Mathematics Test - 26

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

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

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

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

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

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

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

Given:

cot θ (1 + sin θ) = 4m

⇒ \((\frac{\cos \theta} { \sin \theta})\) (1 + sin θ) = 4m ----(1)

Also, cot θ (1 – sin θ) = 4n

⇒ \((\frac{\cos \theta} { \sin \theta})\) (1 – sin θ) = 4n ----(2)

Multiplying (1) and (2), we get,

\(\Rightarrow\left(\frac{\cos ^{2} \theta} { \sin ^{2} \theta}\right)\left(1-\sin ^{2} \theta\right)=16 \mathrm{mn}\left[\because \cos ^{2} \theta=1-\sin ^{2} \theta\right] \)

\(\Rightarrow \frac{\cos ^{4} \theta} { 16 \sin ^{2} \theta}=\mathrm{mn}\) ----(3)

Now,

Squaring and adding equation \((1)\) and \((2)\), we can write,

\( 16\left(m^{2}+n^{2}\right)=\left(\frac{\cos ^{2} \theta} { \sin ^{2} \theta}\right)\left(1+\sin ^{2} \theta+2 \sin \theta+1+\sin ^{2} \theta-2 \sin \theta\right)\)

\(\Rightarrow\left(m^{2}+n^{2}\right)=(\frac{1 }{ 8})\left(\frac{\cos ^{2} \theta} { \sin ^{2} \theta}\right)\left(1+\sin ^{2} \theta\right) \)

Squaring both sides,

\(\Rightarrow\left(m^{2}+n^{2}\right)^{2}=(\frac{1 }{64})\left(\frac{\cos ^{4} \theta} { \sin ^{4} \theta}\right)\left(1+\sin ^{2} \theta\right)^{2}\)

Comparing with equation (3),

\(\Rightarrow\left(\mathrm{m}^{2}+\mathrm{n}^{2}\right)^{2} \neq \mathrm{mn}\)

Also, \(\left(m^{2}+n^{2}\right)^{2} \neq m^{2} n^{2}\)

Squaring and substracting equation \((1)\) and \((2)\), we can write,

\( 16\left(m^{2}-n^{2}\right)=\left(\frac{\cos ^{2} \theta} { \sin ^{2} \theta}\right)\left(1+\sin ^{2} \theta+2 \sin \theta-1-\sin ^{2} \theta+2 \sin \theta\right) \)

\(\Rightarrow\left(m^{2}-n^{2}\right)=(\frac{1} { 4})\left(\frac{\cos ^{2} \theta} { \sin ^{2} \theta}\right)(\sin \theta) \)

Squaring both sides,

\(\Rightarrow\left(m^{2}-n^{2}\right)^{2}=(\frac{1} { 16})\left(\frac{\cos ^{4} \theta} { \sin ^{4} \theta}\right)\left(\sin ^{2} \theta\right)\)

\(\Rightarrow\left(m^{2}-n^{2}\right)^{2}=\left(\frac{\cos ^{4} \theta }{ 16 \sin ^{2} \theta}\right)\)

Comparing with equation (3),

\(\left(m^{2}-n^{2}\right)^{2}=m n\)

Let the events be defined as:

A: Obtaining a sum of 8

B: Getting an even number on both dice

Now cases favourable to A are \((3,5)(5,3)(2,6)(6,2)(4,4)\)

So, \(P(A)=\frac{5}{36}\)

Cases favourable to B: \((2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)\)

P(B)=\(\frac{9}{36}\)

Now, \((2,6)(6,2)\) and \((4,4)\) are common to both events \(A\) and \(B\)

So, \(P(A \cap B)=\frac{3}{36}\)

\(\Rightarrow P ( A \cup B )=\frac{5}{36}+\frac{9}{36}-\frac{3}{36}\)

\(\Rightarrow P ( A \cup B )\)\(=\frac{11}{36}\)

Term of an AP are \(403,397,391 \ldots \)

So, First term \((a) =403\)

Common difference \((d)=397-403=-6\)

Let \(n ^{\text {th }}\) term is the first negative term of the given AP.

As we know,

\(n^{\text {th}}\) term of AP is given by,

\(a_ n=a+(n-1) \times d\)

\(\therefore a_n =403+(n-1) \times-6 \)

\(\Rightarrow a_n=403-6 n+6 \)

\(\Rightarrow a_n=409-6 n\)

Now, we have to find the suitable value of \(n\), so that the value of \((409-6 n)\) is negative.

Checking options:

Taking \(n=68\)

\(\therefore a_{n}=409-6 \times 68\)

\(\Rightarrow a_n=409-408\)

\(\Rightarrow a_n=1\) (which is not negative)

Now, putting \(n=69\), we get

\(\therefore a_{n}=409-6 \times 69\)

\(\Rightarrow a_n=409-414\)

\(\Rightarrow a_n=-5\)

This is the first negative term of the series.

So, \(69^{\text {th}}\) term is the first negative term of given series.

