JEE Main Mathematics Mock - 1
25 Questions MCQ Test JEE Main & Advanced Mock Test Series | JEE Main Mathematics Mock - 1
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In the following question, a Statement of Assertion (A) is given followed by a corresponding Reason (R) just below it. Read the Statements carefully and mark the correct answer-
Assertion(A):If Cr is the coefficient of xr in the expansion of (1 + x)20
Reason(R) : Cr = Cn − r for any positive integer n
The area (in square units) bounded by the curves y2 = 4x and x2 = 4y in the plane is
sin θ is real, then θ =In the following question, a Statement of Assertion (A) is given followed by a corresponding Reason (R) just below it. Read the Statements carefully and mark the correct answer-
Assertion(A) :The inverse of 
does not exist.
Reason(R) :The matrix is non singular.
If the line 3x-4y=λ touches the circle x2+y2-4x-8y-5=0, λ can have the values
The length of the tangent from (0,0) to the circle 2x2 + 2y2 + x - y + 5 = 0 is
The differential equation which represents the family of plane curves y=exp. (cx) is
y = ecx
dy/dx = c. ecx
y' = cy
The fundamental period of the function f(x) = 2 cos 1/3(x - π) is
Consider the planes 3x – 6y – 2z = 15 and 2x + y – 2z = 5.
Statement-1:
The parametric equations of the line of intersection of the given planes are
x = 3 + 14t, y = 1 + 2t, z = 15t.
Statement-2:
The vector 14î+2ĵ+15k̂ is parallel to the line of intersection of given planes
In the following question, a Statement-1 is given followed by a corresponding Statement-2 just below it. Read the statements carefully and mark the correct answer-
Tangents are drawn from the point (17,7) to the circle x2+y2=169.
Statement-1:
The tangents are mutually perpendicular.
Statement-2:
The locus of the points from which mutually perpendicular tangents can be drawn to the given circle is x2+y2=338.
Clearly, m1m2 = - 1.
Hence, the two tangents arc mutually perpendicular.
Statement 1 is true.
Now, the locus of the point of intersection of two mutually perpendicular tangents to the circle x2 + y2 = r2 is the director circle i.e. the circle x2 +y2 = 2r2.
For the given circle r = 13. ..
Its director circle is x2 + y2 = 338.
Hence, statement 2 is true and a cogect explanation of statement as the point (17, 7) lies on the director circle of the circle (i).
The value of a for which the system of equations
a3x+(a+1)3y+(a+2)3z = 0
ax+(a+1)y+(a+2)z = 0
x+y+z = 0
has a non-zero solution, is
The system of equation has a non-zero solution
The pole of the line 2x + 3y − 4 = 0 with respect to the parabola y2 = 4 x is
Total outcomes = 36
Doublet are 6 (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)
Probability of getting doublet = 6/36
= 1/6
Covariance is a quantitative measure of the extent to which the deviation of one variable from its mean matches the deviation of the other from its mean. It is a mathematical relationship that is defined as:
Cov(X,Y) = E[(X − E[X])(Y − E[Y])]
Correlation between two random variables, ρ(X,Y) is the covariance of the two variables normalized by the variance of each variable. This normalization cancels the units out and normalizes the measure so that it is always in the range [0, 1]:
The two opposite vertices of a square on xy-plane are A(-1,1) and B(5,3), the equation of other diagonal (not passing through A and B) is
Given: Here,AB is the diagonal of square.
The vertices of a square A 
Let the mid−point of AB be EThen coordinates of E are 
Therefore equation of other diagonal is
If the normal to the curve y=f(x) at the point (3,4) makes an angle 3π/4 with the positive x-axis, then f'(3)
Given y = f(x)
differentiating w.r.t x
y' = f'(x) which is the slope of the tangent
Hence the slope of the normal is - 1/f'(x) = 3pi/4 = -1
therefore f'(x) = 1
Hence f'(3) = 1
Assume e-4/5 = 2/5. If x, y satisfy, y = ex and the minimum value of (x2 + y2) is expressed in the form of m/n then (2m - n)/5 equals (where m & n are coprime natural numbers)
OP2 = x2 + y2
y = ex, y' = ex,
Let ƒ(x) be non-constant thrice differentiable function defined on (–∞, ∞) such that ƒ(x) = ƒ(6 – x) and ƒ'(0) = 0 = ƒ'(2) = ƒ'(5). If 'n' is the minimum number of roots of (ƒ"(x))2 + ƒ'(x)ƒ"'(x) = 0 in the interval x ∈ [0, 6] then sum of digits of n equals
ƒ(x) = ƒ(6 – x)
⇒ ƒ'(x) = –ƒ'(6 – x) .... (1)
put x = 0, 2, 5
ƒ'(0) = ƒ'(6) = ƒ'(2) = ƒ'(4) = ƒ'(5) = ƒ'(1) = 0
and from equation (1) we get ƒ'(3) = –ƒ'(3)
⇒ ƒ'(3) = 0
So ƒ'(x) = 0 has minimum 7 roots in
x ∈ [0, 6] ⇒ ƒ"(x) has min 6 roots in x ∈ [0,6]
h(x) = ƒ'(x).ƒ"(x)
h'(x) = (ƒ"(x))2 + ƒ'(x) ƒ"'(x)
h(x) = 0 has 13 roots in x ∈ [0, 6]
h'(x) = 0 has 12 roots in x ∈ [0, 6]
If the co-ordinate of the vertex of the parabola whose parametric equation is x = t2 – t + 1 and y = t2 + t + 1, t ∈ R is (a, b) then (2a + 4b) equals
x = t2 – t + 1 .... (1)
y = t2 + t + 1 .... (2)
y – x = 2t & x + y = 2(t2 + 1)
________on elminating 't' we get
⇒ (x + y – 2) = 2(y - x)/22
(x – y)2 = 2(x + y – 2)
Axis : x – y = 0
Tangent at vertex : x + y – 2 = 0
Vertex : (1, 1) = (x, y)
If a, b, c, x, y, z are non-zero real numbers and 
then the value of (a3 + b3 + c3 + abc) equals
x2(y + z)y2(z + x)z2(x + y) = a3b3c3 = x3y3z3
⇒ (x + y) (y + z) (z + x) = xyz
⇒ x2(y + z) + y2(z + y) + z2(x + y) + xyz = 0
⇒ a3 + b3 + c3 + abc = 0
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