BITSAT Mathematics Test - 7 - JEE MCQ

f(0)>0, f(3)>0, 0< <3, D≥0
⇒ 27 + (2λ + 1)3 + 3λ² > 0
⇒ λ² + 2λ + 10 > 0
λ ∈ R

D ≥ 0

Intersection gives λ∈ϕ​

Given the equation
(x−b)(x−c)+(x−c)(x−a)+(x−a)(x−b)=0
⇒3x²−2(a+b+c)x+(ab+bc+ca)=0
Since, roots of the equation are equal, then discriminant D=0
⇒[−2(a+b+c)]2−4×3×(ab+bc+ca)=0
⇒4(a+b+c)²−12(ab+bc+ca)=0
⇒(a+b+c)²−3(ab+bc+ca)=0
⇒[a²+b²+c²+2(ab+bc+ca)]−3(ab+bc+ca)=0
⇒a²+b²+c²−(ab+bc+ca)=0
⇒2a²+2b²+2c²−2(ab+bc+ca)=0
⇒(a−b)²+(b−c)²+(c−a)2=0
This is only possible when a=b=c

Given 

Squaring both sides,
then 

Again, squaring both sides
  ⇒1−x=1+4x−4 √x
⇒ 4√x = 5x
⇒25x²−16x=0

x = 16/25 (∵ x = 0 does not satisfy the given equation)

Given equation (b−c)x²+(c−a)x+(a−b)=0
Putting x= 1, we get
(b−c)+(c−a )+(a−b)=0
So x=1 is a root but both roots are equal so both roots are  1,1
Now sum of roots  1+1= 

⇒2b=a+c
⇒a, b, c 
⇒ a,b,c are in A.P.

α, β are roots of  x²+px+q=0, then
α+β=−p and αβ=q
α+h, β+h are roots of x²+rx+s=0, then
(α+h)+(β+h)=−r and (α+h)(β+h)=s
Clearly difference of roots for both is same.


⇒p²−4q=r²−4s

Given quadratic equation is 
Let α and β be the roots of given equation.
⇒(2x+p+q)r=(x+p)(x+q)
⇒x²+(p+q−2r)x+pq−pr−qr=0
Now, sum of roots  (α+β)= -b/a =−(p+q−2r)
⇒−(p+q−2r)=0 (∵Given that roots are equal in magnitude and opposite in sign)
⇒p+q=2r  ...(1)
Product of roots  (αβ)= c/a =pq−pr−qr
Now, α²+β²=(α+β)²−2αβ
=0−2[pq−pr−qr]=−2pq+2r(p+q)
=−2pq+(p+q)2=p2+q² (∵ from (1)
=p2+q²

Put The equation (1) becomes
y²−5y+4=0  ⇒  (y−1)(y−4)=0  ⇒  y=1,y=4 
When 

When 

Thus the roots are 

Given that, α+β=−2 and α³+β³=−56
So, Quadratic equation whose roots are α and β will be given by
x²−(α+β)x+αβ=0  ...(1)
We are given with the sum of roots, so finding product of roots,
∵ α³+β³=−56
⇒(α+β)(α2+β²−αβ)=−56
⇒α2+β²−αβ= - 28 [α + β = -2]
⇒α2+β²=28+αβ  ...(2)
Now,  (α+β)²=(−2)²
⇒α2+β²+2αβ=4
⇒28+3αβ=4 (from equation (2)

Thus, from equation (1), required quadratic equation is given as
x²−(−2)x+(−8)=0
⇒x²+2x−8=0

Let α,β be the roots of equation,
3m²x²+m(m−4)x+2=0
Given λ is the ratio of the roots.
Then,

Add 2αβ on either sides of the equation.
⇒(α+β)²=3αβ  ⋯(1)
From the given quadratic equation,
Sum of the roots,

Product of the roots = 
Substitute sum and product of the roots in the equation (1).

m≠0⇒(m−4)2=18
⇒m−4=±3√2
∴ m=4±3√2

f(x)=x²−ax−(a−3)=0


D>0  ⇒a²+4(a−3)>0
a²+4a−12>0
(a+6)(a−2)>0   ...(1)
f(1)<0⇒   1−a−a+3<0
⇒2a>4
⇒a>2   ...(2)

⇒ a>2

G={(b,b),(b,c),(c,c),(c,d)}
H={(b,a),(c,b),(d,c)}
(G∪H)={(b,b),(b,c),(c,c),(c,d),(b,a),(c,b),(d,c)}
(G∪H)⁻¹={(b,b),(c,b),(c,c),(d,c),(a,b),(b,c),(c,d)}
(G∪H)⊕(G∪H)⁻¹
=[(G∪H)−(G∪H)⁻¹]∪[(G∪H)⁻¹−(G∪H)]
={(b,a)}∪{(a,b)}
={(b,a),(a,b)}
Hence, number of elements in
(G∪H)⊕(G∪H)⁻¹ is 2.

