AP ICET 2018 Question Paper Shift-2 (2nd May)

Introduce negation to both sides of relation.
p gets changed to "not p"
"And" function gets changed to "Or"
"Not q" gets changed to "q" 
"r" gets changed to "not r" 

Hence, option C is the answer. 

S:$$p\longrightarrow\ q$$

where, p: 10<8 is a mathematically incorrect/false statement
and      q: 2+4=8 is also a mathematically false statement. 

False implying another false statement. This makes S as a true statement. 
S would have been false, if either of p or q had been true and the other one false. 

This can be determined by doing some mapping:
The first member of the 5-member set can take 8 values
The second member of the 5-member set can take 7 values
The third member of the 5-member set can take 6 values
The fourth member of the 5-member set can take 5 values
The fifth member of the 5-member set can take 4 values

hence, the total number of injective functions possible = 8*7*6*5*4 = 6720

Slope of the line = -12/5

Equation of the line:
y - 7 = (-12/5) * (x + 2)
X-intercept is obtained by putting y = 0 in above equation
Hence, 35/12 = x + 2 ==> x = 11/12

Points to remember: The perpendicular bisector will pass through the mid-point of the segment obtained by joining the given points. 
The product of slope of the bisector and the slope of the segment = -1

Firstly, mid-point of the segment = [(-2+1)/2, (3-2)/2] = (-1/2, 1/2) 
hence, perpendicular bisector passes through this point. 

Slope of the segment = (-2-3)/(1+2) = -5/3 
Hence, slope of perpendicular bisector = 3/5

Now, equation of perpendicular bisector:
(y - 1/2) = (3/5)*(x+1/2)
Solving the equation gives us option B

Sin 60 = (root 3)/2
Cos 225 = cos (180 + 45) = - cos 45 = -1/(root 2)
Tan 210 = tan (180 + 30) = tan 30 = 1/(root 3)
Cot 240 = cot (180 + 60) = cot 60 = 1/(root 3) 

Squaring up all terms and adding, we get answer as 23/12

From the given relations, it becomes clear that $$\alpha\ =\frac{\pi}{3},\ \beta\ =\frac{\pi}{6}\ \ $$

Hence, the given relation results in option D

$$\tan\ 2\theta\ +\sec\ 2\theta\ $$
$$=\frac{\left(1+\sin2\theta\ \right)}{\cos\ 2\theta\ }$$
$$\frac{\left(1+2\sin\ \theta\ \cos\ \theta\ \right)}{\cos\ ^2\theta\ -\sin\ ^2\theta\ }$$

Dividing both numerator and denominator by $$\cos\ ^2\theta$$ , we get

$$\frac{\left(\frac{1}{\cos\ ^2\theta}+2\tan\ \theta\ \right)}{1-\tan\ \theta\ }=\frac{\left(1+\tan^2\theta\ +2\tan\ \theta\ \right)}{1-\tan^2\theta\ }$$

Substitute the value of tan from the given data, we get option D as answer.