}+\hat{j}-2 \hat{k})}{|\hat{i}-2 \hat{j}-2 \hat{k}|-2 \hat{i}+\hat{j}-2 \hat{k} \mid} \)

\(\cos \theta=\frac{-2-2+4}{9}=0 \)

\(\Rightarrow \theta=\frac{\pi}{2}\)

\(\begin{aligned} & p=\frac{k \cot 2 \alpha}{\sqrt{\operatorname{cosec}^2 \alpha+\sec ^2 \alpha}} \\ & \Rightarrow q=\frac{\mathrm{k} \sin 2 \alpha}{\sqrt{\sin ^2 \alpha+\cos ^2 \alpha}} \\ & \Rightarrow p=\frac{\mathrm{k}\left(\frac{\cos 2 \alpha}{\sin 2 \alpha}\right)}{\sqrt{\frac{\sin ^2 \alpha+\cos ^2 \alpha}{\sin ^2 \alpha \cos ^2 \alpha}}}=\frac{\mathrm{k} \cos 2 \alpha}{\sin 2 \alpha} \\ & \Rightarrow p=\left(\frac{k}{2}\right) \cos 2 \alpha \\ & \Rightarrow q=k \sin 2 \alpha \\ & \Rightarrow \cos 2 \alpha=(2 p / k) \\ & \Rightarrow \sin 2 \alpha=(q / k) \\ & \Rightarrow \sin 2 \alpha+\cos ^2 2 \alpha=1 \\ & \Rightarrow \frac{4 p^2}{k^2}+\frac{q^2}{k^2}=1 \\ & \Rightarrow 4 p^2+q^2=k^2\end{aligned}\)

Given that:

Direction cosines of a line are \(\left(\frac{1}{k}, \frac{2}{k},-\frac{2}{k}\right)\)

So, \(l=\frac{1}{k}, m=\frac{2}{k}\) and \(n=-\frac{2}{k}\)

We know that:

Sum of squares of the direction cosines of a line is equal to unity, i.e.,

\(l ^{2}+ m ^{2}+ n ^{2}=1\)

\(\Rightarrow \frac{1}{k^{2}}+\frac{4}{k^{2}}+\frac{4}{k^{2}}=1\)

\(\Rightarrow \frac{9}{k^{2}}=1\)

\(\Rightarrow k ^{2}=9\)

\(\therefore k =\pm 3\)

Given,

\(\tan 54^{\circ}=\tan \left(45^{\circ}+9^{\circ}\right)\)

As we know that,

\(\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}\)

\(\tan 54^{\circ}=\tan \left(45^{\circ}+9^{\circ}\right)\)

\(=\frac{\tan 45^{\circ}+\tan 9^{\circ}}{1-\tan 45^{\circ} \tan 9^{\circ}}\)

\(=\frac{1+\tan 9^{\circ}}{1-\tan 9^{\circ}}\)

\(=\frac{1+\frac{\sin 9^{\circ}}{\cos 9^{\circ}}}{1-\frac{\sin 9^{\circ}}{\cos 9^{\circ}}}\)

\(\tan 54^{\circ}=\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}\)

Geometric progression is a sequence of numbers where each term except the first is calculated by multiplying the previous one by a fixed, non-zero number called the common ratio.

Since sin a, sin b and cos a are in Geometric progression.

\(\therefore \sin ^{2} b=\sin a \cos a\)     .....(i)

(Because we know that when three terms are in geometric progression, the square of the second term is the product of the first and third term.)

Also \(x^{2}+2 x \cot b+1=0\) (given)

We will solve the discriminant of this quadratic equation to find its roots.

Formula: \(D=b^{2}-4 a c\)

On comparison we get,

\(a=1\)

\(b=2 \cot b\)

\(c=1\)

Now, Substituting the values of a, b and c in the formula, we get

\(D=4 \cot ^{2} b-4\)

\(D=4\left[\frac{\cos ^{2} b}{\sin ^{2} b}-1\right]\)

\(\because\left\{\frac{\cos }{\sin }=\cot \right\}\)

\(\Rightarrow D=4\left[\frac{\cos ^{2} b-\sin ^{2} b}{\sin ^{2} b}\right]\)

\(\Rightarrow D=4\left[\frac{1-\sin ^{2} b-\sin ^{2} b}{\sin ^{2} b}\right]\)

\(\because\left\{\cos ^{2} \theta+\sin ^{2} \theta=1\right\}\)

\(\Rightarrow D=4\left[\frac{1-2 \sin ^{2} b}{\sin ^{2} b}\right]\)

\(\Rightarrow D=4\left[\frac{1-2 \sin a \cos a}{\sin ^{2} b}\right]\)

By using equation (i)

\(\Rightarrow D=4\left[\frac{\sin ^{2} a+\cos ^{2} a-2 \sin a \cos a}{\sin ^{2} b}\right]\)

\(\Rightarrow D=4\left[\frac{(\sin a-\cos a)^{2}}{\sin ^{2} b}\right]\)

Since, \((a-b)^{2}=a^{2}+b^{2}-2 a b\)

Above equation can also be written by taking all the terms inside the bracket as:

\(\Rightarrow D=\left[\frac{2(\sin a-\cos a)}{\sin b}\right]^{2}\)

Now that we have the value of \(D\), let us see the nature of roots on the basis of discriminant found. And we know.

(i) \(\mathrm{D}>0\) so the roots of the quadratic equation are real and distinct.

(ii) \(\mathrm{D}=\) Othen the roots of the quadratic equation are real and equal.

(iii) \(\mathrm{D}<0\) then the roots of the quadratic equation are unreal or imaginary.

Also we know that the square of any term (positive or negative) is always positive i.e. always greater than 0, Which means \(D>0 \). Also \(\sin b \neq 0\), otherwise the whole fraction will be undefined.

Thus, The roots of the given equation i.e. \(x^{2}+2 x \cot b+1=0\) are real.

Given,

\(x+i y=\sqrt{\frac{a+i b}{c+i d}}\quad\quad\)...(1)

Calculating \(x+i y\)

Replacing \(i\) by\(-i\).

\(x+i y=\sqrt{\frac{a+i b}{c+i d}}\quad\quad\)....(2)

Multiplying (1) and (2).

\((x-i y)(x+i y)=\sqrt{\frac{a-i b}{c+i d}} \times \sqrt{\frac{a-i b}{c-i d}}\)

\((x-i y)(x+i y)\)

Using \((a-b)(a+b)=a^{2}-b^{2}\)

\(=(x)^{2}-(i y)^{2}\)

\(=x^{2}-(i)^{2} y^{2}\)

\(=x^{2}-(-1) y^{2} \quad \quad\) (As \(\left.i^{2}=-1\right)\)

\(=x^{2}+y^{2}\)

\(x^{2}+y^{2}=\sqrt{\frac{a+i b}{c+i d} \times \frac{a-i b}{c-i d}}\)

\(=\sqrt{\frac{(a+i b)(a-i b)}{(c+i d)(c-i d)}}\)

\(=\sqrt{\frac{(a)^{2}-(i b)^{2}}{(c)^{2}-(i d)^{2}}}\)

\(=\sqrt{\frac{a^{2}-i^{2} b^{2}}{c^{2}-i^{2} d^{2}}}\)

Putting \(i^{2}=-1\)

\(=\sqrt{\frac{a^{2}-(-1) b^{2}}{c^{2}-(-1) d^{2}}}\)

\(=\sqrt{\frac{a^{2}+b^{2}}{c+d^{2}}}\)

Thus, \(x^{2}+y^{2}=\sqrt{\frac{a^{2}+b^{2}}{c^{2}+d^{2}}}\)