MHT CET 2023 Question Paper May 11 Shift 2 1

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MHT CET 2023 Question Paper with Answers and

Solution May 11 Shift 2 (Memory-based)
Question 1. The equation of the tangent to the curve y = √(9-2x2), at

the point where the ordinates and abscissa are equal, is?
Answer. y > 0
Question 2. The minimum value of function (1 - x + x2) / (1 + x + x2)
A. ⅓
B. 0
C. 3
D. 1
Answer. A
Question 3. ∫ sin(log x) dx
A. (x/2)[sin(logx) - cos(logx)]
B. cos(logx) - x
C. ∫ (x-1)ex / (x+1)3
D. - cos logx
Answer. A
Question 4. The value of ∫ (1 - cosx).cosec2dx is?
Answer. cos(x) + sin(x) + cot(cos(x)) + C

Solution. To evaluate the integral ∫ (1 - cosx) cosec^2(x) dx, we can
simplify the integrand using trigonometric identities.
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Recall that cosec^2(x) is equal to 1 + cot^2(x), where cot(x) is the

cotangent of x.
∫ (1 - cosx) cosec^2(x) dx = ∫ (1 - cosx) (1 + cot^2(x)) dx
Expanding the expression:
= ∫ (1 - cosx + cot^2(x) - cosx * cot^2(x)) dx
Now, let's evaluate each term separately:
∫ (1 - cosx) dx = x - sin(x) + C₁
∫ cot^2(x) dx can be integrated by using the formula ∫ cot^2(x) dx = -x -

cot(x) + C₂
∫ cosx * cot^2(x) dx can be integrated by substitution. Let's denote cos(x) as

u:
u = cos(x) du = -sin(x) dx
Replacing dx and cos(x) with du and u, respectively, we have:
∫ u * cot^2(x) (-du/sin(x)) = -∫ u cot^2(x) du = -∫ cot^2(x) du
Using the formula mentioned earlier, we know that ∫ cot^2(x) dx = -x - cot(x)

+ C₂. Hence, the integral of -∫ cot^2(x) du will be -( -u - cot(u) + C₂) = u +
cot(u) - C₂.
Putting it all together, the integral becomes:
x - sin(x) + C₁ + (-x - cot(x) + C₂) + (cos(x) + cot(cos(x)) - C₂)
Simplifying:
= x - x + cos(x) + sin(x) + cot(cos(x)) + C₁ - C₂
The final result is cos(x) + sin(x) + cot(cos(x)) + C, where C = C₁ - C₂ is the

constant of integration.
Question 5. The function of f(x)= 2x3-9x2+ 12x+29 is monotonically

increasing in the interval
A. (-∞, 1)
B. (-∞, 1) U (2,∞)
C. (-∞,-∞ )
D. (2,∞)
Answer. B
Solution. To determine whether the function f(x) = 2x^3 - 9x^2 + 12x + 29
is monotonically increasing in an interval, we need to analyze the first
derivative of the function, which is given by:
f'(x) = 6x^2 - 18x + 12
To find the critical points of the function (where the derivative is equal to
zero), we need to solve the equation f'(x) = 0:
6x^2 - 18x + 12 = 0
Dividing both sides by 6, we get:
x^2 - 3x + 2 = 0
Factoring the left-hand side, we get:
(x - 1)(x - 2) = 0
So the critical points are x = 1 and x = 2.
Now we need to analyze the sign of the derivative in the different intervals:
Interval (-∞, 1):
For x < 1, we can choose x = 0 as a test point. Plugging this into the
derivative, we get:
f'(0) = 6(0)^2 - 18(0) + 12 = 12

Since f'(0) > 0, the derivative is positive in the interval (-∞, 1). This means
that the function is monotonically increasing in this interval.
Interval (1, 2):
For 1 < x < 2, we can choose x = 1.5 as a test point. Plugging this into the
derivative, we get:
f'(1.5) = 6(1.5)^2 - 18(1.5) + 12 = -3
Since f'(1.5) < 0, the derivative is negative in the interval (1, 2). This means
that the function is not monotonically increasing in this interval.
Interval (2, ∞):
For x > 2, we can choose x = 3 as a test point. Plugging this into the
derivative, we get:
f'(3) = 6(3)^2 - 18(3) + 12 = 30
Since f'(3) > 0, the derivative is positive in the interval (2, ∞). This means
that the function is monotonically increasing in this interval.
Therefore, the function f(x) = 2x^3 - 9x^2 + 12x + 29 is monotonically

increasing in the interval (-∞, 1) U (2, ∞), which corresponds to option B.
Question 6. For all real x, the minimum value of function

