Mathematics Test - 25

If \(f: \mathrm{R} \rightarrow \mathrm{R}\) be given by \(f(x)=\left(3-x^{3}\right)^{\frac{1}{3}},\)

\(f(x)=\left(3-x^{3}\right)^{\frac{1}{3}}\)

\(\therefore f o f(x)=f(f(x))=f\left(\left(3-x^{3}\right)^{\frac{1}{3}}\right)=\left[3-\left(\left(3-x^{3}\right)^{\frac{1}{3}}\right)^{3}\right]^{\frac{1}{3}}\)

\(=\left[3-\left(3-x^{3}\right)\right]^{\frac{1}{3}}=\left(x^{3}\right)^{\frac{1}{3}}=x\)

\(f o f(x)=x\)

Given: \({ }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}}=2760,{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=23\)

To find: \(\mathrm{r}=\) ?

We know, \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{{ }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}}}{\mathrm{r} !}\)

\(\Rightarrow 23=\frac{2760}{\mathrm{r} !}\)

\(\Rightarrow \mathrm{r} !=\frac{2760}{23}=120\)

\(\Rightarrow \mathrm{r} !=5 \times 24=5 \times 4 \times 6\)

\(=5 \times 4 \times 3 \times 2 \times 1\)

\(\therefore \mathrm{r}=5\)

Given: \(\lim _{x \rightarrow-\infty}\left[\frac{x^{4} \sin \left(\frac{1}{x}\right)+x^{2}}{\left(1+|x|^{3}\right)}\right]\).

We know, when \(x\) tends to \(-\infty, \frac{1}{x}\) tends to \(0\).

We also know, we can write \(x^{4}\) as \(\frac{x^{3}}{\frac{1}{x}} \ldots \) (i) and \(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\ldots\)(ii).

Now, let the value of the limit be \(\mathrm{L}\).

So, \(L=\lim _{x \rightarrow-\infty}\left[\frac{x^{4} \sin \left(\frac{1}{x}\right)+x^{2}}{\left(1+|x|^{3}\right)}\right]\)

Now, from (i), we can write \(L=\lim _{x \rightarrow-\infty}\left[\frac{\frac{x^{3}}{\frac{1}{x}} \sin \left(\frac{1}{x}\right)+x^{2}}{\left(1+|x|^{3}\right)}\right]\)

\(\Rightarrow L=\lim _{x \rightarrow-\infty}\left[\frac{x^{3} \frac{\sin \left(\frac{1}{x}\right)}{\frac{1}{x}}+x^{2}}{\left(1+|x|^{3}\right)}\right]\)

Now, from (ii), we know \(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\)

So, we can write \(L=\lim _{x \rightarrow-\infty}\left[\frac{x^{3}+x^{2}}{\left(1+|x|^{3}\right)}\right]\)

Now, as \(x \rightarrow-\infty\), the value of \(x\) is negative.

So, by the definition of modulus function, we can say \(|x|=-x\).

\(\Rightarrow|x|^{3}=-x^{3}\)

\(\Rightarrow 1+|x|^{3}=1-x^{3}\)

So, we get the limit as \(L=\lim _{x \rightarrow-\infty}\left[\frac{x^{3}+x^{2}}{\left(1-x^{3}\right)}\right]\)

Now, the function consists of polynomials.

So, we will take the variable with the highest power, i.e., \(x^3\), common from the numerator as well as the denominator.

So, we get:

\(L=\lim _{x \rightarrow-\infty}\left[\frac{x^{3}\left(1+\frac{1}{x}\right)}{x^{3}\left(\frac{1}{x^{3}}-1\right)}\right]\)

\(\Rightarrow L=\lim _{x \rightarrow-\infty}\left[\frac{\left(1+\frac{1}{x}\right)}{\left(\frac{1}{x^{3}}-1\right)}\right]\)

Substituting \(x=-\infty\) in the limit, we get:

\(L=\frac{1+\frac{1}{-\infty}}{\frac{1}{-\infty^{3}}-1}\)

\(\Rightarrow L=\frac{1+0}{0-1}\)

