Mathematics Test - 22

Given:

\(\left\{(x, y): 0 \leq x \leq \frac{9}{4}, \quad 0 \leq y \leq 1, \quad x \geq 3 y, \quad x+y \geq 2\right\}\)

Rough sketch of required region is:

\(\therefore\) Required area is:

Area of \(\triangle A C D+\) Area of \(\triangle A B C\)

i.e., \(\frac{1}{4}+\frac{3}{32}=\frac{11}{32}\) sq. units

Let the coordinates of \(Q\) be \((0, a)\).

\(\therefore\) Angle which \(P Q\) makes with \(x\)-axis is same as angle which \(QR\) makes with (x)-axis.

\(\frac{3-a}{2-0}=-\frac{10-a}{5-0}\)

\(\Rightarrow 5(3-a)=-2(10-a)\)

\(\Rightarrow 15-5 a=-20+2 a\)

\(\Rightarrow 7 a=35\)

\(\Rightarrow a=5\)

Thus, coordinates of \(Q\) are \((0,5)\).

Given,

\(5+\left|2^x-1\right|=2^x\left(2^x-2\right)\)

Let \(2^x-1=t\)

Then \(2^x=t+1\)

\(5+|t|=(t+1)(t+1-2)\)

\(\Rightarrow 5+|t|=(t+1)(t-1)\)

\(\Rightarrow 5+|t|=t^2-1\)

\(\Rightarrow|t|=t^2-1-5\)

\(\Rightarrow|t|=t^2-6\)

When \(t>0, t^2-t-6=0\)

\(\Rightarrow t=3\) or \(-2\)

\(t=-2(\) rejected \()\)

When \(t<0, t^2+t-6=0\)

\(\Rightarrow t=-3\) or \(2(\) both rejected \()\)

\(\therefore 2^x-1=3\)

\(\Rightarrow 2^x=4\)

\(\Rightarrow 2^x=2^2\)

\(\Rightarrow x=2\)

Thus, there is only one number of real roots of the equation \(5+\left|2^x-1\right|=2^x\left(2^x-2\right)\)

Given,

\(2 x+y-2 z+6=0\) and \(4 x+2 y-4 z-6=0\) are two planes.

Here, we can rewrite the equation of plane \(2 x+y-2 z+6=0\) as \(4 x+2 y-4 z+12=0\) by multiplying both the sides of \(2 x+y-2 z+6=0\) with \(2\).

As we can see that, the plane \(4 x+2 y-4 z+12=0\) and \(4 x+2 y-4 z-6=0\) are parallel planes.

As we know that, the distance between two parallel planes \(a x+b y+c z+d_1=0\) and ax \(+\) by \(+\mathrm{cz}+\mathrm{d}_2=0\) is given by: \(D=\left|\frac{d_1-d_2}{\sqrt{a^2+b^2+c^2}}\right|\)

Here, \(a=4, b=2, c=-4, d_1=12\) and \(d_2=-6\).

\(\Rightarrow D=\left|\frac{d_1-d_2}{\sqrt{a^2+b^2+c^2}}\right|\)

\(\Rightarrow D=\left|\frac{12-(-6)}{\sqrt{4^2+2^2+(-4)^2}}\right|\)

\(\Rightarrow D=\left|\frac{18}{\sqrt{16+4+16}}\right|\)

\(\Rightarrow D=\left|\frac{18}{6}\right|=3\)

Concept:

If f(x) isevenfunction then f(-x) = f(x)

If f(x) isoddfunction then f(-x) = -f(x)

Properties of definite integral,

If \(f(x)\) is even function then \(\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x\)

If \(f ( x )\) is odd function then \(\int_{- a }^{ f } f ( x ) dx =0\)

Let \(I=\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \frac{\sin ^{5} x \cos ^{3} x}{x^{4}} d x\)

And, \(f(x)=\frac{\sin ^{5} x \cos ^{3} x}{x^{4}}\)

Replaced \(x\) by \(-x_{s}\)

\(\Rightarrow f(-x)=\frac{\sin ^{5}(-x) \cos ^{3}(-x)}{(-x)^{4}}\)

As we know \(\sin (-\theta)=-\sin \theta\) and \(\cos (-\theta)=\cos \theta\)

\(=\frac{-\sin ^{5} x \cos ^{3} x}{x^{4}}\)

\(\Rightarrow f(-x)=-f(x)\)

So, \(f(x)\) is odd function

Therefore, \(I=0\).

Given, \(\sqrt{1-{x}^{2}}+\sqrt{1-{y}^{2}}={a}\),

Differentiating both sides,

\(\frac{1}{2 \sqrt{1-x^{2}}} \cdot \frac{d}{d x}\left(1-x^{2}\right)+\frac{1}{2 \sqrt{1-y^{2}}} \cdot \frac{d}{d x}\left(1-y^{2}\right)=\frac{d a}{d x}\)

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

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

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

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

Given:

\(\left(1+4 x+4 x^{2}\right)^{5} \)

As we know,

\((a+b)^{2}=(a^{2}+b^{2}+2ab)\)

\((1+4x+4x^{2})=\left[(1+2 x)^{2}\right]^{5}\)

\(=(1+2 x)^{2\times5}\)

\(=(1+2 x)^{10}\)

For Middle term: \(r=\frac{10}{2}=5\)

As we know,

\({}^{n}C_{r}(p)^{n}(q)^{n-r}\)

After putting the value,

\(p=1, q=2x, r=5\)

Middle Term \(={ }^{10} C_{5}(1)^{10}(2 x)^{5} \)

\(=\frac{10 !}{5 ! 5 !} \times 32 x^{5} \)

\(= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5!\times5!}\)

\(= \frac{10 \times 9 \times 8 \times 7 \times 6}{5\times 4 \times 3 \times 2!}\)

\(=36\times7\times32x^{5}\)

\(=252\times32x^{5}\)

\(=8064 x^{5}\)

\(\therefore\) The coefficient of the middle term in the expansion of \(\left(1+4 x+4 x^{2}\right)^{5}\) is \(8064\).

Given:

\(f: R \rightarrow R\) be defined as \(f( x )=2 x -1\) and \(g : R -\{1\} \rightarrow R\) be defined as \(g ( x )=\frac{ x -\frac{1}{2}}{ x -1}\).

\(f( g ( x ))=2 g ( x )-1=2\left(\frac{2 x -1}{2( x -1)}\right)-1\)

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

Range of \(f( g ( x )= R -\{1\}\)

Range of \(f( g ( x ))\) is not onto and \(f( g ( x ))\) is one-one.

So, \(f( g ( x ))\) is one-one but not onto.