The equation is not valid for 1 to 9. 30 to 49
(i.e) for 9 + 20 = 29 numbers
Valid for 10 to 29 and 50 to 100 = 71 numbers
n(A) = 71, n(S) = 100
\(\mathrm{P}(\mathrm{A})=\frac{71}{100}=0.71\)

\(\int \frac{x^{2}+\cos ^{2} x}{x^{2}+1} \operatorname{cosec}^{2} x d x \)
\(=\int \frac{x^{2}+\left(1-\sin ^{2} x\right)}{x^{2}+1} \times \frac{1}{\sin ^{2} x} d x \)
\(=\int \frac{\left(x^{2}+1\right)-\sin ^{2} x}{\left(x^{2}+1\right) \sin ^{2} x} d x \)
\(=\int\left(\frac{1}{\sin ^{2} x}-\frac{1}{x^{2}+1}\right) d x \)
\(=\int\left(\operatorname{cosec}^{2} x-\frac{1}{x^{2}+1}\right) d x \)
\(=-\cot x-\tan ^{-1} x+c \)

\(u =\frac{2}{3} x^{3}+5 \)
\(\frac{d u}{d x} =\frac{2}{3}\left(3 x^{2}\right)=2 x^{2} \)
\(y =\frac{1}{4} u^{4} \)
\(\frac{d y}{d x} =\frac{1}{4}\left(4 u^{3}\right) \frac{d u}{d x}=u^{3}\left(2 x^{2}\right) \)
\(=\left(\frac{2}{3} x^{3}+5\right)^{3}\left(2 x^{2}\right) \)
\(=\left(\frac{2 x^{3}+15}{3}\right)^{3} \times 2 x^{2}=\frac{\left(2 x^{3}+15\right)^{3}}{27}\left(2 x^{2}\right) \)

We know that,
\(\sin \alpha-\cos \alpha =\sqrt{2}\left[\frac{1}{\sqrt{2}} \sin \alpha-\frac{1}{\sqrt{2}} \cos \alpha\right] \)
\(=\sqrt{2}\left[\cos \frac{\pi}{4} \sin \alpha-\sin \frac{\pi}{4} \cos \alpha\right] \)
\(=\sqrt{2} \sin \left(\alpha-\frac{\pi}{4}\right) \)
\(\therefore \lim _{\alpha \rightarrow \frac{\pi}{4}} \frac{\sin \alpha-\cos \alpha}{\alpha-\frac{\pi}{4}} =\lim _{\alpha-\frac{\pi}{4} \rightarrow 0} \frac{\sqrt{2} \sin \left(\alpha-\frac{\pi}{4}\right)}{\left(\alpha-\frac{\pi}{4}\right)}=\sqrt{2} \)

\(\vec{m}=\frac{|\vec{b}+2 \vec{a}|}{1+2}=\frac{2 \vec{a}+\vec{b}}{3}\)

\(\text { Given } \text {matrix } A=\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right] \text { is unit matrix }\)
\(|A|=1 \)
\(-\alpha^{2}-\beta \gamma=1 \)
\(1+\alpha^{2}+\beta \gamma=0 \)

\(n^{\text {th }} \text { term }=1-\frac{1}{2^{n}}=1-2^{-n}\)

\(\text { In }(-\infty, 2), \frac{|x-2|}{x-2} \text { is negative }\)
\(\operatorname{In}[2, \infty), \frac{|x-2|}{x-2} \geq 0\)

\(M \cap C=70 \text { which is } 10 \% \text { of } M \text { and } 14 \% \text { of } C\)
\(M =700, C=500\)
\(M \cup C=700+500-70=1130\)