f(x)=1-x+x2/1-x+x
A. 1/3
B. 0
C. 3
D. 1
Answer.1/3
Question 7. If the line (x - 1)/2 = (y+1)/3 = (z-2) /4 = 𝜆 meets the plane,

x+2y+3z = 15 at a point P, then the distance of P from the origin is?
A. 7/2
B. 9/2
C. √5/2
D. 2√5
Answer. B
Question 8. If cosx + cosy - cos(x+y) = 3/2 then,
A. x+y = 0
B. x = 2y
C. x =y
D. 2x =y
Answer. C
Question 9. If the vertices of a triangle are (-2,3) , (6,-1) and (4,3), then
the co-ordinates of the circumcentre of the triangle are?
A. (1,1)
B. (-1,-1)
C. (-1,1)
D. (1,-1)
Answer. D
Question 10. If cos-1√p + cos-1√(1-p) + cos-1√(1-a) = 3𝜋/4, then q is?
A. 1/2
B. 1/√2
C. 1
D. 1/3
Answer. A
Question 11. In △ABC b=√3, c=1 angle A = 30, then largest angle?
Answer. 120
Question 12. If the area of the parallelogram with a and b as two

adjacent sites 16 sq, units, then the area of the Parallelogram having
3a+2b and a+3b as two adjacent sides in sq.units is
A. 96
B. 112
C. 144
D. 128
Answer. B
Question 13. dy/dx + y/x = sinn

Answer. xy+cosy-sinx=c
Question 14. x=5/1-21,value of x3+x2-x122

Answer. x2-2x+1= -4
Question 15. Equation of tangent to the curve y = √(9-2x2) where x=y.

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MHT CET 2023 Question Paper with Answers and

Question 1. The equation of the tangent to the curve y = √(9-2x2), at

Question 2. The minimum value of function (1 - x + x2) / (1 + x + x2)

Question 4. The value of ∫ (1 - cosx).cosec2dx is?

Answer. cos(x) + sin(x) + cot(cos(x)) + C

Recall that cosec^2(x) is equal to 1 + cot^2(x), where cot(x) is the

∫ (1 - cosx) cosec^2(x) dx = ∫ (1 - cosx) (1 + cot^2(x)) dx

Expanding the expression:

= ∫ (1 - cosx + cot^2(x) - cosx * cot^2(x)) dx

Now, let's evaluate each term separately:

∫ cot^2(x) dx can be integrated by using the formula ∫ cot^2(x) dx = -x -

∫ cosx * cot^2(x) dx can be integrated by substitution. Let's denote cos(x) as

Replacing dx and cos(x) with du and u, respectively, we have:

∫ u * cot^2(x) (-du/sin(x)) = -∫ u cot^2(x) du = -∫ cot^2(x) du

Using the formula mentioned earlier, we know that ∫ cot^2(x) dx = -x - cot(x)

Putting it all together, the integral becomes:

x - sin(x) + C₁ + (-x - cot(x) + C₂) + (cos(x) + cot(cos(x)) - C₂)

= x - x + cos(x) + sin(x) + cot(cos(x)) + C₁ - C₂

The final result is cos(x) + sin(x) + cot(cos(x)) + C, where C = C₁ - C₂ is the

Question 5. The function of f(x)= 2x3-9x2+ 12x+29 is monotonically

f'(x) = 6x^2 - 18x + 12

Dividing both sides by 6, we get:

Factoring the left-hand side, we get:

So the critical points are x = 1 and x = 2.

Interval (-∞, 1):

f'(0) = 6(0)^2 - 18(0) + 12 = 12

Interval (1, 2):

f'(1.5) = 6(1.5)^2 - 18(1.5) + 12 = -3

Interval (2, ∞):

f'(3) = 6(3)^2 - 18(3) + 12 = 30

Therefore, the function f(x) = 2x^3 - 9x^2 + 12x + 29 is monotonically

Question 6. For all real x, the minimum value of function

Question 7. If the line (x - 1)/2 = (y+1)/3 = (z-2) /4 = 𝜆 meets the plane,

Question 8. If cosx + cosy - cos(x+y) = 3/2 then,

Question 10. If cos-1√p + cos-1√(1-p) + cos-1√(1-a) = 3𝜋/4, then q is?

Question 12. If the area of the parallelogram with a and b as two

Question 13. dy/dx + y/x = sinn

Question 14. x=5/1-21,value of x3+x2-x122

Question 15. Equation of tangent to the curve y = √(9-2x2) where x=y.

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