GATE Mathematics Mock Test - 5 - GATE Mathematics MCQ

Required percentage = (6699/9004) × 100 = 0.744 × 100 ≈ 75%

Then, the second pipe can fill the reservoir in x – 10 hours.

Now, according to the question,

1/x + 1/x-10 = 1/12

⇒ 12(x – 10 + x) = x² – 10x

⇒ 24x – 120 = x² – 10x

⇒ x² – 34x + 120 = 0

⇒ x² – 30x - 4x + 120 = 0

⇒ (x – 30)(x – 4) = 0

⇒ x = 30, 4

But, x ≠ 4 (∵ x – 10 should be positive)

∴ x = 30 hours

x – 10 = 20 hours

Since fuel consumption/litre is asked and not total fuel consumed, only average speed is relevant. Maximum efficiency comes at 45 km/hr. So, least fuel was consumed per litre in lap Q.

Let 

Therefore , option(a) is incorrect.

For option (b) , let 

But 

Therefore option (b) is correct

For optin (c0 and (d)

Also in T.

Therefore option (c) and (d) are incorrect.

If σ is the n cycle then σⁱ is also n cycle iff(i, n) = 1

Now (12, 11) = 1

⇒ σ¹¹ is 12 cycle.

Number of elements of order d in Z_n where ���� is (d).

Therefore, number of elements of order p in Z_p²_q = (p) = p-1

The matrix form of the given system of equations is,

The given system of equations will have a unique solution if and only if the coefficient matrix is non-singular.

Performing , we get,

Performing , we get

Therefore the coefficient matrix will be non-singular if and only if,

Thus the given system will have a unique solution if 

Showing that given equations are inconsistent in this case. Thus if  

= -3.5 no solution exists.

Hence, the correct answer is −3.5

0

We compute using the first row cofactor expansion

= 8(10 − 63)−(6 − 28) + 6(27 − 20)

= 8(−53) − (−22) + 6(7)

= −424 + 22 + 42 

= -360.

Analytic Method : since 

Replacing x by 2x and y by 0, then 

= f’(0) = –1 x ∈ R   (given) 

Integrating, we get f(x) = –x + c 

Putting x = 0, then f(0) = 0 + c = 1     (given)

∴ c = 1 

then f(x) = 1 – x 

∴ f(2) = 1 – 2 = –1 

Since let 

By L hospital rule 

Consider 

Given a_n = 3^-n a_n-1

Let 

 convergent if |z| < √3 and divergent if |z| > √3

So, radius of convergence is √3.

The characteristic equation is given by r² + 2r + 6 = 0

using y(0) = 2, we have c₁ = 2

Now y'(0) = α, we have  

Therefore

Smallest positive value of x for which y= 0 is x(α) = 

Let y = mx², then

depends on the value of m. Hence, ,  f(x,y) do not exist.

Hence, f isnotcontinuous at (0,0). Therefore, f is not differentiable at (0,0).

For v₂= 0, means that it is partial derivative with respect tox, which exists at (0,0).

Hence, directional derivative exists for everyunit vector .

Re-wrirte the equation as

Compare with general form y" + py' +Qy = 0, we have p(t) = 

Now using W (1) = 1, we have 1 =ce i.e. c = 1/e

⇒ W(t) = t²e^t-1

The length of th curve y = x^3/2, 0 ≤ x ≤4 is given by = 

Given 

Let us consider A_n = 

then A_n is open for each n ∈ ℕ. But

 is not open.

Since   do not exist

Continuous at x = 2

f(2) = 5·2 +1=11

also 

and 

Note that

So 

So f (x) is not continuous at x = 2 hence. fix) is not differentiable at x=2

R.H.D of f(x) atx = 2, is

hence the right derivative of fix) at x =2 does not exist.

Let H be any subgroup of order 12 in S₄

Let H ≠ A₄

and let if possible that H contains and odd permutation thus H has 6 odd and 6 even permutation

 ⇒ H ∩ A₄ is a subgroup of A₄ of order 6 

⇒ A₄ has subgroup of order 6 contradiction by a well know theorem 

Hence A₄ is the only subgroup of S₄ of order 12. 

We have

⇒ for every closed path .

Given 

⇒ -e^-y = e^x + c

⇒ -e^-1 = e¹ + c                          (as y(1) = 1)

⇒ c = -(e^-1 + e¹)

∴ -e^-y = e^x -(e + e^-1)

∴ -e^-y(-1) = e^-1 - e^-1 -e, at x = 1

⇒ e^-y(-1) = e

⇒ e^-y(-1) = e

⇒ -y(-1) = 1

⇒ y(-1) = -1