\(x=a\sec ^{ 3 }{ \theta } \)
\(\Rightarrow \frac { dx }{ d\theta } =3a\sec ^{ 2 }{ \theta } .\frac { d }{ d\theta } (\sec { \theta } )=3a\sec ^{ 2 }{ \theta } .\sec { \theta } \tan { \theta } =3a\sec ^{ 3 }{ \theta } \tan { \theta } \)
\(y=a\tan ^{ 3 }{ \theta } \)
\(\Rightarrow \frac { dy }{ d\theta } =a.3\tan ^{ 2 }{ \theta } .\frac { d }{ d\theta } (\tan { \theta } )=3a\tan ^{ 2 }{ \theta } .\sec ^{ 2 }{ \theta } \)
\(\frac { dy }{ dx } =\frac { dy }{ d\theta } /\frac { dx }{ d\theta } =\frac { 3a\tan ^{ 2 }{ \theta } \sec ^{ 2 }{ \theta } }{ 3a\sec ^{ 3 }{ \theta } \tan { \theta } } =\frac { \tan { \theta } }{ \sec { \theta } } =\frac { \sin { \theta } }{ \cos { \theta \times \frac { 1 }{ \cos { \theta } } } } =\sin { \theta } \)
\(\therefore \frac { dy }{ dx } \quad at\quad \theta =\frac { \pi }{ 3 } =\sin { \frac { \pi }{ 3 } } =\frac { \sqrt { 3 } }{ 2 } \)

\(\int { \frac { dx }{ \sqrt { 1-{ 81x }^{ 2 } } } } =\int { \frac { dx }{ \sqrt { 1-\left( 9x \right) ^{ 2 } } } } =\frac { sin^{ -1 }\left( 9x \right) }{ 9 } +c\)

We know that, y = e^x is the inverse function of y = log_ex and hence y = e^x is the reflection of y -= log_ex about y = x.

Here a₁ = a₂ = 1,an = a_n-1 + a_n-2 (n≥3)
Putting n = 3, a₃ = a₂ + a₁ = 1 + 1= 2
Putting n = 4, a4 = a₃ + a₂ = 2 + 1= 3
Putting n=5, a₅ = a₄ + a3 = 3 + 2 = 5
Putting n = 6, a₆ = a₅ + a₄ = 5 + 3 = 8
First six terms of the sequence are 1, 1, 2, 3, 5, 8.

\(cot(\frac{-15\pi}{4})=-cot(\frac{15\pi}{4})=-cot(4\pi-\frac{\pi}{4})=cot\frac{\pi}{4}=1\)

The acute angle \(\theta\) between two lines with slopes m₁ and m₂ is
\(\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|\)
Let \(m_1=\frac{1}{2}\) and m₂=m,\(\theta=\frac{\pi}{4}\)
\(\Rightarrow\tan\frac{\pi}{4}=\left|\frac{\frac{1}{2}-m}{1+\frac{1}{2}m}\right|\)
\(\Rightarrow1=\left|\frac{1-2m}{2+m}\right|\)
\(\Rightarrow1=\pm\frac{1-2m}{2+m}\Rightarrow2+m=\pm(1-2m)\)
If 2+m=1-2m\(\Rightarrow\)3m=-1\(\Rightarrow m=\frac{-1}{3}\)
If 2+m=-1+2m
m=3
Hence, the slope of the other line is 3 or \(-\frac{1}{3}\)

Given Central angle \(\theta\) = 41^o
⇒ \(\theta\) = 41 x \(\frac { \pi }{ 180 } =\frac { 41\pi }{ 180 } \) radians
and radius of the circle = 10 ft
Let l cm be the length of the arc, subtending the angle then
\(\theta =\frac { 1 }{ r } \Rightarrow l={ r }^{ \theta }\)
∴ \(l=10\times \frac { 41\pi }{ 180 } =10\times 41\times \frac { 22 }{ 7\times 180 } \)
∴ \(l=10\times \frac { 41\times 11 }{ 7\times 9 } =\frac { 451 }{ 63 } \)
 = 7.158 = 7.16 ft

log \({{(\sqrt{\sqrt{625}+11})(\sqrt{64})}\over{\sqrt{\sqrt [ 5 ]{ 3125 }+\sqrt{\sqrt [ 3 ]{ 343 } } }}}\)
\(=\log{{\sqrt{25+11}(8)}\over{\sqrt{5}+7}}\)
\(\left[ \because \sqrt{625} \ =25 \\ \sqrt [ 5]{ 3125 } ={5}^{5\times1/5}=5\\ \sqrt [ 3 ]{ 343 }=(7^3) ^{1/3}=7\right]\)        
\(=\log\left({{\sqrt{36}\times8}\over{\sqrt{12}}} \right)\)
\(=\log\left({{6\times8}\over{2\sqrt{3}}}\right)=\log{{3\times8}\over{\sqrt{3}}}\)
\(=\log8+{{1}\over{2}}\log\ 3\)
\(=0.9031+{{1}\over{2}}(0.4771)\)
= 1.1416

