y = x² e^x sin x
u = x² , v = e^x and w = sin x
u' = 2 . x, v' = e^x and w' = cos x
y' = uvw' + vwu' + uwv'
= (x² e^x) cos x + (e^x sin x) (2x) + (x²sin x)e^x
= x² e^x cos x + 2 . x e^x sin x + x² e^x sin x
= xe^x{x cos x + 2 sin x + x sin x}

= \(\int { \left[ a{ sec }^{ 2 }\left( bx+c \right) +\cfrac { q }{ { e }^{ 1-mx } } \right] } dx\) 
= \(a\int { { sec }^{ 2 }\left( bx+c \right) dx+q } \int { { e }^{ mx-1 }dx } \) 
= \(a.\cfrac { 1 }{ b } tan\left( bx+c \right) +q.\cfrac { 1 }{ m } { e }^{ mx-1 }dx\) 
= \(\cfrac { a }{ b } tan\left( bx+c \right) +\cfrac { q }{ { me }^{ 1-mx } } +c\)

Let \( f(x)=\frac{x-2}{x^2-x-2}=\frac{x-2}{(x-2)(x+1)}=\frac{1}{x+1} \)
\( \therefore \lim _{x \rightarrow 2} \frac{x-2}{\dot{x}^2-x-2}=\lim _{x \rightarrow 2} \frac{1}{x+1}=\frac{1}{3}=0 . \overline{3} \)

 

Angle between the lines is given by tanθ=\(\frac { \pm 2\sqrt { { h }^{ 2 }-ab } }{ a+b } \)
In this problem, tanθ =\(\frac { \pm 2\sqrt { 16{ a }^{ 2 }{ b }^{ 2 }-({ a }^{ 2 }3{ b }^{ 2 })({ b }^{ 2 }-3{ a }^{ 2 }) } }{ { a }^{ 2 }-3{ b }^{ 2 }+{ b }^{ 2 }-3{ a }^{ 2 } } \)
=\(\frac { \pm 2\sqrt { 16{ a }^{ 2 }{ b }^{ 2 }-{ a }^{ 2 }{ b }^{ 2 }+3{ b }^{ 4 }+3{ a }^{ 4 }-9{ a }^{ 2 }{ b }^{ 2 } } }{ -2{ a }^{ 2 }-2{ b }^{ 2 } } \)
=\(\frac { \pm 2\sqrt { 3{ a }^{ 4 }+3{ b }^{ 4 }+6{ a }^{ 2 }{ b }^{ 2 } } }{ -2({ a }^{ 2 }+{ b }^{ 2 }) } =\pm \sqrt { 3 } \)
tanθ = 60° [If we take the acute angle]

Given that: f(x) = 1 + 3 cos 2x
We know that -1 ≤ cos 2x ≤ 1
⇒ -3 ≤ 3 cos 2x ≤ 3 ⇒ -3 + 1 ≤ 1 + 3 cos 2x ≤ 3 + 1
⇒ -2 ≤ 1 + 3 cos 2x ≤ 4 ⇒ -2 ≤ f(x) ≤ 4
Hence the range of f = [-2, 4]

\({\tan}^{-1}\left( {2 \over 11} \right)\)

 

 

 

 

x² - 2x - 5 = 0

By taking a = 100, b = 2 and n = 4 in the binomial expansion of (a - b)ⁿ we get
98⁴ = (100-2)⁴
= ⁴C₀100⁴ - ⁴C₁ 100³2 + ⁴C₂ 100²2²- ⁴C3 100¹2³ + ⁴C₄ 100⁰2⁴
= 100000000 - 8000000 + 240000 - 3200 + 16
= 92236816.

tan 120° = tan (180° - 60°)
= - tan(60°) =\(-\sqrt 3\)
(or) write tan 120° as tan (90° + 30°) and find the value.

cos 135° = cos (90° + 45°)
= - sin(45°) = \(-\frac{1}{\sqrt 2}\)
(or) cos 135° = cos (180° - 45°)
= -cos(45°) = \(-\frac{1}{\sqrt 2}.\)

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

There are 6 periods in each working day of a school.
Since each subject is allowed at least one period, we first select one subject for the left out period. This can be done in 5C₁ ways.
Now, six subject can be arranged in \(=\frac { 6! }{ 2! } \) ways
Hence, total number of permutations = 5C₁ \(\times\)\(\frac { 6! }{ 2! } \)ways

= 1800.

sin 75^o - sin 35^o = \(2\cos { \left( \frac { 75+35 }{ 2 } \right) } .\sin { \left( \frac { 75-35 }{ 2 } \right) } \)
= 2 cos(55^o) sin 20^o

\(\frac{1}{cos\theta}=\frac{13}{5} ⇒ cos\theta=\frac{5}{13}\)

AB =\(\sqrt{13^2-5^2}\)
= \(\sqrt{169-25}\)
= \(\sqrt{144}=12\)
Since θ lies in the IV quadrant, only cos θ-. and sec θ are positive
sin θ = \(\frac{-12}{13}, cos θ=\frac{5}{13}, tanθ=\frac{-12}{5}, cosecθ=\frac{-13}{12} and\quad cotθ=\frac{-5}{12}\)

The relation R is defined by xRy if x + 2y = 1 for x, y \(\in \)  N.
Reflexivity : Let x, y \(\in \)  N
xRx \(\Rightarrow\)  x + 2x = 1 \(\Rightarrow\) 3x = 1 \(\Rightarrow\) x = \(\frac { 1 }{ 3 } \notin N\)
\(\therefore\)  R is reflexive.
Symmetricity: xRy \(\Rightarrow\) yRx for x, y \(\in \)  N
xRy \(\Rightarrow\)  x + 2y = 1 which is not possible for any values of x, Y \(\in \) N
\(\therefore\) R is not symmetric
Transitivity: xRy and yRz \(\Rightarrow\) xRz.
xRy and yRz are not possible for any values of x, y, z \(\in \) N
\(\therefore\)  R is not transitive.
\(\therefore\) R is neither reflexive, nor symmetric and not transitive.

