Mathematics Test - 15

Given:

\(f(x)=\sin x+\cos x\)

\(f^{\prime}(x)=\frac{d}{d x}(\sin x+\cos x)\)

\(f^{\prime}(x)=\cos x-\sin x\)

Now putting \(f^{\prime}(x)=0\)

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

\(\sin x=\cos x\)

\(\tan x=1\)

\(x=\frac{\pi}{4}\) and \(\frac{5 \pi}{4} \quad\{\because 0 \leq x \leq 2 \pi\}\)

For \(x \in\left(0, \frac{\pi}{4}\right), f^{\prime}(x)>0\) so, \(f(x)\) is increasing.

For \(x \in\left(\frac{\pi}{4}, \frac{5 \pi}{4}\right), f(x)<0\) so, \(f(x)\) is decreasing.

For \({x} \in\left(\frac{5 \pi}{4}, 2 \pi\right), f^{\prime}({x})>0\) so, \(f(x)\) is increasing.

\(\therefore x \in\left(0, \frac{\pi}{4}\right)\) and \(\left(\frac{5 \pi}{4}, 2 \pi\right)\)

\(\frac{d y}{d x}+y \cos x=3 \cos x \)

\( I F=e^{\int \cos x d x} \)

\( \Rightarrow I F=e^{\sin x}\)

Now, \(y \times( IF )=\int Q ( IF ) dx\)

\( \Rightarrow y \times e^{\sin x}=\int 3 \cos x \times e^{\sin x} d x \)

\( \Rightarrow y e^{\sin x}=\int 3 e^{\sin x} \cos x~ d x\)

Integrating, let \(\sin x=t \Rightarrow \cos x d x=d t\)

\( \Rightarrow y e^{\sin x}=\int 3 e^t d t \)

\( \Rightarrow y e^{\sin x}=3 e^t+c \)

\( \Rightarrow y e^{\sin x}=3 e^{\sin x}+c\)

(where \(c\) is integration constant)

The general equation of a sphere is (x - a)²+(y-b)² +(z - c)² = r², where (a, b, c) represents the center of the sphere,'r' represents the radius, and x, y, and z are the coordinates of the points on the surface of the sphere.

Equation of sphere, x² + y²+z² + 2ux + 2vy + 2wz + d = 0 also it passes through (1,0,0), (0,1,0) and (0,0,1).

When the sphere passes through (1,0,0), we get

x² + y² + z² + 2ux + 2vy + 2wz + d = 0

⇒12+0+0+2u +0+0+d=0

⇒ 1+ 2u + d = 0

⇒2u d=-1

When the sphere passes through (0,1,0), we get

x² + y²+z² + 2ux + 2vy + 2wz + d = 0 → 1+2v + d = 0

⇒0+12+0+0+2v+0+d=0

⇒ 2v+d=-1 Similarly, when the sphere passes through (0,0,1), we get

x² + y²+z² + 2ux + 2vy + 2wz + d = 0

⇒ 0+0+1+0+0+2w+ d = 0

⇒1+2w+d=0

⇒2w+ d=−1

Clearly all the equation depends on the diameter, so there can be infinite number of sphere passing through these points.

\(I\) = \(\int \frac{d x}{\sqrt{(x+1)(x+2)}}\)

Let, \(u^2= x + 1\)

\(⇒ 2u du = dx\)

\(I=\int \frac{2 {u}}{{u} \sqrt{\left({u}^{2}+1\right)}} {du}\)

\({I}=2 \int \frac{1}{\sqrt{{u}^{2}+1}} {du}\)

\(I=2 \ln | \sqrt{{u}^{2}+1}+{u} |+{c}\)

\(I=2 \ln |\sqrt{{x}+1+1}+\sqrt{{x}+1}|+{c}\)

\({I}=2 \ln |\sqrt{{x}+2}+\sqrt{{x}+1}|+{c}\)

Given,

\(I=\int\left(e^{x \log a}+e^{a \log x}+e^{a \log a}\right) d x\)

It can be written as;

\(I=\int\left(e^{\log a^{x}}+e^{\log x^{a}}+e^{\log a^{a}}\right) d x\)

We know that \(e^{\log z}=z\)

\(I=\int\left(a^{x}+x^{a}+a^{a}\right) d x\)

\(=\int a^{x} d x+\int x^{a} d x+\int a^{a} d x\)

\(\text { As } \int x^{n} d x=\frac{1}{n+1} x^{n+1}+c\)

\(I=\frac{a^{x}}{\ln a}+\frac{x^{a+1}}{a+1}+a^{a} x+C\)

Given:

\(\mathrm{I}=\int_{0}^{1.5}\left[\mathrm{x}^{2}\right] \mathrm{d} \mathrm{x}\)

\(\mathrm{I}=\int_{0}^{1}\left[\mathrm{x}^{2}\right] \mathrm{d} \mathrm{x}+\int_{1}^{1.5}\left[\mathrm{x}^{2}\right] \mathrm{dx}\)

For \(0 \leq \mathrm{x}<1,[\mathrm{x}]=0\)

