SRMJEEE Maths Mock Test - 5 - JEE MCQ

Any point on the first line is (4r+k, 2r+1,r−1), and any point on the second line is (r′+k+1 ,−r′,2r′+1) for some values of r and r'. The lines are intersecting if these two points coincide i.e

4r + k = r′ + k + 1, 2r + 1 = −r′, r − 1 = 2r′ + 1 for some r and r'

⇒ 4r − r′ = 1, 2r + r′ = −1, r − 2r′ = 2

Now, 4r − r′ = 1, 2r + r′ = −1 ⇒ r = 0, r′ = −1 which satisfy r − 2 r′ = 2.

⇒ The given lines are intersecting for all real values of k.

We know that the word joining two simple statements to form a compound statement is called the connective.
Now, the given statement is
′′2+7>9 or  2+7<9′′
Two mathematical inequality cannot occur together. Either 2 + 7 can be greater than 9 or less than 9.
Hence, the connective word is 'or'.

A statement is mathematically acceptable if either true or false but not both simultaneously. An order or request are not statements. So, "Please do me a favour" is not a statement.

Given:

Let 
So,

And,

Thus,

Hence, this is required solution.

Here,

Let V be the volume of the milk, then

Putting the value of r,

Differentiate on both sides with respect to t,

Given, the value of h = 6m and 
Put these values.

Thus,

Hence, this is required solution.

sinC=sin(A+B)=sinAcosB+cosAsinB

Here,


Therefore, 3 such values of x are possible.
Hence, this is the required solution.

Given:

Again, consider 

From both the above results,
P = Q
Hence, this is the required solution.

We know that

Number of ways = (10 + 1) (9 + 1) (7 + 1) -1
= 879

∴ Given expression is 1 + 1 + 1 = 3

Total number of bijection from set of n elements to itself = n!

(x - a) (x - 10) + 1 = 0
∴ (x - a) (x - 10) = -1
∴ x - a = 1
and x - 1 0 = - 1 ,
or x - a = 1 and x - 10 = 1
∴ a = 8 ora = 12

Degree of the determinant is
n + (n + 2) + (n + 3) = 3n + 5 
and on R .H.S., degree = 2
3n + 5 = 2
⇒ n = -1

∵ 4cosAcosB+ 4sinAsinBsinC= 4

⇒cos(A − B)≥ 1
⇒cos(A − B)= 1
True if A=B  ...(2)
Put (2) in (1)
sinC= 1
C= 90°

Here,

Hence, this is the required solution.

The centre and radius of a circle x² + y² + 2gx + 2fy + c = 0 are (−g, −f) and

Hence, the centre of x² + y² − 4x −2y − 11 = 0 is A(2, 1) and the radius

If P(α, β) be the centre of C₂ of radius r₂ = 1

We know that, if two circles with centres A and P and radii r₁ and r₂ touches each other externally, then distance between their centres AP is equal to the sum of their radii i.e. AP = r₁ + r₂

The locus is obtained by replacing (α, β) by (x, y)

Hence, the locus is x² + y² − 4x − 2y − 20 = 0

Given, vertex = (0, 0), point = P(2, 3) and axis is along X axis.
Since point (2, 3) lies in the first quadrant,
Therefore, equation of the parabola will be of the form y2 = 4ax , which passes through P(2, 3).
Therefore,
Therefore, required equation is

Hence, this is the required solution.

As we know that maximum value of

According to the question,

Thus,
Maximum value of the given expression is 17.
Hence, this is required solution.

∵L.H.S.=(b + c)cos A + (c + a)cos B + (a + b)cos C
=bcos A +c cos A+ c cos B +a cos B +a cos C +b cos C
=(b cos A + a cosB ) + (c cos A + a cosC) + (c cos B + cos C)
=a + b + c = R.H.S
Hence L.H.S.=R.H.S.

New mean

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Let 
⇒
⇒

If a=4 cm, b=6 cm and c=8 cm,
then we have
2s = a + b + c 
⇒s=9 cm
By Heron's formula, we have

Now, we know that

Given equation is,

Semi-perimeter 

Given, sides of the triangle ABC are in AP.
Let sides are a, b, & c in which a is the smallest side.
Since, sides are in AP, So we have
2b = a+c⇒a=2b−c  ...(1)
Now, using cosine rule, we have

Using equations (1) & (2), we have