The graphs figures (i) and (ii) look identical until we compare the scales on the y-axis. The scale in figure(ii) is four times as large, reflecting the multiplication of the original function by 4 (i).
The effect looks different when the functions are plotted on the same scale as in figure(iii).
The graph of y = (4x)² - 1 is shown in figure (iv). Can you spot the difference between figure (i) and figure (iv)? In this case, x-scale has now changed, by the same factor of 4 as in the function (figure (iv)).
To see this, note that substituting \(x={1\over 4}\) into (4x)² - 1 produces 12 - 1, 1, exactly the same as substituting x = 1 into the original function (figure (i)), When plotted on the same set of axes (as in figure (v» the parabola y = (4x)² - 1 looks thinner.
Here, the x-intercepts are different, but y-intercepts are the same.

Let there be n sides of the polygon. We know that the number of diagonals of n sided polygon is \(\frac { n(n-3) }{ 2 } \)
⇒ Given \(\frac { n(n-3) }{ 2 } =90\)
⇒ n²-2n = 180
⇒ n²-3n-180 = 0
⇒ (n-15) (n+12) = 0
⇒  n = 15 or n = -12
⇒ There are 15 sides for the polygon which has 90 diagonals.

We have log (1 + x) = \(x-\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 2 } }{ 3 } -\frac { { x }^{ 4 } }{ 4 } +...\)
\(\therefore \log { \left( 1+4x \right) } =4x-\frac { { \left( 4x \right) }^{ 2 } }{ 2 } +\frac { { \left( 4x \right) }^{ 3 } }{ 3 } -\frac { { \left( 4x \right) }^{ 4 } }{ 4 } +\frac { { \left( 4x \right) }^{ 5 } }{ 5 } -\frac { { \left( 4x \right) }^{ 6 } }{ 6 } +...\)  
\(=4x-\frac { { 16x }^{ 2 } }{ 2 } +\frac { { 64x }^{ 3 } }{ 3 } -\frac { 256x^{ 4 } }{ 4 } +\frac { { 1024x }^{ 5 } }{ 5 } -\frac { { 4096x }^{ 6 } }{ 6 } +...\) 
\(=4x-{ 8x }^{ 2 }+\frac { { 64x }^{ 3 } }{ 3 } -{ 64x }^{ 4 }+\frac { { 1024x }^{ 5 } }{ 5 } -\frac { { 2048x }^{ 6 } }{ 3 } +...\)
The series is valid only when \(\left| 4x \right| <1\)
\(\Rightarrow \left| x \right| <\frac { 1 }{ 4 } \)
Hence, This series is valid only in the interval \(-\frac { 1 }{ 4 }

LHS = sin 12^o sin 48^o sin 54^o
= \(\frac{1}{2}\) [cos (12 - 48^o) - cos (12 + 48^o)] sin 54^o
= \(\frac{1}{2}\) [cos (-36)^o - cos 60^o] sin 54^o
= \(\frac { 1 }{ 2 } \left[ \frac { \sqrt { 5 } +1-2 }{ 4 } \right] \left[ \frac { \sqrt { 5 } +1 }{ 4 } \right] =\frac { 1 }{ 2 } \left[ \frac { \sqrt { 5 } -1 }{ 4 } \right] \left[ \frac { \sqrt { 5 } +1 }{ 4 } \right] \)
= \(\frac { 1 }{ 32 } \left[ { \left( \sqrt { 5 } \right) }^{ 2 }-{ 1 }^{ 2 } \right] =\frac { 1 }{ 32 } \left( 5-1 \right) =\frac { 4 }{ 32 } =\frac { 1 }{ 8 } \) = RHS
Hence proved.

