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KEAM Paper 2 Mock Test - 5 - JEE MCQ


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13 Questions MCQ Test KEAM Mock Test Series 2024 - KEAM Paper 2 Mock Test - 5

KEAM Paper 2 Mock Test - 5 for JEE 2024 is part of KEAM Mock Test Series 2024 preparation. The KEAM Paper 2 Mock Test - 5 questions and answers have been prepared according to the JEE exam syllabus.The KEAM Paper 2 Mock Test - 5 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for KEAM Paper 2 Mock Test - 5 below.
Solutions of KEAM Paper 2 Mock Test - 5 questions in English are available as part of our KEAM Mock Test Series 2024 for JEE & KEAM Paper 2 Mock Test - 5 solutions in Hindi for KEAM Mock Test Series 2024 course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt KEAM Paper 2 Mock Test - 5 | 13 questions in 180 minutes | Mock test for JEE preparation | Free important questions MCQ to study KEAM Mock Test Series 2024 for JEE Exam | Download free PDF with solutions
KEAM Paper 2 Mock Test - 5 - Question 1

If one root of the equation ax2 + bx + c = 0 be n times the other root, then

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KEAM Paper 2 Mock Test - 5 - Question 2

Let n(U) = 700, n(A) = 200, n(B) = 300 and n(A ∩ B) = 100,
Then n(Ac∩Bc) =

Detailed Solution for KEAM Paper 2 Mock Test - 5 - Question 2

n(Ac ∩ Bc) = n(U) - n(A ∪ B)
= n(U) - [n(A) + n(B) - n(A ∩ B)]
= 700 - [200 + 300 - 100] = 300.

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KEAM Paper 2 Mock Test - 5 - Question 3

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KEAM Paper 2 Mock Test - 5 - Question 4

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KEAM Paper 2 Mock Test - 5 - Question 5

In a ΔABC, a = 13 cm, b = 12 cm and c = 5 cm. The distance of A from BC is

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KEAM Paper 2 Mock Test - 5 - Question 6

The lengths of sides of a triangle are in the ratio 5 : 12 : 13 and its area is 270 cm2. The respective lengths of sides of the triangle (in cm) are

Detailed Solution for KEAM Paper 2 Mock Test - 5 - Question 6

Let the lengths of sides of the triangle be 5x, 12x, 13x;  Obviously, the triangle is right-angled.
Hence, area of the Δ = ½ (12x) (5x) ⇒ 30x2 = 270 ⇒ x = 3 
Hence, the lengths of sides (in cm) are 15, 36 and 39.

KEAM Paper 2 Mock Test - 5 - Question 7

If the mean of a binomial distribution is 25, then its standard deviation lies in the interval

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KEAM Paper 2 Mock Test - 5 - Question 8

The solution of the equation x2 dy/dx = x2 + xy + y2  is

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KEAM Paper 2 Mock Test - 5 - Question 9

Assuming that f is continuous everywhere, (1/c)  is equal to

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KEAM Paper 2 Mock Test - 5 - Question 10

If (√8+i)50 = 349 (a + ib), then a2 + b2 is

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KEAM Paper 2 Mock Test - 5 - Question 11

The equation of the normal to the ellipse x2/a2 + y2/b2 = 1 at the positive end of the latus rectum is

Detailed Solution for KEAM Paper 2 Mock Test - 5 - Question 11

The equation of the normal at (x1, y1) to the given ellipse is

Here x1 = ae and y1 = b2/a 
So the equation of the normal at positive end of the latus rectum is 

KEAM Paper 2 Mock Test - 5 - Question 12

 is equal to

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KEAM Paper 2 Mock Test - 5 - Question 13

If  then f(2x) − f(x) is not divisible by

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