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Mathematics Time: 2 hours

Note: Question number 1 to 8 carnes 2 maries eaCh, 9 to 16 carries 4 marks eaCh and 17 to 18 carnes 6 marks each.

Q1. A persoogoes to office either by car, soooter, bus or traon probability of whoch be1ng y y y andy respectively Probabil~y lhal he reaches office late, If he takes car, scooter, bus or train Is % i ~ and -ij- respectively Gwen that he reaChed office In tome, then what IS the probability that he travelled by a car

Sol Let C, S, B. T be the events of the person goong by car, soooter, bus or train respectively.

Q2.

Sol

Q3,

' 1 3 2 1 Goven that P(C) = 7. P(S) = 7 . P(B) = 7 . P(T) = 7 Let [ be the event of 1 he person reaChing the office In time => P( c )=I.. P(f)= ~ P(c)= ~. P( c )= ! C 9 S 9 ' B 9 T 9

p(f) P(C) 1 7 (c) c --=-P-- 79 1 [ - P{C) !x Z.+ ;! Y ~ +~~ ~+~ , ! 7 7 9 7 9 7 9 9 7

Rnd the range or values oft for wh1ch 2 sin 1 = 1 - 22x + sx . t e [-!, !!:.]

3x - 2x - 1 2 2

Lety =2 sont 1- 2x + 5x2

so, y= 2 3x - 2x - 1

=> (3y- 5)x2 - 2x(y - 1)- (y + 1) = 0 since X e R - {t -i}. SOO l- Y- 1 "'0

1+ .../5 1- ./5 or y ., -2- andy ,; -

2-

orsin t Hence ra.v.eof t os - - - - u - -[ " " ] [3" "] . ., 2 ' 10 10 ' 2 . Circles with radii 3, 4 and 5 touch each other externally If P Is the point of intersection of tangents to these circles at their points of oontacl Find the distance of P from the points or contact

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Sol Let A, B, C be the centre of lhe three clrcl~. Clearly the point P IS the 1n-centre of the .6ABC, and hence r= t. = ~s(s - a)\s-b)(s-o) = /(s - a)(s - b)(s - o)

s s v s Now 2s = 7 + 8 + 9 = 24 => s = 12.

Hence r =P~;3 =./5 . Q4. F"md the equallon of the plane containing lhe line 2x- y + l - 3 = 0, 3x t y + z = 5 and at a d1stance of

.~ From the point (2, 1, - 1 ). Sol Let the equation of plane be (3:\. + 2)x + (i. - 1 )y + (A. + 1 )z - 5). - 3 = 0

GA. -r 'l + i. - 1- 1.. - 1- 51- - 3 1 => 2 + lim f(x,) - f(x2 ) < lim I x, - x2 1 => If' (x)l < o => f' (x) = 0

x,-x, > r (x)" 2 is the curve. Hence the equation of tangent IS y - 2 = 0.

Q6. If total number or runs scored 1n n matches IS ( "; 1) (2"1 - n - 2) where n > 1, and the runs scored In the I 4h2 =~ 2. (k - 1)2 => k - 1 " 2h =locus lsy = 2x + 1, y = - 2x+ 1.

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QB. Evaluate l e~} ' 2(x12 t y12) 81x 2 81y 2

Hence 1 - 1 1 9(x12 ~ v/)2 4(x12 + y/)2

x2 y2 l x2+y2 ' 2 => the reqUired locus ts - - - =

9 4 9

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Q11. Find the equauon of the common tangent in 1" quadrant to the circle X: + y2 = 16 and the eUipse X~ y~ 25

+"4 = 1 Also find the length of the intercept ofthe tangent belween the coordinate axes

Sol Let the equations of tangents to the given circle and the ellipse respectively be y = mx + 4~1+ m2 and y = mx + ,Jr2-5m-::-2 -+-4 Since both of these represent the same common tangent,

4~1 + m' = ~2f>m2 + 4 => 16{1 + m1 =25m2 + 4 =>m = t .3..