Let \(\overrightarrow{BC}=\overrightarrow{a},\overrightarrow{AC}=\overrightarrow{b},\overrightarrow{BA}=\overrightarrow{c}\)

Then \(|\overrightarrow{a}|=a,|\overrightarrow{b}|=b,\) and \(|\overrightarrow{c}|=c\)
Since \(\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}\)
We have \(\overrightarrow{a}=\overrightarrow{c}+\overrightarrow{b}\) and angle between \(\overrightarrow{c}\) and \(\overrightarrow{b}\) is (180-A)
\(=|\overrightarrow{a}|^2=|\overrightarrow{c}+\overrightarrow{b}|^2=|\overrightarrow{c}|^2+|\overrightarrow{b}|^2+2|\overrightarrow{c}||\overrightarrow{b}|cos (180-A)\)
\(=c^2+b^2+2cb \ cos (180-A)\)
\(=b^2+c^2-2cb \ cos A\)    \([\because cos(180-A)=-cos \ A]\)

Let I = \(\int { \frac { { x }^{ 2 }{ tan }^{ -1 }\left( { x }^{ 3 } \right) }{ 1+{ x }^{ 6 } } } dx=\int { \frac { { x }^{ 2 }{ tan }^{ -1 }\left( { x }^{ 3 } \right) }{ 1+\left( { x }^{ 3 } \right) ^{ 2 } } } \)dx
put x³ = t \(\Rightarrow\) 3x² dx = dt \(\Rightarrow\) x² dx = \(\frac { dt }{ 3 } \)
I = \(\int { \frac { tan^{ -1 }(t) }{ 1+{ t }^{ 2 } } } .\frac { dt }{ 3 } =\frac { 1 }{ 3 } \int { \frac { tan^{ -1 }(t) }{ 1+{ t }^{ 2 } } } dt\)
Let tan^-1 (t) = u \(\Rightarrow\) \(\frac { dt }{ 1+{ t }^{ 2 } } \) = du
I = \(\frac { 1 }{ 3 } \int { udu } =\frac { 1 }{ 3 } \left( \frac { { u }^{ 2 } }{ 2 } \right) +c=\frac { { u }^{ 2 } }{ 6 } +c=\frac { 1 }{ 6 } \left[ { tan }^{ -1 }\left( t \right) \right] ^{ 2 }+c\) [\(\therefore\)u = tan^-1 (t)]
= \(\frac { 1 }{ 6 } \left[ { tan }^{ -1 }\left( x^{ 3 } \right) \right] ^{ 2 }+c\)  [\(\because\) t = x³]