\(LHS=\frac { (cos\theta -cos3\theta )(sin8\theta +sin2\theta ) }{ (sin5\theta -sin\theta )(cos4\theta -cos6\theta ) } \)
\(=\frac { 2sin\left( \frac { \theta +3\theta }{ 2 } \right) sin\left( \frac { 3\theta -\theta }{ 2 } \right) .2sin\left( \frac { 8\theta +2\theta }{ 2 } \right) cos\left( \frac { 8\theta -2\theta }{ 2 } \right) }{ 2cos\left( \frac { 5\theta +\theta }{ 2 } \right) sin\left( \frac { 5\theta -\theta }{ 2 } \right) .2sin\left( \frac { 4\theta +6\theta }{ 2 } \right) sin\left( \frac { 6\theta -4\theta }{ 2 } \right) } \)
\(=\frac { sin2\theta .sin\theta .sin5\theta .cos3\theta }{ cos3\theta .sin2\theta .sin5\theta .sin\theta } =1=RHS\)

Perpendicular distance from origin to the line is
4x + 3y - 20 = 0 is
\(\left(\frac{-20}{\sqrt{16+9}}\right)=\frac{20}{\sqrt{25}}=\frac{20}{5}=4 \text { units }\)
Area of the square = 4 \(\times\) 4 = 16 sq. units.

\(1+1 \mid 1+2\lfloor 2+3\lfloor 3+\ldots \ldots . n\lfloor n \quad=\lfloor n+1\)
\(\text { Let } \mathrm{n}=1, \quad \text { L.H.S }=1+1=2\)
\(\text { R.H.S }=\lfloor 2=2\)
It is true for n = 1, In fact it is true for n = 0 also let us assume that it is true for n=k
\(1+1 \mid 1+2\lfloor 2+3\lfloor 3+\ldots \ldots . n\lfloor n=\lfloor k+1\)
\(1+1 \mid 1+2\lfloor 2+3\lfloor 3+\ldots \ldots k \mid k+(k+1)\lfloor k+1\)
\(=\lfloor k+1+(k+1)\lfloor k+1=\lfloor k+1[1+k+1]\)
\(\lfloor k+1 (k+ 2)\)
\(\lfloor k+2\)
It is true for (k + 1)
Also by mathematical induction, it is true for all value of \(n \geq 0, n \in Z\)

\(\mathrm{AM}=16, \quad \mathrm{GM}=8, \quad \mathrm{HM}=?\)
\(\frac{a+b}{2}=16 \Rightarrow a+b=32\)
\(\sqrt{a b}=8 \Rightarrow a b=64\)
\(\therefore \mathrm{HM} =\frac{2 a b}{a+b} \)
\(=\frac{2(64)}{32}= 4 \)

\(\frac{\tan 140^{\circ}-\tan 130^{\circ}}{1+\tan 140^{\circ} \tan 130^{\circ}} =\tan \left(140^{\circ}-130^{\circ}\right) \)
\(=\tan 10^{\circ} \ldots \ldots(1) \)
\(\tan 40^{\circ} =\lambda \)
\(\tan 80^{\circ}=\frac{2 \tan 40^{\circ}}{1-\tan ^{2} 40^{\circ}} =\frac{2 \lambda}{1-\lambda^{2}} \)
\(\tan \left(90^{\circ}-10^{\circ}\right) =\frac{2 \lambda}{1-\lambda^{2}} \)
\(\cot 10^{\circ} =\frac{2 \lambda}{1-\lambda^{2}} \)
\(\Rightarrow \tan 10^{\circ} =\frac{1-\lambda^{2}}{2 \lambda} \)
\(\frac{\tan 140^{\circ}-\tan 130^{\circ}}{1+\tan 140^{\circ} \tan 130^{\circ}} \)
\(=\frac{1-\lambda^{2}}{2 \lambda}\)

\(\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R} \text { is defined by }\)
\(\mathrm{f}(x)=1-|x|\)
\(\text { The range is }(-\infty, 1] \text { as } f(-\infty)=-\infty\)
\(f(0)=1\)
\(f(\infty) =-\infty\)

\(sin(-\frac{11\pi}{3})=-sin\frac{11\pi}{3}=-sin\frac{11}{3}\times180\)
\(=-sin\frac{11}{3}\times\frac{360}{2}=-sin360(\frac{11}{6})\)
\(=-sin360(2-\frac{1}{6})\)
\(=-[sin(360\times2)-\frac{360}{2}]\)
\(=-(-sin60)=\frac{\sqrt 3}{2}\)

\({\sqrt{3}-1\over2\sqrt{2}},{\sqrt{3}+1\over2\sqrt{2}},{\sqrt{3}-1\over \sqrt{3}+1},4\)

There are 8 girls and 7 boys in the mathematics club.
The number of ways of selecting 6 members in that half of them girls (3 girls and 3 others) is ⁸C₃ \(\times\) ⁷C₃ = 56 \(\times\) 35 = 1960.