../3 The tangent is at a point in the first quadrant => m < 0 => m = -Js . so that the equabon of the common tangent is y = - }Jx-t4~ It meets the coordinate axes at A ( 2.fi, 0) and a( 0, 4~) => AB=~

.,f3 Q12. If length of tangent at any polfll on the curve y = f(x) intercepted between the point and the x-axis Is or

length 1 Flncl the equation of the curve,

so1 Lengthortangent = v}+(:rl =>1 = v'[1+(: r] dy v R

= - = R => I - dy -X+C dx 1 - y~ y Wnling y = s1n e, dy = cos l:l dl'f and mtegrating, we get the equation ol the curve as

J1 - y2 + ln~ 1 - J~-y2 >=X+ C. Q13. FJnd the area bounded by the curves x2 = y, x2 = -y and y2 = 4x- 3.

Sol The reg10n bounded by lhe g1ven curves x2 = y, x'=- y andy'= 4x - 3 is symmetrical about the x-axls. The parabolas x' = y andy' = 4X - 3 touch at the po1nt (1 1 J Moreover the vertex or lhe cu!Ve

y2 = 4x - 3isat (% 0) Hence the area of the reg10n

= 2[J x~dx - j .J4x - 3 ctx] 0 3/-4

[( )(3 )' 1( ''')' ] [1 1] 1 = 2 - - - (4x - 3) = 2 - - - = - . sq unll$ 3 0 6 31. 3 6 3

X

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Q14. lr one of the vertlces or the square c1roumsctiblng the e~role IZ - 11 = ../2 Is 2 + J3 L Find the other vertices of square.

Sol. Since centre of circle I.e, (1, OJ Is also the rriG-polnt of dtagonals of square => z, + z, = Zo => Zz = -.fii

2 z. - 1 .. /'

and --= e"" Zj -1

::::> other vertices are z.. z.o = (1- J3) + i and (1 + ,/3)-1.

Q16. If F (x - y) = I (x). g (YJ - f (y). g (x) and 9 (X - y) = g (X) g (y) + I ()(). I (y) for all X, y I! R If right hand derivative at x = 0 exists for r (x). Find derivative or g (x) at x = o.

Sol. I (X - y) = f (x) g (y) - f (V) 9 (X) -- (1) Putx = yIn (1), we get f (0) = 0 puty = 0 in (1), we get 9 (0)" 1 Now, I' (Ol = lim f(O+ h) - 1(0) = lim f(O)g(- h) - g(O)f(-h) - 1(0)

-o- h ~o h = IJm f(-h) (: f (O) = O)

~o - h ,. llfn f(O- h) - f(O) -o- - h

= r (O/. Hence r (x) is d1fferertiable at x = o. Put y = X in g lX - y) = 9 (X). 9 (y) + f (x). f (y) Also f' (X) + 9 (X) = 1 => g' (l 2~ (OJ g (0) =-2f (OJ f (0) = 0 => g' (0) = 0,

Q16. If p(x) be a polynomial of degree 3 satisfying p(-1) = 10, p(1) = - 6 and p(x) has maximum at x = - 1 and P'(X) has minima at x = 1. Find the distance between the local maximum and local mimmum of the CUIVe.

Sol Let the polynomial De P (x) = ax3 + bx' +ex + d According to given conditions P(-1)= -a + b-e + d = 10 P(1) =a + b+ C+d =~ AlsoP' (- 1) = 3a - 2b+c = 0 and P'" (1) = 6a + 2b = 0 = 3a + b = 0 Solving for a, b, c. d we get P (x) = x3 - 3x' - 9x + 5 => P' (x)= 3x'- 6X -9= 3(x+ 1)(x - 3) => x = - 1 is the point of maximum and x = 3ls the point or minimum. Hence distance between (-1, 10) and (3, -22) is 4./65 units

Q17. f(X) Is a dillerenliable function and g (x) Is a double differentiable function such that Jf (X)I s 1 and f (x) = g ()(). II f (OJ + g' (OJ = 9. Prove that there exiSts some c e (- 3, 3) such that g (c). g"(c)< 0.