A = {(1, 1) (2, 2) (3, 3) 94, 4) (5, 5) (6, 6)}
B = {(4, 6) (6, 4) (5, 5) (6, 5) (5, 6) (6, 6)}
A\(\cap \)B = {(6, 6) (5, 5)}
\(\therefore\) P(A) = P(1, 1) +P(2, 2)+P(3, 3)+P(4, 4) +P(5, 5) +P96, 6)
= P(1). P(1)+P(2).P(2) + P(3).P(3) +P(4).P(4) +P(5).P(5)+P(6).P(6)
\(=\frac { 2 }{ 10 } \times \frac { 2 }{ 10 } +\frac { 2 }{ 10 } \times \frac { 2 }{ 10 } \times \frac { 1 }{ 10 } \times \frac { 1 }{ 10 } +\frac { 3 }{ 10 } \times \frac { 3 }{ 10 } +\frac { 1 }{ 10 } \times \frac { 1 }{ 10 } +\frac { 1 }{ 10 } \times \frac { 1 }{ 10 } \)
\(=\frac { 4 }{ 100 } +\frac { 4 }{ 100 } +\frac { 1 }{ 100 } +\frac { 9 }{ 100 } +\frac { 1 }{ 100 } +\frac { 1 }{ 100 } =\frac { 20 }{ 100 } =\frac { 1 }{ 5 } \)
P(B) = P(4, 6) + p(6, 4) + P(5, 5) + P(6, 5) + P(5, 6) + P(6, 6)
\(=\frac { 3 }{ 10 } \times \frac { 1 }{ 10 } +\frac { 1 }{ 10 } \times \frac { 3 }{ 10 } +\frac { 1 }{ 10 } \times \frac { 1 }{ 10 } +\frac { 1 }{ 10 } \times \frac { 1 }{ 10 } +\frac { 1 }{ 10 } \times \frac { 1 }{ 10 } +\frac { 1 }{ 10 } \times \frac { 1 }{ 10 } \)
\(=\frac { 3 }{ 100 } +\frac { 3 }{ 100 } +\frac { 1 }{ 100 } +\frac { 1 }{ 100 } +\frac { 1 }{ 100 } +\frac { 1 }{ 100 } =\frac { 10 }{ 100 } =\frac { 1 }{ 10 } \)
\(P(A\cap B)\) = P95, 5) + P(6, 6) = P(5).P(5)+P(6).P(6)
\(=\frac { 1 }{ 10 } \times \frac { 1 }{ 10 } +\frac { 1 }{ 10 } \times \frac { 1 }{ 10 } =\frac { 1 }{ 100 } +\frac { 1 }{ 100 } =\frac { 2 }{ 100 } =\frac { 1 }{ 50 } \)
\(\therefore\) P(A\(\cap \)B) = P(A) \(\times\) P(B)
\(\frac { 1 }{ 50 } =\frac { 1 }{ 5 } \times \frac { 1 }{ 10 } =\frac { 1 }{ 50 } \)
Hence A and B are independent events. 

Let y = \({ \left( \sin { x } \right) }^{ { \cos { ^{ -1x } } } }\)
Taking logarithm on both sides we get,
\(\log { y } =\cos ^{ -1 }{ x.\log { (\sin { x } ) } } \)
Differentiating both sides with respect to 'x', we get
\(\frac { 1 }{ y } \frac { dy }{ dx } =\cos ^{ -1 }{ x. } \frac { d }{ dx } (\log { \sin { x } } )+\log { \sin { x. } \frac { d }{ dx } } (\cos ^{ -1 }{ x } )\)
\(\Rightarrow \frac { 1 }{ y } \frac { dy }{ dx } =\cos ^{ -1 }{ x. } \frac { 1 }{ \sin { x } } .\cos { x } +\log { (\sin { x) } } \left( \frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } \right) =\cos ^{ -1 }{ x. } \cot { x } -\frac { \log { \sin { x } } }{ \sqrt { 1-{ x }^{ 2 } } } \)
\(\therefore \frac { dy }{ dx } =y\left[ \cos ^{ -1 }{ x. } \cot { x } -\frac { \log { \sin { x } } }{ \sqrt { 1-{ x }^{ 2 } } } \right] \Rightarrow \frac { dy }{ dx } ={ \left( \sin { x } \right) }^{ { \cos { ^{ -1x } } } }\left[ \cos ^{ -1 }{ x. } \cot { x } -\frac { \log { \sin { x } } }{ \sqrt { 1-{ x }^{ 2 } } } \right] \)

Let A'and B be the availability of first and second fire engine respectively, then A and B are independent.
Then \(P(A)=P(B)=0.96 \)
\(P(\bar{A})=P(\bar{B})=1-0.96=0.04\)
(i) P(a fire engine is available when needed)
\(=P(A \cap \bar{B})+P(\bar{A} \cap B)+P(A \cap B)\)
\(=P(A) \cdot P(\bar{B})+P(\bar{A}) \cdot P(B)+P(A) \cdot P(B)\)
\(=0.96 \times 0.04+0.04 \times 0.96+0.96 \times 0.96\)
\(=0.96(0.04+0.04+0.96) \)
\(=0.96 \times 1.04 \)
\(=0.9984\)
(ii) P (Neither is available when needed)
\(=P(\bar{A} \cap \bar{B}) \)
\(=P(\bar{A}) \cdot P(\bar{B})\)
\(=0.04 \times 0.04\)
\(=0.0016\)