For \(1 \leq \mathrm{x}<2,[\mathrm{x}]=1\)

\(\therefore \mathrm{I}=\int_{0}^{1}\left[0^{2}\right] \mathrm{d} \mathrm{x}+\int_{1}^{1.5}\left[1^{2}\right] \mathrm{d} \mathrm{x}\)

\(\mathrm{I}=\int_{1}^{1.5} \mathrm{dx}\)

\(\mathrm{I}=[\mathrm{x}]_{1}^{1.5}\)

\(\mathrm{I}=1.5-1=0.5\)

Given:

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

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

Now, by integrating both the sides we get

\(\Rightarrow \int y^{-2} d y+\int x^{-2} d x=0\)

\(\Rightarrow-\frac{1}{y}-\frac{1}{x}=-C_{1}\)

\(\Rightarrow \frac{1}{y}+\frac{1}{x}=C_{1}\)

\(\Rightarrow x+y=C_{1}(x y)\)

\(\Rightarrow C(x+y)=x y,\) where \(C=\frac{1}{c_{1}}\)

Let \(y=\tan ^{-1} \frac{x}{\sqrt{\left(a^{2}-x^{2}\right)}} \ldots\) (i)

Put \(x=a \sin \theta\)

\(\theta=\sin ^{-1}\left(\frac{x}{a}\right)\)

Then,

\(\tan y=\frac{a \sin \theta}{\sqrt{a^{2}\left(1-\sin ^{2} \theta\right)}}\)

As we know,

\(\cos \theta=\sqrt{1-\sin ^{2} \theta}\)

\(\frac{\sin \theta}{\cos \theta}=\tan \theta\)

\(\frac{d}{d x} \sin ^{-1} x=\frac{1}{\sqrt{a^{2}-x^{2}}}\)

\(\tan y=\frac{a \sin \theta}{a \cos \theta}\)

\(\tan y=\tan \theta\)

\(y=\theta=\sin ^{-1} \frac{x}{a}\)

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

\(=\frac{1}{\sqrt{a^{2}-x^{2}}}\)

Given,

\(3 a x^{2}+2 b x+c=0\)

If \(\alpha, \beta\) are the roots of quadratic equation \(a x^{2}+b x+c=0\), then

Sum of roots \(=\alpha+\beta=\frac{-b}{a}\)

Product of roots \(=\alpha \times \beta=\frac{c}{a}\)

Comparing with standard quadratic equation \(a x^{2}+b x+c=0\), we get \(a=3 a, b=2 b\)

Roots are in the ratio \(3: 2\).

Let, roots are \(3 \alpha\) and \(2 \alpha\).

Sum of roots \(=3 \alpha+2 \alpha=\frac{-2 b}{3 a}\)

\(\Rightarrow 5 \alpha=\frac{-2 b}{3 a}\)

\(\Rightarrow \alpha=\frac{-2 b}{15 a}\)

\(\Rightarrow \alpha^{2}=\frac{4 b^{2}}{15 \times 15 a^{2}}\)

Now, product of roots \(=3 \alpha \times 2 \alpha=\frac{c}{3 a}\)

\(\therefore 6 \alpha^{2}=\frac{c}{3 a}\)

\(\Rightarrow 6\left(\frac{4 b^{2}}{15 \times 15 a^{2}}\right)=\frac{c}{3 a}\)

\(\Rightarrow 6 \times \frac{4 b^{2} \times 3 a}{15 \times 5 a}=c\)

\(\Rightarrow 2 \times 4 b^{2}=25 a c\)

\(\Rightarrow 8 b^{2}=25 a c\)

The given first-order ordinary differential equation is \(\left(1-x^2\right) \frac{d y}{d x}-x y=1\).

Dividing both sides by \(1- x ^2\), we can get it in the standard form.

\( \Rightarrow \frac{ dy }{ dx }+\left(\frac{- x }{1- x ^2}\right) y =\frac{1}{1- x ^2} \)

\(\therefore P =\frac{- x }{1- x ^2} .\)

Let's calculate \(\int Pdx\).

\(\int P d x=\int \frac{-x}{1-x^2} d x\)

Substituting \(1-x^2=t\), so that \(-2 x d x=d t\), we get:

\(\Rightarrow \int Pdx =\frac{1}{2} \int \frac{1}{ t } dt\)

Using \(\int \frac{1}{x} d x=\log x+C\) and ignoring the constant \(C\), we get:

\(\Rightarrow \int Pdx =\frac{1}{2} \log t\)

Back substituting \(1-x^2=t\), we get:

\( \Rightarrow \int Pdx =\frac{1}{2} \log \left(1- x ^2\right) \)

\( \Rightarrow \int Pdx =\log \sqrt{1- x ^2}\)

Now, the integrating factor will be:

\(F = e ^{\int P dx } \)

\( \Rightarrow F = e ^{\log \sqrt{1- x ^2}} \)

\( \Rightarrow F =\sqrt{1- x ^2} \) which is the required integrating factor.