Sol Let us suppose I hat both 9 (X) and 9" (x) are posftive for all x e (- 3, 3) S1noe r7 (OJ+ g' (OJ= 9 and - 1 ~ r (x) ~ 1, 2../2 , g (OJ , 3 From f (x) = g ()

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Moreover, go (x) is assumed to be positive => the curve y = g (~) is open upwards.

If g (x) Is decreasrng, lhen for some value of x J g(x)dx >area of the rectangle (0- (-3))2 .J2

- :::> r (x) > 2 .J2 " 3 - 1 i.e. r (x) > 1 which is a contradiction

If g (x) Is Increasing, for some value or x J g(x)dx > area or lhe rectangle (3 - 0))2 ./2

=> r (X) > 2 ../2 " 3 - 1 I.e r (x) > 1 which is a oonlradlctlon .

II g(x) is minimum at x : o then J g(x)dx > area ollhe rectangle (3-0)2 ./2

- 3

= r (x) > 2 ./2 " 6 - 1 i.e r (x) > 1 which is a contradiction Hence g (x) and g (x) caMOt be both positive throughOut the Interval (-3, 3). Simrlarty we can prove that g(x) and g(x) cannot be both negative throi(Jhout the interval (-3, 3) Hence there is atleast one value of o e (- 3, 3) where g (x) and 9" (x) are or opposite sign = g (c) rt (c) < 0.

Altemate: l J g(x)dx = J f'(x)dx = r (3)- r (0) 0 0

l

= J g(x)dx < 2 0

0

-- (1)

In the same way J g(x)dx < 2 ...... (2) . )

' 0 = J g(X)dx + J g(X):lX < 4 .... (3)

0 -3

From (1(0))2 + (9 (0))2 = 9 we get 2 ../2 0 ...... (7) From equation (6) end (7) 9 (c) rt (c)< 0 for some c e (-3, 3).

',(0, 3) ' ' '

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Case II: -3 < g (0) < -2 .fi l et g (x) IS coocave downward v x (- 3, 3) then the area

l g(x)dx +~ g(x)dx " s.fi. which is a cootradiction from equation (3). : . g (x) will be concave up.vard for some c e (-3, 3) l e. g (c) " o .. (6) also at that po1nt c g (c) will be less than -2 .f2 => g (c) < 0 . (9) From equation (8) and (9) g (c) g" (c)< o for some c e (-3, 3).

(3, 0)

Q18. If[:: : ~][':~;)]=[!~ :~:]. f(x) Is a quadratic function and its max1m1.111 value occurs at a 4ea 4e 1 1(2) 3c2 ~3c

point V A IS a pomt of Intersection of y = f (x) with x-ax1s and point B iswoh that chord AB subtendS a nght angle at V. Find the area enclosed by f (x) and chord AB.

Sol Let we have 4a2 f(-1) + 4a 1(1) + f (2)= 3a' +3a (1) 4b2 f(-1) + 4b 1(1) + f (2) = 3b2 + 3b (2) 4e2 f (-1) + 4cf (1) + 1(2) = 3c2 ... 3c (3) Consider a quadrabo equabon 4x2 f (-1) + 4X f ,1) + f(2) = 3x2 + 3x orj4f (-1)-3J x +(41(1)-3Jx+f(2)= 0 (4) As equation (4) has three roots I.e. x " a, b, c. It Is an identity.

3 3 =>I HJ=-;;-. 1(1) = '4 and f(2)=0

(4 x2 ) => r (x) " - (5)

4 l et point A be (-2, 0) and B be (21, - t2 + 1) Now as AB subtends a rtght angle at the venex v(o,11 v (0, 1) 1 - 12 - x- =-1= 1= 4 2 21

=> B = (6, - 15)

J8 (4-x1 3x+6L 125 .

:. Area= _2

- 4 - -t - 2-r = 3 sq. units