Let I = \(\int \sqrt{(x-3)(5-x)}dx\)
\(=\int \sqrt{8x-x^2-15}dx\)
\(=\int \sqrt{1^2-(x-4)^2}dx\)
\(={x-4\over 2}\sqrt{1^2-(x-4)^2}+{1\over2}sin^{-1}({x-4\over 1})+c\)
Therefore, I \(={x-4\over 2}\sqrt{8x-x^2-15}+{1\over2}sin^{-1}({x-4})+c\)

\(lim_{x\rightarrow0}{e^x-e^{-x}\over sin x}\)\(=lim_{x\rightarrow0}{e^x-1-e^{-x}+1\over {sin x\over x}\times x}\)
\(=lim_{x\rightarrow0}{{e^x-1\over x}-{(e^{-x}+1)\over x}\over {sin x\over x}}\)
\(={(lim_{x\rightarrow 0}{e^x-1\over x})-(lim_{x\rightarrow 0}{e^{-x}-1\over x})\over (lim_{x\rightarrow 0}{sin x\over x})}\)  \((\because lim_{x\rightarrow 0}{e^x-1\over x})=1\)
\({1-(-1)\over 1}={2\over1}=2\)

\(\begin{vmatrix} log_364 &log_43 \\ log_38 & log_49 \end{vmatrix}\times \begin{vmatrix} log_23 & log_83 \\ log_34 & log_34 \end{vmatrix}\) = \(\left| \begin{matrix} { log }_{ 3 }64.{ { log }_{ 2 }3+log }_{ 4 }3.{ log }_{ 3 }4 & { log }_{ 3 }64.{ log }_{ 8 }3+{ log }_{ 4 }3.{ log }_{ 3 }4 \\ { log }_{ 3 }8.{ log }_{ 2 }3+{ log }_{ 4 }9.{ log }_{ 3 }4 & { log }_{ 3 }8.{ log }_{ 8 }3+{ log }_{ 4 }9.{ log }_{ 3 }4 \end{matrix} \right| \)
= \(\left| \begin{matrix} { log }_{ 2 }64+1 & { log }_{ 8 }64+1 \\ { log }_{ 2 }8+{ log }_{ 3 }9 & 1+{ log }_{ 3 }9 \end{matrix} \right| [\therefore { log }_{ y }x.{ log }_{ x }y=1]\)
= \(\left| \begin{matrix} { log }_{ 2 }{ 2 }^{ 6 }+1 & { log }_{ 8 }{ 8 }^{ 2 }+1 \\ { log }_{ 2 }{ 2 }^{ 3 }+{ log }_{ 3 }{ 3 }^{ 2 } & 1+{ log }_{ 3 }{ 3 }^{ 2 } \end{matrix} \right| [\therefore { log }_{ x }x=1]\)
= \(\left| \begin{matrix} 6+1 & 2+1 \\ 3+2 & 1+2 \end{matrix} \right| =\left| \begin{matrix} 7 & 3 \\ 5 & 3 \end{matrix} \right| \) = 21 - 15 = 6

Let,
p(n) : = 1 + 2 + 3 +....+n = \({n(n+1)\over2}\) .
Substituting the value of n = 1, in the statement we get, P(1) = \({1(1+1)\over2}\) = 1. Hence, P(1) is true.
Let us assume that the statement is true for n = k. Then
P(k) = 1 + 2 + 3 + .... + k = \({k(k+k)\over2}\)
We need to show that P(k + 1) is true. Consider.
P(k+1) = \(\underbrace { 1+2+3+......+k } \) + (k + 1) = \({k(k+1)\over2}+(k+1)\)
That is, P(k + 1) = \({k(k+1)+2(k+1)\over2}={(k+1)(k+2)\over2}\)
This implies, P(k + 1) is true. The validity of P(k + 1) follows from that of P(k). Therefore by the principle of mathematical induction, for all integers n \(\ge\),
 1 + 2 + 3 + .... + n = \({n(n+1)\over2}\)