\(\Rightarrow L=-1\)

Let, \(z=\frac{1-\mathrm{i}}{1+\mathrm{i}}\)

or \(\mathrm{z}=\frac{1-\mathrm{i}}{1+\mathrm{i}} \times \frac{1-\mathrm{i}}{1-\mathrm{i}}=\frac{1+\mathrm{i}^{2}-2 \mathrm{i}}{1-\mathrm{i}^{2}}\)

⇒ z = -\(\frac{2i}{2}\) = -i

x + iy = -i

so, x = 0 and y = -1

We know that, \(\operatorname{Arg}(z)=\tan ^{-1}\left(\frac{y}{x}\right)\)

\(\operatorname{Arg}(z)=\tan ^{-1}\left(\frac{-1}{0}\right)=-\tan ^{-1}\left(\frac{1}{0}\right)\)

\(\operatorname{Arg}(z)=-\tan ^{-1}(\infty)=-\frac{\pi}{2}\)

Since, equation of plane through intersection of planes

\(x+y+z=1\)      .......... (1)

\(2 x+3 y-z+4=0\)         ........... (2)

Equation of the plane is:

\(\left(A_{1} x+B_{1} y+C_{1} z-d_{1}\right)+\lambda\left(A_{2} x+B_{2} y+C_{2} z-d_{2}\right)=0\)       .......... (3)

Using equation (1) and (2) in the equation (3)

\(\Rightarrow(x+y+z-1)+\lambda(2 x+3 y-z+4)=0\)

\(\Rightarrow(x+y+z-1)+(2 \lambda x+3 \lambda y-\lambda z+4 \lambda)=0\)

\(\Rightarrow(x+2 \lambda x)+(y+3 \lambda y)+(z-\lambda z)+(-1+4 \lambda)=0\)

\((1+2 \lambda) x+(1+3 \lambda) y+(1-\lambda) z+(-1+4 \lambda)=0\) ........ (4)

But the above plane is parallel to \(y\) -axis then its normal plane is perpendicular to y-axis.

\(\Rightarrow((1+2 \lambda) \hat{\mathbf{i}}+(1+3 \lambda) \hat{\mathbf{j}}+(1-\lambda) \hat{\mathrm{k}}) \cdot(0 \hat{\mathbf{i}}+\hat{\mathbf{j}}+0 \hat{\mathbf{k}})=0\)

\(\Rightarrow[(1+2 \lambda) \times 0]+[(1+3 \lambda) \times 1]+[(1-\lambda) \times 0]=0\)

\(\Rightarrow 1+3 \lambda=0\)

\(\Rightarrow 1=-3 \lambda\)

\(\therefore \lambda=-\frac{1}{3}\)

By substituting \(\lambda=-3\) in equation (4),

\(\Rightarrow\left(1+2\left(-\frac{1}{3}\right)\right) \mathrm{x}+\left(1+3\left(-\frac{1}{3}\right)\right) \mathrm{y}+\left(1-\left(-\frac{1}{3}\right)\right) \mathrm{z}+\left(-1+4\left(-\frac{1}{3}\right)\right)=0\)

\(\Rightarrow\left(1-\frac{2}{3}\right) \mathrm{x}+(1-1) \mathrm{y}+\left(1+\frac{1}{3}\right) \mathrm{z}+\left(-1-\frac{4}{3}\right)=0\)

\(\Rightarrow\left(\frac{3-2}{3}\right) \mathrm{x}+0 \mathrm{y}+\left(\frac{3+1}{3}\right) \mathrm{z}+\left(\frac{-3-4}{3}\right)=0\)

\(\Rightarrow\left(\frac{1}{3}\right) \mathrm{x}+\left(\frac{4}{3}\right) \mathrm{z}-\left(\frac{7}{3}\right)=0\)

So, the equation of required plane is

\(\therefore \mathrm{x}+4 \mathrm{z}-7=0\)

Now, we need to substitute the points given in the option to find which point is satisfying the above equation. The point which satisfies the equation is the point through which the plane is passing.

