Mhtcet Tricks

MHT-CET Preparation Sachin Gadkar - Vidyalankar Classes, Mumbai. April 16, 2009

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Suggestions while preparing for MHT-CET • MHT-CET is a problem solving exam.So solving lot of problems is very

important.

• Its not that dicult like other Engg. exams- IITJEE or AIEEE. So the

competitive edge is to master time management.

• Try to solve a problem for about a minute, if you can't then move to the

next problem. This way you will solve all easy problems rst & fast. You will come back to those unsolved problem after the rst cycle. This way you come into exam habit of solving paper in cycles.

Few days from exam -week or two • Don't read any new topic, that will only make you panic, lower your

condence.

• Though you might not be condent about 5-10% of syllabus you have

100% of the rest 90-95% in your hands. Make the most of what you know. Don't worry about things which you don't know!

• Go through all the formulae, tricks & shortcuts that you might have learnt

during your preparation.

• Stay away from friends those come and ask you near the exam hall - Have

you done those kinds of problem? Inshort keep your condence high!

While in the exam hall • Since time in the only important factor for MHT-CET, Solve the paper in

cycles. Means solve all the easy problems rst without stoping for more than a minute on a particular problem. This will give double advantage of solving all easy problems correctly in less time by which you will be left with lot of time at the end to try those problems which are moderately dicult or are from topics that you might not have mastered.

• Don't be topic specic, means don't stick around your favourite topic,

since problems in topics which you don't like might be easy.

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Matrices Tricks • Since det(AB) = det(A) det(B) we get det(A−1 ) = A with det(A) 6= 0

1 det(A)

, for any matrix

• For any matrix A of order 2x2, A2 − (sum of diagonal elements)A + det(A) · I

1. If A =



1 3

2 4

=

0



then det(A−1 ) is

(a) 2 (b) -2 (c) 12 (d) − 12 Solution : det(A−1 ) =

2. Matrix A = (a) (b) (c) (d)



1 3

2 4



1 det(A)

= − 21 . Hence (d) is the solution.

such that A2 + 5A = αA + βI then

α = 10, β = 10 α = 2, β = 10 α = 10, β = 2 α = 2, β = 2

Solution : We know that A2 − 5A − 2I

=

0 · · · · · · · · · (1)

Now we need to nd A2 + 5A and compare it with αA + βI A2 + 5A = (5A + 2I) + 5A = 10A + 2I (from (1)) Comparing we get α = 10 & β = 2. Hence the correct alternative is (c).

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Limits & Dierentiation Tricks 0

(n)

(x) (a) (a) is of the form 00 then lim fg(x) = fg0 (a) and you need = · · · = fg(n) (a) x→a to continue the process till you get some nite number, zero or innity. ∞ Same can be extended to problems taking form 0 · ∞, ∞ , 1∞

• If

f (a) g(a)

d • To nd dx (f (x)g(x) ) - Solve this problem as uv rule by rst assuming f (x) constant & g(x) as variable next assuming f (x) as variable & g(x)

as constant.

1. lim

x→0

(a) (b) (c) (d)

sin x+sin2 x x+x2

=

0 1 2 none of these

x = 00 . Hence we can get Solution :We see that for x = 0 we get sin x+sin x+x2 the limit by dierentiating both Numerator & Denominator. Therefore, 2

lim

x→0

sin x+sin2 x x+x2

=

cos x + 2 sin x cos x 1 + 2x cos 0 + 2 sin 0 cos 0 = 1+2·0 =1

Solution is (b). 2. lim √

x→ 3

(a) (b) (c) (d)

x2 −3 x4 +x2 +1

=

0

√

3

3 ∞

0 0 Solution : Here x4x+x−3 2 +1 = 13 6= 0 .Hence the dierentiation technique cannot be used. Just the numerator becomes zero hence just substitute the value. Note all techniques we apply in limits is when the denominator becomes zero. Hence the correct alternative is (a). 2

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3.

d sin x ) dx (x

(a) (b) (c) (d)

=

xsin x ( sinx x − cos x log x) xsin x (sin x − cos x log x) xsin x ( sinx x + cos x log x) xsin x (sin x + cos x log x)

Solution :Dierentiate as if applying uv rule by rst keeping x constant and sin x as variable & viceversa. d(xsin x ) dx

=

sin x · xsin x−1 + xsin x · log x · cos x

=

xsin x (

sin x + cos x · log x) x

The solution is (c)

Dierential Equations • Degree and order of a dierential equation.

 Remove radicals, fractional powers & denominators if any dn y  See the highest dierentiated term i.e. dx n so n is called the order  And now see the power on this highest dierentiated term, this is the degree of the dierential equation. • Number of arbitrary independent constants = Order of the dierential

equations

• If a general solution satises a Dierential Equation, then even a particular

solution too satises the same D.E

  2 3/2  2 2 dy d y 1. The order and degree of DE 1 + dx = a2 dx are respectively 2

(a) (b) (c) (d)

2,2 3,2 2,4 6,4

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Solution : First step is to remove the radicals and denominators. On  

2

3



4

d y dy = a4 dx removing the radicals we get 1 + dx . Now we 2 search  2  for the term that has the highest dierentiated y term. i.e. d y So the order of the DE is 2. And the power over this term is dx2 4, so the Degree of the DE is 4. The correct alternative is (c) 2

2. The dierential equation of y = ae2x−1 + b−2x−1 is (a) (b)

d2 y dx2 = 4 dy dx = 4y

(c)

d2 y dx2

(d)

2

d y dx2

= 4y = −4y

Solution :Since this general solution satises one of the dierential equations in the alternatives. Even a particular solution will satisfy that DE. So lets assume a = 1 & b = 0 we get y = e2x−1 . Now we dierentiate it twice since we had two arbitrary constants at the beginning. we get dy dx d2 y dx2 d2 y dx2

=

2e2x−1

=

4e2x−1

=

4y

This is the required DE of the given general solution. The correct alternative is (c)

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