The equation is x² = -2x(1+3k)+7(3+2k) = 0
The roots are real and equal
⇒ Δ = 0 (i.e) b²- 4ac = 0
Here a = 1, b = -2(1+ 3k), c = 7(3 + 2k)
So b²-4ac = 0 ⇒ [-2(1+3k)]² - 4(1) (7) (3 + 2k) = 0
(i.e) 4(1 + 3k)²-28(3 + 2k) = 0
(÷ by 4) (1 + 3k)² - 7(3 + 2k) = 0
1 + 9k²+ 6k - 21-14k = 0
9k²- 8k - 20 = 0
(k-2) (9k + 10) = 0
⇒ k - 2 > 0 or 9k + 10 = 0
⇒ k = 2 or k = \(\frac { -10 }{ 9 } \)
To solve the quadratic inequalities ax²+ bx+ c < 0 (or) ax²+ bx + c > 0

Given that:
2x+y=5.....(i)
x+3y+8=0...(ii)
3x+4y=7 ....(iii)
Equation of any line passing through the point of intersection of equation (i) and (ii) is
(2x+y-5)+λ(x+3y+8)=0 ...(iv)     (λ=constant)
⇒ 2x+y-5+λx+3λy+8λ=0
⇒ (2+λ)x+(1+3λ)y-5+8λ=0
Slope of line m₁ (say) = \(\frac { -(2+\lambda ) }{ 1+3\lambda } \)   \(\left[ \because m=\frac { -a }{ b } \right] \)
Now slope of line 3x + 4y = 7 is
m₂(say) = -\(\frac { 3 }{ 4 } \)
If equation (iii) is parallel to equation (iv) then m₁ = m₂
⇒ \(\frac { -(2+\lambda ) }{ 1+3\lambda } =-\frac { 3 }{ 4 } \)
⇒ \(\frac { 2+\lambda }{ 1+3\lambda } =\frac { 3 }{ 4 } \) ⇒ 8+4λ=3+9λ
⇒ 9λ-4λ=5  ⇒ 5λ=5  ⇒  λ=1
On putting the value of A.in equation (iv) we get
(2x+y-5)+1(x+3y+8)=0
⇒ 2x+y-5+x+3y+8=0  ⇒ 3x+4y+3=0
Hence, the required equation is 3x+4y+3=0

Let AB be the chord and O be the centre of the circular park.
Let\(\angle\)AOB = \(\theta\)
Area of the segment = Area of the sector - Area of \(\triangle\)OAB

\(={1\over2}r^2\theta-{1\over2}r^2sin \theta\)
\(=({1\over 2}\times 4^2)[\theta -sin \theta] \) \(=8[\theta -sin \theta]...(i) \)
But cos \(\theta ={4^2+4^2-4^2\over 2(4)(4)}={1\over2}\)
Thus, \(\theta ={\pi\over3}\)
From (i), area of the segment to be allotted for the veterinary hospital
\(=8[{\pi\over3}-{\sqrt{3}\over2}]={4\over3}[2\pi-3\sqrt{3}]m^2\)

