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X^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0
A+b+c=0
A^2+b^2+c^2=-2(ab+bc+ca)
=-2p>0
P<0. Only possible answers -1
Among options
abc=q

Areas √12*4*3*5=6*2√5=12√5
R=12√5/5=12/√5
Or
Tan(p/2)=✓4*3/12*5=1/√5
R=12*1/√5

4p^2-48<0
-2√3<p<2√3
4q^2-48<=0
Same range
21p^2/2q^2 will be maximum when q minimum or p maximum
Maximum integer p=3
Minimum integer q=1
21(3)^2/2(1)=10.5=94.5

-11^45/180
-(11^3)^15*/180
-(71)^15/180
-(71^2)^7*71*/180
-71/180
180-71=109
71^2=5041
180*28=5040

Radius of the circles that touchs
Both circle 1/√r=√2. R=1/2
1/2:2. 1:4
1:5. 5----4
1--4/5. Area 16/25

323=17*19
16!+91/17 =16+91/17=5 reminder
16!+91/19= 9+91/19=5 reminder
17a+5=19b+5
a=0. B=0
Reminder 5

323 = 17*19
16! + 91) mod 17 = 16 + 6 = 5

17! mod 19 = 1
17*16! mod 19 = (19k+1)
17 = 2k +1. k = 8
17*16! mod 19 = (19*8+1)= 153
16! mod 19 = 9

(16! + 91) mod 19 = 100 == 5

17a + 5 = 19b + 5
a=b=0

Remainder = 5 ✅

Cofficient of x^5 we can make x^5 by. X^2*x*x*x
Possible combinations
(1)(1)(1)(1)(2)=2
2+8-11+3=2
2x^5. So 2

Can be 1. If x^2-7x+11=1
X^2-7x+10=0. X=5,2
X^2-13x+42=0
X=6,7.
(-1)^something. Where something=even number
X^2-7x+12=0
X=4,3 f(x)=x^2-13x+42
At x=3. F(3)=12,. F(4)=6 both even
So total 6 solutions

5^40*2^50*3^100
(5*2)^40*2^10*3^100
40 zeroes
6^10*3^90=x
Logx=10log6+90log3
=10log2+100log3
=10*(.301)+100(.477)
=3.01+47.7
=50+++
Number of digits=51
Total 51+40=91

M1)

y² + 3x²(y² - 10) = 517

±10

(y² - 10)(3x² + 1) = 507

Factorise 507 and equate

x = 2, y = 7

23 ans


M2)

Using options

x + y + xy ± 1

(x + 1)(y + 1) - 1

Put options equal to this expression and check

(x + 1)(y + 1) - 1 = 23

(x + 1)(y + 1) = 24 = 3*8

x = 2, y = 7

Given equation satisfies

So 23 is ans.

Consider a 3 digit number having all three digit are different & number is divisible by 7. If we reverse the digit then this number is also divisible by 7. Then how many such types of number are possible, and find those numbers?

Ans- two, 952 & 861

Try this question

Number 100x + 10y + z

(100x + 10y + z) % 7 = 0

(100z + 10y + x) % 7 = 0

99(x - z) % 7 = 0

99 is not divisible by 7 , means (x - z) is divisible by 7.

Possible values of (x,z) = (1,8),(2,9),(8,1),(9,2)

Numbers : 168, 259, 861, 952

differentiate

3cosx - 3csc^2xcotx = 0

csc^3x = 1

x = pie/2

put

4

Arithmetic Mean >= Geometric Mean

here

(sinx + sinx + sinx + csc^3x)/4 >= (sinx*sinx*sinx*csc^3x)^1/4

(3sinx + csc^3x) >= 4

min. value = 4

(x^2 + y^2)/2 >= (xy)

48.5 >= xy

xy must be less than 48.5

so xy cant be 64

2&rh/2&r(h+r) = 1232/3080
5 = 3080
h/h+r = 1/2. 5
h/r= 2/3
22/7*2*6x^2= 1232
6x^2/7 = 28
X^2 = 14*14/6
Volume =
22/7*3x*3x*2x
22*42/root6*42/root6*28/root6

4312root6✅

[-3,3]. C can never be
If it is -3/2. 4x=120 x=30 doesn't satisfy a rejected
Now angle is 4 times there is a maximum possibllity it will be negative so only option d remain

6*200 + 12*199 + 18*198 + ......... + 1200*1

6(1*200 + 2*199 + 3*198 + ..... + 200*1)

formula = 1/6 * n(n + 1)(n + 2) where n = 200

put

n(n + 1)(n + 2)

200*201*202

8120400