Mathematics Test - 18

It is given that

\(\frac{d}{d x}\left(\frac{x^{4}+x^{2}+1}{x^{2}-x+1}\right)=a x+b\) ----(1)

Now,

\(\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)=\left(x^{2}+1\right)^{2}-x^{2} \quad\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]\)

\(=x^{4}+2 x^{2}+1-x^{2}=x^{4}+x^{2}+1\)

So, factor of \(x^{4}+x^{2}+1\) is \(\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)\)

\(\mathrm{LHS}=\frac{d}{dx}\left(\frac{x^{4}+^{2}+1}{x^{2}-x+1}\right)\)

\(=\frac{d}{dx}\left[\frac{\left(x^{2}+x+1\right) \times\left(x^{2}-x+1\right)}{\left(x^{2}-x+1\right)}\right]=\frac{d}{dx}\left(x^{2}+x+1\right)=2 x+1\)

Compare with equation \(1^{\text {st }}\)

∴ a = 2 and b = 1

Given:

\(\vec{F}=5 \hat{i}+3 \hat{j}+2 \hat{k}\) displaces a particle by \(\vec{S}=3 \hat{i}-\hat{j}+2 \hat{k}\)

As we know that, \(W=\) Force \(\cdot\) Displacement \(=\vec{F}. \vec{S}\)

\( W=\vec{F} \cdot \vec{S}=(5 \hat{i}+3 \hat{j}+2 \hat{k}) .(3 \hat{i}-\hat{j}+2 \hat{k})\)

\( W=15-3+4=16\) units

The number of ways in which r distinct objects can be selected from a group of \(n\) distinct objects, is: \({ }^{n} C_{r}=\frac{n !}{r !(n-r) !}\)

\(\mathrm{n} !=1 \times 2 \times 3 \times \ldots \times \mathrm{n}\)

\(0 !=1\)

There are a total of 4 ingredients. A salad can be made by using any combination of all or some of these.

The required number of possibilities is:

(Number of salads with 1 ingredient) \(+\) (Number of salads with 2 ingredients) \(+(\)Number of salads with 3 ingredients) \(+(\)Numer of salads with 4 ingredients)

\(={ }^{4} C_{1}+{ }^{4} C_{2}+{ }^{4} C_{3}+{ }^{4} C_{4}\)

\(=\frac{4 !}{1 !(4-1) !}+\frac{4 !}{2 !(4-2) !}+\frac{4 !}{3 !(4-3) !}+\frac{4 !}{4 !(4-4) !}\)

\(=4+6+4+1\)

\(=15\)

If a line \(y=m x+c .\) (1)

Touches the ellipse

\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { then } \)

\(c^{2}=a^{2} m^{2}+b^{2} \ldots(2)\)

Given line \(x \cos \alpha+y \sin \alpha=p\)

Let,

\(x \cos \alpha+y \sin \alpha=0 \)

\(\Rightarrow y \sin \alpha=-x \cos \alpha+b \ldots(3) \)

\(\Rightarrow y=-x \frac{\cos \alpha}{\sin \alpha}+\frac{b}{\sin \alpha}\)

On comparing with (1), we get

\(m=-\frac{\cos \alpha}{\sin \alpha} \text { and } c=\frac{b}{\sin \alpha}\)

Then from equation (2)

\(c^{2}=a^{2} m^{2}+b^{2} \)

\(\Rightarrow \frac{b^{2}}{\sin ^{2} \alpha}=a^{2}\left(\frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}\right)+b^{2} \)

\(\Rightarrow b^{2}=a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha\)

Comparing it with given line, we get

\(p^{2}=a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha\)

Given:

\(|\overrightarrow{\mathrm{a}}|=\alpha,|\overrightarrow{\mathrm{b}}|=\alpha\) and \(|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=\alpha\)

As we know,

\(|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{ab} \cos \theta}\)

