Mathematics Test - 72

Given:

\(\underset{{{x \rightarrow 0}}}{\lim}\frac{-3 x^{2}-7 x+8}{7 x^{2}+2 x+2}=k\)

As we know that, if \(\underset{{{x \rightarrow 0}}}{\lim} f(x)\) does not result into indeterminate form, then we use direct substitution in order to find the limits.

Here, also we can see that \(\underset{{{x \rightarrow 0}}}{\lim}\frac{-3 x^{2}-7 x+8}{7 x^{2}+2 x+2}\) does not result into any indeterminate form.

So, we can substitute \(x=0\) in the expression \(\frac{-3 x^{2}-7 x+8}{7 x^{2}+2 x+2}\) in order to find the value of \(k\).

\(\Rightarrow\underset{{{x \rightarrow 0}}}{\lim}\frac{-3 x^{2}-7 x+8}{7 x^{2}+2 x+2}=k\)

\(\Rightarrow \frac{-3 (0)^{2}-7 (0)+8}{7 (0)^{2}+2 (0)+2}=k\)

\(\Rightarrow \frac{8}{2}=k\)

\(\Rightarrow 4=k\)

or, \(k=4\)

Equality of complex numbers.

Two complex numbers \(z_{1}=x_{1}+i y_{1}\) and \(z_{2}=x_{2}+i y_{2}\) are equal if and only if \(x_{1}=x_{2}\) and \(y_{1}=y_{2}\)

Or \(\operatorname{Re}\left(z_{1}\right)=\operatorname{Re}\left(z_{2}\right)\) and \(\operatorname{Im}\left(z_{1}\right)=\operatorname{Im}\left(z_{2}\right)\).

\(\Rightarrow z=\frac{1-i}{i}\)

Multiplying numerator and denominator by \(\hat{j}\)

\(\Rightarrow z=\frac{1-i}{i} \times \frac{i}{i}\)

\(=\frac{i-i^{2}}{i^{2}}\)

\(=\frac{i+1}{-1}\)

\(=-1-i\)

\(\operatorname{Re}(z)=-1\)

\(\operatorname{lm}(z)=-1\)

Given,

\(f: R \rightarrow R\) is defined as \(f(x)=x^{4}\).

Let \(x, y \in R\) such that:

\(f(x)=f(y)\)

Then,

\( x^{4}=y^{4}\)

\( x=\pm y\)

\(\therefore f\left(x_{1}\right)=f\left(x_{2}\right)\) does not imply that \(x_{1}=x_{2}\)

For instance \(f(1)=f(-1)=1\)

\(\therefore f\) is not one-one.

Consider an element \(-2\) in co-domain \(R\).

It is clear that, there does not exist any \(x\) in domain \(R\) such that:

\(f(x)=-2\)

\(\therefore f\) is not onto.

Therefore, function \(f\) is neither one-one nor onto.

Since f(x) is an odd periodic function with period 2.

\(\therefore f(-x)=-f(x)\) and \(f(x+2)=f(x)\)

\(\therefore f(2)=f(0+2)=f(0)\)

and \(f(-2)=f(-2+2)=f(0)\)

Now, \(f(0)=f(-2)=-f(2)=-f(0)\)

\(\Rightarrow 2 f(0)=0\), i.e.,\(f(0)=0\)

\(\therefore f(4)=f(2+2)=f(2)=f(0)=0\)

Thus, \(f(4)=0\)

Given,

The objective function,

\(Z=5 x+7 y\)

The Shaded region is bounded and has coordinate of corner points as \((0,0),(7,0),(3,4)\) and \((0,2)\).

At \( (0,0) \),

\(Z=5 \times 0+7 \times 0=0\)

At \( (7,0) \),

\(Z=5 \times 7+7 \times 0=35\)

At \( (3,4) \),

\(Z=5 \times 3+7 \times 4=43\)

At \( (0,2) \),

\(Z=5 \times 0+7 \times 2=14\)

So, the maximum value of \(\mathrm{Z}\) is \(43\) at \((3,4)\).