It is given that \(\frac{\tan x}{2}=\frac{\tan y}{3}=\frac{\tan x}{5}=k\) (say).
\(\therefore \tan x=2 k,\) \(\tan y=3 k\) and \(\tan z=5 k\).
It is also given that \(x+y+z=\pi\) \(\Rightarrow x+y=\pi-z\) \(\Rightarrow \tan (x+y)=\tan [\pi+(-z)]\)
\(\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}\), \(\tan (n \pi+\theta)=\tan \theta\), \(\tan (-\theta)=-\tan \theta\)
\(\Rightarrow \frac{\tan x+\tan y}{1-\tan x \tan x}=\tan (-z)=-\tan z\)
\(\Rightarrow \tan x+\tan y=-\tan z+\tan x \tan y \tan z\)
\(\Rightarrow \tan x+\tan y+\tan z=\tan x \tan y \tan z\)
Substituting the values in terms of \(k\) from the above result, we get
\(2 k+3 k+5 k=(2 k)(3 k)(5 k)\) \(\Rightarrow 10 k=30 k^{3}\) \(\Rightarrow {k}^{2}=\frac{1}{3}\)
Now, \(\tan ^{2} x+\tan ^{2} y+\tan ^{2} z\)
\(=(2 k)^{2}+(3 k)^{2}+(5 k)^{2}\)
\(=4 k^{2}+9 k^{2}+25 k^{2}\) \(=38 {k}^{2}\) \(=\frac{38}{3}\)

Let

\(f(x)=e^{3 x+b}\)

\(\Rightarrow f(x+h)=e^{a(x+h)+b}\)

\(\frac{d}{d x}(f(x))=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)

\(=\lim _{h \rightarrow 0} \frac{e^{3(x+h)+b}-e^{(2 x+b)}}{h}\)

\(=\lim _{h \rightarrow 0} \frac{e^{a x+b} e^{a x}-e^{a x+b}}{h}\)

\(=\lim _{h \rightarrow 0} e^{a x+b}\left\{\frac{\left(e^{a h}-1\right)}{a h}\right\} \times a\)

\(=a e^{a x+b}\)

\(\left[\right.\) Since, \(\left.\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1\right]\)

So,

\(\frac{d}{d x}\left(e^{a x+b}\right)=a e^{a x+b}\)

Given: 

\(y=\sin x+\cos x\) in the interval \(0

Let the required area be \(\mathrm{A}\).

Using the formula of the area under the curve as:

\(\mathrm{A}=\left|\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x}) \mathrm{dx}\right|\)

\(=\left|\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}\right|+\left|\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{g}(\mathrm{x}) \mathrm{d} \mathrm{x}\right|\)

\(\Rightarrow \mathrm{A}=\left|\int_{0}^{\frac{\pi}{2}}(\sin \mathrm{x}+\cos \mathrm{x}) \mathrm{d} \mathrm{x}\right|\)

\(=\left|\int_{0}^{\frac{\pi}{2}}(\sin \mathrm{x}) \mathrm{dx}\right|+\left|\int_{0}^{\frac{\pi}{2}}(\cos \mathrm{x}) \mathrm{dx}\right|\)

\(=\left|[-\cos \mathrm{x}]_{0}^{\frac{\pi}{2}}\right|+\left|[\sin \mathrm{x}]_{0}^{\frac{\pi}{2}}\right|\)

Using the value of \(\cos 0=1, \sin 0=0, \cos \frac{\pi}{2}=0\) and \(\sin \frac{\pi}{2}=1\)

\(\Rightarrow \mathrm{A}=\left|-\cos \frac{\pi}{2}+\cos 0\right|+\left|\sin \frac{\pi}{2}+\sin 0\right|\)

\(=|0+1|+|1-0|\)

\(=1+1\)

\(=2\)

Given that,

\(\Rightarrow \frac{ xdy }{ dx }+3 y =4 x ^3\)

Now,

\(\Rightarrow \frac{ dy }{ dx }+\frac{3 y }{ x }=4 x ^2\)

By comparing with \(\frac{ dy }{ dx }+ Py = Q\)

\( \Rightarrow P=\frac{3 } x \text { and } Q=4 x^2 \)

\( \Rightarrow \text { I.F. }=e^{\int P d x}=e^{\int \frac{3}{x} d x} \)

\(\Rightarrow \text { I.F. }=e^{3 \ln x} \)

\( \Rightarrow \text { I.F. }=e^{\ln x^3}\)

\( \Rightarrow \text { I.F. }=x^3\left(\because e^{\ln x}=x\right)\)

Now general solution will be,

\( \Rightarrow y \cdot( I \cdot F \cdot)=\int( Q \cdot( I \cdot F \cdot)) dx + c \)

\( \Rightarrow y \cdot\left( x ^3\right)=\int\left(4 x ^2 \cdot\left( x ^3\right)\right) dx + c\)

\( \Rightarrow x ^3 \cdot y =\int 4 x ^5 dx + c \)

\( \Rightarrow x ^3 \cdot y =4 \frac{ x ^6}{6}+ c \)

\( \Rightarrow x^3 \cdot y=\frac{2}{3} \cdot x^6+c\)