(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.

Gold medal can be awarded to anyone of the 8 candidates in 8 ways.
Silver medal can be awarded to anyone of the remaining 7 candidates in 7 ways.
Bronze medal can be awarded to anyone of the remaining 6 candidates in 6 ways.
∴ Total numbers of ways of awarding the prize
= 8 \(\times\) 7 \(\times\) 6 = 336

sin 50^o + sin 40^o = \(2\sin { \left( \frac { 50+40 }{ 2 } \right) } .\cos { \left( \frac { 50-40 }{ 2 } \right) } \)
= 2 sin 45^o cos 5^o 
= \(2\frac { 1 }{ \sqrt { 2 } } \cos { { 5 }^{ o } } =\sqrt { 2 } \cos { { 5 }^{ o } } \)

The one's place can be filled up in 5 ways using 1,2,3,4,5 and tens place can be filled up in 4 ways.
∴ Number of two digit numbers using the digits 1,2,3,4,5 is 4 x 5 = 20.

LHS = \(\sin { 4\alpha } =sin2(2\alpha)\)
= \(2\left( \frac { 2\tan { \alpha } }{ 1+\tan ^{ 2 }{ \alpha } } \right) \left( \frac { 1-\tan ^{ 2 }{ \alpha } }{ 1+\tan ^{ 2 }{ \alpha } } \right) \)
= \(4\tan { \alpha } .\frac { 1-\tan ^{ 2 }{ \alpha } }{ { \left( 1+\tan ^{ 2 }{ \alpha } \right) }^{ 2 } } \) = RHS
Hence proved.

Given equation is x⁴ = 16
\(\Rightarrow\)   x⁴ = 16 = 0
\(\Rightarrow\) (x²)²- (4)² = 0
\(\Rightarrow\) (x² + 4)(x²- 4) = 0   [ \(\because\) a²- b² = (a + b)(a - b) ]
\(\Rightarrow\) x = -4,
\(\Rightarrow\) x² = 4 When
\(\Rightarrow\) x² = -4,
x = \(\pm\sqrt{-4}=\pm2\) i where \(i=\sqrt{-1}\) when
\(\Rightarrow\)  x² = 4,
x = \(\pm\sqrt{4}=\pm2\)
Hence the four roots of the given equation are x = 2, -2, 2i, -2i 

(A\(\times\) B) = {(1,4) (1,5) (1,6) (1,7) (2,4) (2,5) (2,6) (2,7) (3,4) (3,5) (3,6) (3,7)}
(B\(\times\)A) = {(4,1) (4,2) (4,3) (5,1) (5,2) (5,3) (6,1) (6,2) (6,3) (7,1) (7,2) (7,3)}
LHS = (A\(\times\)B)\(\cap \)(B\(\times\)A) = { }....(1)
(A\(\cap \)B) = { }, (B\(\cap \)A) = { }
\(\therefore\) RHS = (A\(\cap \)B) \(\times\) (B\(\cap \)A) = { }.....(2)
From (1) and (2), LHS = RHS

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 A = { x \(\in \) N: x is an even prime number}
\(\Rightarrow\) A = {2} 
\(\Rightarrow\) A is a finite set.

In (a+b)ⁿ, the general terms is T_r+1 = nC_r a^n-r b^r
To find the Coefficient of 4th term , put r = 3 in (1)
∴ T₄  = nC₃ a^n-3 b³
To find the Coefficient of 13^th them
Put r = 12 in (1)
T₁₃ = nC₁₂ a^n-12 b¹²
Given nC³ = nC₁₂
3 +12  = n
n = 15

Given equation is (2x + 1)²- (3x + 2)² = 0
\(\Rightarrow\)   (2x+1+3x+2)(2x+1-3x-2) = 0      [ \(\because\) a²- b² = (a + b) (a - b) ]
\(\Rightarrow\)   (5x + 3) (-x - 1) = 0
\(\Rightarrow\)  (5x + 3) (x + 1) = 0
\(\Rightarrow\)   x = \({{}-3\over{5}}\) or -1
\(\therefore\)  Solution set is \(​​​​\left\{-1,{{-3}\over5} \right\} \)