Option (A):

(-3, 0, -1) ⇒ -3 + 4(-1) – 7 ≠ 0

Option (B):

(-3, 1, 1) ⇒ -3 + 4(1) – 7 ≠ 0

Option (C):

(3, 3, -1) ⇒ 3 + 4(-1) – 7 ≠ 0

Option (D):

(3, 2, 1) ⇒ 3 + 4(1) – 7 = 0

Therefore, the plane passes through the point (3, 2, 1).

The dot product of normal and the given vector will be zero.

1. \((\hat{\imath}+\hat{\jmath}+\hat{k}) \cdot(\hat{\imath}+\hat{\jmath}-\hat{k})=1+1-1 \neq 0\)

2. \((\hat{\imath}+\hat{\jmath}+\hat{k}) \cdot(\hat{\imath}-\hat{\jmath}+\hat{k})=1-1+1 \neq 0\)

3. \((\hat{\imath}+\hat{\jmath}+\hat{k}) \cdot(\hat{i}-\hat{\jmath}-\hat{k})=1-1-1 \neq 0\)

1. Both variance and standard deviation are measures of variability in the population.

This statement is correct.

2.The standard deviation is the square root of the variance.

So, the given statement is not correct.

From the top of a lighthouse, \(70 \) m high with its base at sea level, the angle of depression of a boat is \(15^{\circ}\)

Consider the base is "b'.

\(\tan \theta=\frac{70}{x}\)

\(\tan 15^{\circ}=\frac{70}{x}\) ..(1)

\( \tan \left(45^{\circ}-30^{\circ}\right)=\frac{70}{x}\)

We know that,

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

\(\tan \left(45^{\circ}-30^{\circ}\right)=\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \cdot \tan 30^{\circ}}\)

\(\tan \left(15^{\circ}\right)=\frac{1-\frac{1}{\sqrt{3}}}{1+1. \frac{1}{\sqrt{3}}}\)

\(\tan \left(15^{\circ}\right)=\frac{\sqrt{3}-1}{\sqrt{3}+1}\)

By Rationalisation, we get

\( \tan 15^{\circ}=2-\sqrt{3}\)

From equation (1), we get

\( \frac{70}{x}=2-\sqrt{3}\)

\(\mathrm{x}=\frac{70}{2-\sqrt{3}}\)

\( \mathrm{x}=\frac{70}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}}\)

\(x=70(2+\sqrt{3})\) m

Given, 

\(m\left[\begin{array}{ll}-3 & 4\end{array}\right]+n\left[\begin{array}{ll}4 & -3\end{array}\right]=\left[\begin{array}{ll}10 & -11\end{array}\right]\)

\(\Rightarrow\left[\begin{array}{lll}-3 & m+4 n & 4 m-3 n\end{array}\right]=\left[\begin{array}{ll}10 & -11\end{array}\right]\)

By equating above matrices, we get

\(-3 m+4 n=10 \Rightarrow 12 m-16 n=-40 \ldots \ldots\)...(i)

\(4 m-3 n=-11 \Rightarrow 12 m-9 n=-33 \ldots \ldots\) (ii)

Solving equation (i) and (ii) , we get,

\(-7 n=-7\)

\(\therefore n=1\) and \(m=-2\)

Given:

Number of shake hands \(= 780\)

Let the total number of students is \(y\)

\(\Rightarrow{ }^{y} C_{2}=780\)

\(\Rightarrow \frac{y !}{2 !(y-2) !}=780\)

\(\Rightarrow \frac{\mathrm{y}(\mathrm{y}-1)}{2}=780\)

\(\Rightarrow \mathrm{y}(\mathrm{y}-1)=1560\)

\(\Rightarrow \mathrm{y}^{2}-\mathrm{y}-1560=0\)

Using discriminant method:

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

\(=(-1)^{2}-4(1)(-1560)\)

\(=1+6240\)

\(=6241\)

\(X=\frac{(-b+\sqrt{D})}{2 a}\)

\(=\{\frac{-(-1)+\sqrt{6241}} {2}\}\)

\(=\frac{(1+79)}2\)

\(=\frac{80}2\)

\(=40\)

\(\therefore\) Number of students is 40.