Let P(h, k) be the locus of the point and A(4, 0) B(-4, 0) are the given points.
Given PA + PB = 10
\(\sqrt { { (h-4) }^{ 2 }+{ (k-0) }^{ 2 } } +\sqrt { { (h+4) }^{ 2 }+{ k-0) }^{ 2 } } =10\)
\(\sqrt { { (h-4) }^{ 2 }+{ k }^{ 2 } } =10-\sqrt { { (h+4) }^{ 2 }+{ k }^{ 2 } } \)
Squaring both sides we get,
\({ (h-4) }^{ 2 }+{ k }^{ 2 }=100+[({ h }+4)^{ 2 }+{ k }^{ 2 }]-20\sqrt { { (h+4) }^{ 2 }+{ k }^{ 2 } } \)
\(\Rightarrow \ { h }^{ 2 }-8h+16+{ k }^{ 2 }=100+{ h }^{ 2 }+16+8h+{ k }^{ 2 }-20\sqrt { { (h+4) }^{ 2 }+{ k }^{ 2 } } \)
\(\Rightarrow \ { h }^{ 2 }-8h+16+{ k }^{ 2 }-100+{ h }^{ 2 }+16+8h+{ k }^{ 2 }-20\sqrt { { (h+4) }^{ 2 }+{ k }^{ 2 } } \)
\(\Rightarrow -16h-100=-20\sqrt { { (h+4) }^{ 2 }+{ k }^{ 2 } } \)
Dividing by -4, we get
4h + 25 = \(\\ 5\sqrt { ({ h+4) }^{ 2 }+{ k }^{ 2 } } \)
Squaring both sides we get,
\(\Rightarrow\) (4h + 25)² = 25(h²+ 8h + 16 + k²)
\(\Rightarrow\) 16h² + 625 + 200h = 25h²+ 400 + 200h + 25k²
\(\Rightarrow\)  9h² + 25k² = 225
\(\Rightarrow\) 9h²+ 25k² = 225
Dividing by 225, we get
\(\Rightarrow \ \frac { { h }^{ 2 } }{ 25 } +\frac { { k }^{ 2 } }{ 9 } =1\)
\(\therefore \) Locus of (h, k) is \(\frac{x^2}{25}+\frac{y^2}{9}=1\)

Let S_n = 8 + 88 + 888 + 8888 + .... upto n terms
= 8 (1 + 11 + 111 + 1111 + ....) upto n terms
\(={8\over9}(9 + 99 + 999 + ...)\)
\({ S }_{ n }=\frac { 8 }{ 81 } \left[ \left( { 10 }^{ n }-1 \right) -9n \right] \) [multiplying and dividing by 9]
\(={8\over9}[10 -1) + (100 -1) + (1000 -1) + ...]\)
\(S_n={8\over 9}[(10^1 +10^2 +10^3 + ... +10^n)-(1+1+1+ ... +1n\ terms)]\)
In 10 + 10² + 10³ + ... + 10ⁿ, a = 10, r= 10, and it forms a G.P.
\(∴\ S_n={a(r^n-1)\over r-1}=10{(10^n-1)\over 10-1}={10\over 9}(10^n)-1\) and 1 + 1 + 1 ... + upto n terms = n
Substituting these values in (1) we get
\(S_n={8\over 9}\left[ 10(10^n-1)n\over 9\right]\)
\(S_n={8\over 81}[(10^n-1)-9n]\)

Let E₁ and E₂ be the events of choosing a boy and a girl respectively from the class.
Given that the number of boys to the number of girls = 1: 2
\(\therefore P({ E }_{ 1 })=\frac { 1 }{ 1+2 } =\frac { 1 }{ 3 } \) and P(E₂) = \(\frac { 2 }{ 1+12 } =\frac { 2 }{ 3 } \)
Let A be the event that a student chosen will get first class
Given P(A/E₁) = 0.28 and P(A/E₂) = 0.25
\(\therefore\)  By theorem of total probability, 
P(A) = P(E₁).P(A/E₁) + P(E₂).P(A/E₂)
\(\Rightarrow \quad =\frac { 1 }{ 3 } \times 0.28+\frac { 2 }{ 3 } \times 0.25\)
\(=\frac { 28 }{ 300 } +\frac { 50 }{ 300 } =\frac { 78 }{ 300 } =0.26\) 

\(\lim _{x \rightarrow 2} \frac{x^{2}-3 x+2}{x^{2}-x-2}=\lim _{x \rightarrow 2} \frac{(\not x-2)(x-1)}{(\not x-2)(x+1)}=\lim _{x \rightarrow 2} \frac{(x-1)}{(x+1)}=\frac{2-1}{2+1}=\frac{1}{3}\)

\(\int {1\over \sqrt{1-25 x^2}} dx=\int {1\over \sqrt{1-(5x)^2}}dx={1\over5}sin^{-1}(5x)+c\)