Given:  \(f(x)=e^{-2 x}\) and \(g(x)=0\) 

\(\mathrm{x}_{1}=-1\) and \(\mathrm{x}_{2}=1\)

Let us plot a rough graph of given curves and shade the bounded area to be calculated as:

So, area under the curve \(=\left|\int_{x_{1}}^{x_{2}}\{f(x)-g(x)\} d x\right|\)

\(=\left|\int_{-1}^{1}\left\{e^{-2 x}-0\right\} d x\right|\)

\(=\left|-\frac{e^{-2 x}}{2}\right|_{-1}^{1}\)

\(=\frac{-e^{-2}-\left(-\mathrm{e}^{2}\right)}{2}\)

\(=\frac{e^{2}-\mathrm{e}^{-2}}{2}\)

Given,

\(f(x)=x^{3}+2 x^{2}+5 x+2 \cos x\)

\(f^{\prime}(x)=3 x^{2}+4 x+5-2 \cdot \sin x\)

\(=3\left(x+\frac{2}{5}\right)^{2}+\frac{11}{3}-2 \cdot \sin x\)

\(\Rightarrow f^{\prime}(X)>0, \forall x\)

\(f(x)\) is increasing for all \(x \in R\)

Also, \(f(0)=2 \)

\(\Rightarrow f(x)=0\)

So, \(f(x)\) has no solution.

As we know,

The equation of the line that passes through the two points \(P\left(x_{1}, y_{1}, z_{1}\right)\) and \(Q\left(x_{2}, y_{2}, z_{2}\right)\) is given by the formula,

\(\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}\)

The line passing through the points \((1,2,-1)\) and \((3,-1,2)\) is,

\(\frac{x-1}{3-1}=\frac{y-2}{-1-2}=\frac{z+1}{2+1} \)

Let, \(\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+1}{3}=\lambda \)

\(\therefore x=2 \lambda+1, y=-3 \lambda+2 \text { and } z=3 \lambda-1\)

Line meets the yz-plane,

So, \(x =0 \)

\(\Rightarrow 2 \lambda+1=0\)

\(\Rightarrow 2 \lambda=-1\)

\(\Rightarrow \lambda=\frac{-1 }{ 2}\)

Now,

\(y=-3 \lambda+2\)

\(\Rightarrow y=-3 \times \frac{-1}{2}+2\)

\(\Rightarrow y =\frac{3}{2}+2\)

\(\Rightarrow y=\frac{7 }{ 2}\)

Now,

\(z=3 \lambda-1\)

\(\Rightarrow z=3 \times \frac{-1}{2}-1\)

\(\Rightarrow z =\frac{-3}{2}-1\)

\(\Rightarrow z=\frac{-5 }{ 2}\)

\(\therefore\) Point \((x, y, z)\) is \(\left(0, \frac{7}{2},-\frac{5}{2}\right)\).

Given,

\(2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)\)

As we know that, \(2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right),-1 \leq x \leq 1\)

\(2 \tan ^{-1}(\cos x)=\tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^{2} x}\right)\)

As we know that, \(\sin ^{2} x+\cos ^{2} x=1\)

\(\Rightarrow 2 \tan ^{-1}(\cos x)=\tan ^{-1}\left(\frac{2 \cos x}{\sin ^{2} x}\right)\)

\(\Rightarrow \tan ^{-1}\left(\frac{2 \cos x}{\sin ^{2} x}\right)=\tan ^{-1}(2 \operatorname{cosec} x)=\tan ^{-1}(\frac{2}{\sin x})\)

\(\Rightarrow \frac{2 \cos x}{\sin ^{2} x}=\frac{2}{\sin x} \)

\(\Rightarrow \cos x \sin x-\sin ^{2} x=0\)

\(\Rightarrow \sin x(\cos x-\sin x)=0\)

\(\Rightarrow \sin x=0 \text { or } \cos x-\sin x=0 \)

\(\Rightarrow x=0 \text { or } \frac{\pi}{4}\)

As we can see that for \(x=0\) the given equation does not exist Thus, \(x=\frac{\pi }{4}\) is the only solution for the given equation.

The standard equation of the circle is

\(( x - h )^2+( y - k )^2= r ^2\)

Where centre is \(( h , k\) ) and radius is \(r\).

Now,

The family of circle having centre \((0,0)\) and radius \(r\) is

\(x^2+y^2=r^2\)

\(\because\) There is only one constant \(r\).

Differentiating w.r.t. \(x\)

\(\Rightarrow 2 x+2 y \frac{d y}{d x}=0 \)

\(\Rightarrow \frac{d y}{d x}=-\frac{x}{y}\)