Given relation is aRb if a is not a sister of b.
Let a, b, C \(\in \)  A.
Reflexivity : aRa \(\Rightarrow\)  a is not a sister of a
\(\therefore\) R is reflexive.
Symmetricity: aRb \(\Rightarrow\)  bRa
a is not a sister of b \(\Rightarrow\) b is not a sister of a.
\(\therefore\) R is symmetric.
Transitivity : aRb and bRC \(\Rightarrow\) aRC
a is not a sister of b, b is not a sister of C [Eg : Mother is not a sister of daughter, daughter is not a sister of chithi, but mother is a sister of chithi.]
\(\Rightarrow\) a is a sister of C.
\(\therefore\) R is not transitive.
\(\therefore\) R is reflexive, symmetric and but not transitive.

Given 23x<100.
(i) When x is a natural number
23x < 100
\(\Rightarrow\) x < 100/23
\(\Rightarrow\) x < 4.347
\(\Rightarrow\) x = 1,2,3
\(\therefore\) Solution set is { 1, 2, 3, 4}
(ii) When x is an integer.
x < 4.347
\(\Rightarrow\) x = { ...... -4, -3, -2, -1, 0, 1, 2, 3 }
Hence solution set is { ...... -4, -3, -2, -1, 0, 1, 2, 3 }

Let sec-1\(\left( -\sqrt { 2 } \right) \) = y
⇒  -\(\sqrt { 2 } \) = sec y
⇒ sec y = -sec\(\frac { \pi }{ 4 } \)
⇒ sec y = sec\(\left( \pi -\frac { \pi }{ 4 } \right) \)      [∵ sec is negative in the II quad]
⇒ y = \(\frac { 3\pi }{ 4 } \)
Thus, the principal of sec-1(\(\sqrt { 2 } \)) is \(\frac { 3\pi }{ 4 } \) .

Let C = {x \(\in \) Z : x is even and < 10}
\(\Rightarrow\) C = {2, 4, 6, 8} 
\(\Rightarrow\) C is a finite set.

\(\left( 5^{ 3 } \right) ^{ \frac { 2 }{ 3 } }={ 5 }^{ 3\times \frac { 2 }{ 3 } }\) 
= 5² = 25  \(\left[ \because ({ a }^{ m })^{ n }={ a }^{ mn } \right] \)

Given 2|x + 1|- 6 ≤ 7
⇒ 2|x + 1| < 7+6
⇒ 2|x + 1| < 13
\(⇒\ |x+1|\le{13\over 2}\)
This means \(|x+1|\le{13\over2}\)

\(⇒\ {-13\over 2}-1\le x\le{13\over 2}-1\)
\(⇒\ {-15\over 2}\le x\le {11\over 2}\)
\(\therefore\) The Solution set is \(\left[ \frac { -15 }{ 2 } ,\frac { 11 }{ 2 } \right] \)

\(P(B)=0.5, P(A \cup B)=0.7, P(A \cap B)=0.2\)
\(\text {WKT } P(A \cup B)=P(A)+P(B)-P(A \cap B)\)
\(0.7=P(A)+0.5-0.2\)
\(0.7=P(A)+0.3\)
\(\therefore P(A)=0.4 \text {. }\)
To prove that A and B are independent
\(\text { i.e, } P(A \cap B)=P(A) \cdot P(B) \text { is true. }\)
\(P(A \cap B) =0.2 \)
\(P(A) \cdot P(B) =0.4 \times 0.5=0.20\)
\(\text {From (1) and (2), } P(A \cap B)=P(A) \cdot P(B)=0.2\)
Hence A and B are independent.

A -B =\(\begin{bmatrix} 4 & 6 & 2 \\ 0 & 1 & 5 \\ 0 & 3 & 2 \end{bmatrix}\)-\(\begin{bmatrix} 0 & 1 & -1 \\ 3 & -1 & 4 \\ -1 & 2 & 1 \end{bmatrix}\)=\(\begin{bmatrix} 4 & 5 & 3 \\ -3 & 2 & 1 \\1 & 1 & 1 \end{bmatrix}\)
(A - B)^T =\(\begin{bmatrix} 4 & -3 & 1 \\ 5 & 2 & 1 \\3 & 1 & 1 \end{bmatrix}\) ..(1)
A^T - B^T =\(\begin{bmatrix} 4 & 0 & 0 \\ 6 & 1 & 3 \\ 2 & 5 & 2 \end{bmatrix}\)-\(\begin{bmatrix} 0 & 3 & -1 \\ 1 & -1 & 2 \\ -1 & 4 & 1 \end{bmatrix}\)=\(\begin{bmatrix} 4 & -3 & 1 \\ 5 & 2 & 1 \\3 & 1 & 1 \end{bmatrix}\).....(2)
From (1) and (2), (A - B)^T= A^T - B^T.

Observe that \(\sqrt{x^2-10x+25}=\sqrt{(x-5)^2}=|x-5|\)