\(f(x)=-x^2+2\)
\(f(x+h)=-(x+h)^2+2=-x^2-h^2-2 x h+2\)
\(f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)
\(=\lim _{h \rightarrow 0} \frac{-x^2-h^2-2 x h+2+x^2-2}{h}\)
\(=\lim _{h \rightarrow 0} \frac{+h(-h-2 x)}{h}\)
= -0 - 2x
\(f^{\prime}(x)=-2 x\)

Dividing by x, we get 
\({1-x^3\over 3x+2}={{1\over x}-x^2\over 3+{2\over x}}\rightarrow -\infty \ as x\rightarrow \infty\)
Therefore the limit does not exist.

The direction ratios of \(3\hat{i}+4\hat{j}-6\hat{k}\)  are 3, 4, -6.
The direction cosines are \({x\over r},{y\over r},{z\over r},\) where r = \(\sqrt{x^2+y^2+z^2}\) .
Therefore, the direction cosines are \({3\over \sqrt{61}},{4\over \sqrt{61}},{-6\over \sqrt{61}}\)

(i) f( m) = f( n) \(\Rightarrow\) m² = n² \(\Rightarrow\) m = n since \(m,\ n\in N.\) Thus f is one-to-one. But, non-perfect square elements in the co-domain do not have pre-images and hence not onto.
(ii) Two different elements in the domain have same images and hence f is not one-to-one. Clearly the range of f is a proper subset of R. Thus it is not onto.

For every subset C of {6, 7, 8, 9, 10}, let B = C \(\cup \) {1, 2, 3, 5}. Then A - B = {4}. In other words, for every subset C of {6, 7, 8, 9, 10}, we have a unique set B so that A - B = {4}.
So number of sets \(B \subseteq X\) such that A - B = {4} and the number of subsets of {6, 7, 8, 9, 10} are the same.
So the number of sets \(B \subseteq X\) such that A - B = {4} is 2⁵ = 32.

Given \({ \left( \frac { 4x }{ 5 } -\frac { 5 }{ 2x } \right) }^{ 9 }\)
Here n = 9, x = \(\frac{4x}{5}\) and a = \((\frac{-5}{2x})\)
\(\therefore { T }_{ r+1 }={ 9C }_{ r }{ \left( \frac { 4x }{ 5 } \right) }^{ 9-r }{ \left( \frac { -5 }{ 2x } \right) }^{ r }\)
\(={ 9C }_{ r }.\frac { { 4 }^{ 9-r } }{ { 5 }^{ 9-r } } .{ x }^{ 9-r }{ \left( -1 \right) }^{ r }.\frac { { 5 }^{ r } }{ { 2 }^{ r }.{ x }^{ r } } \)
\(={ \left( -1 \right) }^{ r }9Cr\frac { { 12 }^{ 18-3r } }{ { 5 }^{ 9-2r } } .\frac { { 5 }^{ r } }{ { 2 }^{ r } } .{ x }^{ 9-2r }\)
\({ T }_{ r+1 }={ \left( -1 \right) }^{ r }9Cr\frac { { 12 }^{ 18-3r } }{ { 5 }^{ 9-2r } } .{ x }^{ 9-2r },0\le r\le 9.\)

The permutations of vowels A, E, I, O, U in English alphabets taking 3 at a time and starting with A are AEI, AIE, AEO, AOE, AEU, AUE, AIO, AOI, AIU, AUI, AOU and AUO.
Clearly, there are 12 permutations.

If the equation of two lines are in general form as a₁ x + b₁ y₁ + c = 0 and a₂x + b₂y + c₂ = 0
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } }\ or\ { a }_{ 1 }{ b }_{ 2 }={ a }_{ 2 }{ b }_{ 1 }\)
Given lines are 3x + 2y + 9 = 0 and 12x + 8y - 15 = 0
\(\frac { 3 }{ 12 } =\frac { 2 }{ 8 } \)
\(\Rightarrow \frac { 1 }{ 4 } =\frac { 1 }{ 4 } \)
Hence the given lines are parallel.