\( \alpha=\sqrt{\alpha^{2}+\alpha^{2}+2(\alpha)(\alpha) \cos \theta}\)

\( \alpha^{2}=2 \alpha^{2}+2 \alpha^{2} \cos \theta\)

\(-1=2 \cos \theta\)

\( \cos \theta=-\frac{1}{2}\)

Now,

\(|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}-2 \mathrm{ab} \cos \theta}\)

\(|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{\alpha^{2}+\alpha^{2}-2(\alpha)(\alpha) \cos \theta}\)

\(\because \cos \theta=-\frac{1}{2}\)

\(|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{2 \alpha^{2}-2 \alpha^{2}\left(\frac{-1}{2}\right)}\)

\(|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{2 \alpha^{2}+\alpha^{2}}\)

\(|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=\sqrt{\mathrm{3}} {\alpha}\)

We have to find the value of \(\frac{\mathrm{dy}}{\mathrm{dx}}\),

It is given that \(y=\tan ^{-1}\left(\frac{3-2 \tan \sqrt{x}}{2+3 \tan \sqrt{x}}\right)\)

\(\Rightarrow y=\tan ^{-1}\left[\frac{2\left(\frac{3}{2}-\tan \sqrt{x}\right)}{2\left(1+\frac{3}{2} \tan \sqrt{x}\right)}\right]=\tan ^{-1}\left[\frac{\left(\frac{3}{2}-\tan \sqrt{x}\right)}{\left(1+\frac{3}{2} \tan \sqrt{x}\right)}\right]=\tan ^{-1}\left(\frac{3}{2}\right)-\tan ^{-1} \tan \sqrt{x}\)

\(\Rightarrow y=\tan ^{-1}\frac{3}{2}-\sqrt{x} \quad\left(\because \tan ^{-1}(\tan x)=x\right)\)

Differentiating, both sides, we get

\(\Rightarrow \frac{d y}{d x}=0-\frac{1}{2 \sqrt{x}}=\frac{-1}{2 \sqrt{x}}\)

We know that:

Suppose a set of \(n\) objects has \(n_{1}\) of one kind of object, \(n_{2}\) of a second kind, \(n_{3}\) of a third kind, and so, on with \(n=n_{1}+n_{2}+n_{3}+\ldots+n_{k}\), then the number of distinguishable permutations of the \(n\) objects is:

\(=\frac{n !}{n_{1} ! \times n_{2} ! \times n_{3} ! \ldots \ldots . . n_{k} !}\)

If there are m ways to choose an object one and n ways to choose an object two then the number of ways of selecting objects one and two are given by m × n.

If there are m ways to choose an object one and n ways to choose an object two then the number of ways of selecting objects one or two is given by m + n.

Given:

Total no. of toys = 5

Total no. of children = 3

Each should get one toy.

Selection can be done as follows: (2, 2, 1) or (1, 1, 3)

\(={ }^{5} {C}_{2} \times{ }^{3} {C}_{2} \times{ }^{1} {C}_{1} \times \frac{3 !}{2 !}+{ }^{5} {C}_{1} \times{ }^{4} {C}_{1} \times{ }^{3} {C}_{3} \times \frac{3 !}{2 !}\)

\(=(10 \times 3 \times 1 \times 3)+(5 \times 4 \times 1 \times 3)\)

\(=90+60\)

\(=150\)

It is given that,

f(x) = cos x – x

Differentiating with respect to x, we get

⇒ f’(x) = -sin x – 1

As we know that,

-1 ≤ sin x ≤ 1, for x ∈ R

⇒ -1 ≤ -sin x ≤ 1, for x ∈ R

⇒ -1 - 1 ≤ -sin x - 1 ≤ 1 - 1, for x ∈ R

⇒ -2 ≤ f’(x) ≤ 0, for x ∈ R

∴ f’(x) ≤ 0, for x ∈ R

Hence f(x) is decreasing in x ∈ R or x ∈ (-∞,∞).