For any x∈R, we have x−x+ √2 = √2 an irrational number.
⇒ xRx for all x. So, R is reflexive.
R is not symmetric, because (2–√, 1)∈R but (1, 2–√)∉R
Also, R is not transitive because (√2, 1)∈R , (1, 2√2)∈R but (√2, 2√2)∉R.

For R₁ let a=1+√2, b=1−√2, c=814
aR₁b⇒a²+b²=(1+√2)²+(1−√2)²=6∈Q
bR₁c⇒b²+c²=(1−√2)²+ =3∈Q
aR₁c⇒a²+c²=(1+√2)²+ =3+4√2∉Q
∴R₁ is not transitive.
For R₂ let a=1+√2, b=√2, c=1−√2
aR₂b⇒a²+b²=(1+√2)²+(√2)²=5+2√2∉Q
bR₂c⇒b²+c²=(√2)²+(1−√2)²=5−2√2∉Q
aR₂c⇒a²+c²=(1+√2)²+(1−√2)²=6∈Q
∴R₂ is not transitive.

R={(a,b):a,b∈N,a−b=3}
={[(n+3),n]:n∈N}
={(4,1),(5,2),(6,3),...}

(a,b) R (c,d)⇔a+d=b+c
⇒(a,a) R (a,a)⇒R is reflexive
∵ a+a=a+a
Next, Let (a,b) R (c,d)⇒a+d=b+c
⇒c+b=d+a⇒(c,d) R (a,b)
⇒ R is symmetric.
Next, Let (a,b) R (c,d) and  (c,d) R (e,f)
⇒ a+d=b+c and c+f=d+e
⇒ a+d+c+f=b+c+d+e
⇒ a+f=b+e⇒(a,b) R (e,f)
⇒ R is transitive ⇒ R is an equivalence relation.

A={a₁, a₂, a₃, a₄, a₅} subset P,Q
∴ ∴ P∩Q = ϕ so each element has three possibilities i.e. P,Q or null set.
∴∴3 × 3 × 3 × 3 × 3 = 3⁵ = 243

Let  U={0,1,2,3,4,5,6,7,.....}
X={0,3,6,9,.....}
Y={0,5,10,15,.....}
X−Y={3,6,9,12,18,21,24,27,33,...}
 ={0,3,6,9,12,15,18,....}∩{1,2,3,4,6,7,8,9,....}
={3,6,9,12,18,.....}
X−Y=

Let A={1,2,3,…n}
Total subsets =2ⁿ
Number of even integers in  1, 2, 3,……n = 
Number of odd integers in = n - 
⇒ Number of subsets having even integers = 
⇒ Number of subsets having atleast one odd integers

(A−B)∪B=(A∩B^C)∪B
=(A∪B)∩(B^C∪B)
=(A∪B)∩(U)
=A∪B
Hence,
[(A−B)∪B]^C=(A∪B)^C=A^C∩B^C

The given curves are x2+y2=64...(i) and y2=4x...(ii)
Solving both the equations, we get
x2+4x−64=0
x=−2±2√17
Since the intersection point is lying in the first quadrant, so
x=−2 + 2√17
So, the required area is 

On integrating, we get

As f(x), is polynomial function, so it is continuous and differentiable in [0,1]
Here f(0)=11, f(1)=1−4+8+11=16
f′(x)=3x²−8x+8

⇒3c²−8c+3=0

The graph of cosecx in interval 

but tan x, x² and |x−1| are increasing in the given interval 

is the rate at which volume of  water is increasing.


Given when x = 2
cm/sec.
The rate at which the depth of water is increasing is 15/13cm/sec.

Let  f(x)=x2−xsinx−cosx
f′(x)=2x−xcosx−sinx+sinx
=x(2−cosx)
We know that, (2−cosx) is positive for all real numbers, now it depends on x.
So, f(x) is increasing if x>0 and
f(x) is decreasing if  x<0
f(0)=−1
f(∞)=∞
f(−∞)=∞

So, it is clear that graph cuts the x - axis at two points, so number of solutions are two.

f(x)=ax⁴+ bx³+ cx²+dx+e
∵ f(0)=f(3)=e
∴ f′(x) become vanishes in (0,3)

On differentiating the given curves respectively

And
m₁×m₂=−1
Hence, the two curves cut at right angles.

Given,  x=a(1+cosθ), y=asinθ

Equating of normal at the given point is

It is clear that in the given options normal passes through the point (a,0).

I3z - 11 = 3 Iz - 2I, z = x + yi, x, y ∈ R
= 9x² + 9y² - 36 x
⇔ 30x = 35
⇒ x = 7/6

When f(x) is divided by x - a, the remainder = f(a) In this case, a = -2k and f (x) = kx²+1