JEE Mathematics Previous Year Questions MCQs Set 06

Answer: (a) \( \frac5{11} \)

Explanation:
1. 1 first appears on throw \( 2k \) with probability \( \left(\frac56\right)^{2k-1}\frac16 \).
2. Sum over \( k\ge1 \): \( \frac{(5/6)(1/6)}{1-(5/6)^2}=\frac{5/36}{11/36}=\frac5{11} \).

Teacher's Note:
1. This is an infinite geometric series with ratio (5/6) squared.
2. Take only even numbered trials.

Answer: (d) \( x^2+y^2=4 \)

Explanation:
1. The ellipse is \( \frac{x^2}4+\frac{y^2}3=1 \), with \( a=2 \).
2. The foot of the perpendicular from a focus to any tangent lies on the auxiliary circle \( x^2+y^2=a^2 \).
3. So the pedal curve is \( x^2+y^2=4 \).

Teacher's Note:
1. The feet of focal perpendiculars to tangents lie on the auxiliary circle.
2. Here a squared equals 4.

Answer: (a) \( \sqrt{13}\,a \)

Explanation:
1. AB is the polar of \( (2a,3a) \): \( 3a\,y=2a(x+2a) \), i.e. \( x=\frac{3y-4a}{2} \).
2. Substituting in \( y^2=4ax \): \( y^2-6ay+8a^2=0 \), so \( y=2a \) or \( 4a \).
3. The points are \( (a,2a) \) and \( (4a,4a) \); \( AB=\sqrt{9a^2+4a^2}=\sqrt{13}\,a \).

Teacher's Note:
1. The chord whose pole is given is the polar (chord of contact) of that point.
2. Solve it with the parabola to get the end points.

Answer: (d) \( \frac12-\frac{1}{\sqrt2} \)

Explanation:
1. AP gives \( 2b=a+c \), so \( 3b=\frac32 \Rightarrow b=\frac12 \) and \( a+c=1 \).
2. GP gives \( b^4=a^2c^2 \Rightarrow ac=\pm\frac14 \). If \( ac=\frac14 \), then \( a=c=\frac12 \), contradicting \( a\lt c \). So \( ac=-\frac14 \).
3. a and c are roots of \( t^2-t-\frac14=0 \), i.e. \( t=\frac{1\pm\sqrt2}{2} \). The smaller is \( a=\frac12-\frac{1}{\sqrt2} \).

Teacher's Note:
1. Use the AP condition first to find the middle term.
2. Reject the root pair that violates the ordering.

Answer: (c) \( (x^2-y^2)y'=2xy \)

Explanation:
1. \( a=\frac{x^2+y^2}{2y} \). Differentiating the original: \( 2x+2yy'-2ay'=0\Rightarrow x+yy'=ay' \).
2. Substituting a: \( 2y(x+yy')=(x^2+y^2)y' \).
3. \( 2xy=(x^2-y^2)y' \).

Teacher's Note:
1. Eliminate the arbitrary constant a using both the equation and its derivative.
2. Simplify to the standard form.

Answer: (b) \( M(n)+1 \) (a multiple of \( n \) plus 1)

Explanation:
1. Let \( T=a^{n-1}-a^{n-2}b+\dots+b^{n-1} \). For odd n, \( a^n+b^n=(a+b)T \).
2. By Fermat, \( a^n+b^n\equiv a+b \pmod n \), and \( a+b \) is coprime to n, so \( T\equiv1 \).
3. The required sum is \( a^{n-1}+b^{n-1}-T\equiv1+1-1=1 \pmod n \), i.e. \( M(n)+1 \).

Teacher's Note:
1. Use Fermat's little theorem: a to the n-1 is 1 modulo n.
2. Relate the alternating sum to the factorisation of a^n + b^n.

Answer: (d) \( \sqrt x-1 \) for \( x\ge0 \)

Explanation:
1. The reflection in \( y=x \) is the inverse function.
2. From \( y=(x+1)^2 \) with \( x>-1 \), \( x+1=\sqrt y \), so \( x=\sqrt y-1 \).
3. Hence \( g(x)=\sqrt x-1 \), for \( x\ge0 \).

Teacher's Note:
1. The mirror image about y = x is the inverse function.
2. Choose the branch consistent with the domain of f.

Answer: (c) 225

Explanation:
1. For each prime, the exponents \( (i,j) \) of p and q must have maximum equal to the exponent e in the LCM. The number of such pairs is \( 2e+1 \).
2. For \( r^2 \): 5, for \( t^4 \): 9, for \( s^2 \): 5.
3. Total \( =5\times9\times5=225 \).

Teacher's Note:
1. Count exponent pairs prime by prime and multiply.
2. The number of pairs with maximum e is 2e + 1.

Answer: (c) \( \alpha=a^2+b^2,\ \beta=2ab \)

Explanation:
1. \( A^2=\begin{bmatrix} a^2+b^2 & 2ab\\ 2ab & a^2+b^2\end{bmatrix} \).
2. Comparing entries, \( \alpha=a^2+b^2 \) and \( \beta=2ab \).

Teacher's Note:
1. Multiply row by column and compare corresponding entries.
2. The diagonal entries of A squared are equal for this symmetric form.

Answer: (a) one

Explanation:
1. Collect coefficients: \( x^2:\ a^2-a-6=0 \), \( x:\ a^2-9=0 \), constant: \( a^2-2a-3=0 \).
2. The first gives \( a=3 \) or \( -2 \); the second gives \( a=\pm3 \); the third gives \( a=3 \) or \( -1 \).
3. Only \( a=3 \) satisfies all three, so there is exactly one value.

Teacher's Note:
1. For an identity, every coefficient must vanish.
2. Find the common root of the three equations.

Answer: (b) -2

Explanation:
1. The coefficient determinant is \( (\alpha+2)(\alpha-1)^2 \).
2. \( \alpha=1 \): all three equations become \( x+y+z=0 \), consistent with infinitely many solutions.
3. \( \alpha=-2 \): adding the equations gives \( 0=3(-3) \), which is false, so there is no solution.

Teacher's Note:
1. Find the values of alpha where the determinant vanishes.
2. Check consistency for each value separately.

Answer: (c) 4

Explanation:
1. Suppose the word has \( 4+k \) letters: 4 identical and \( k \) others, all different.
2. Number of words \( =\frac{(4+k)!}{4!}=210 \Rightarrow (4+k)!=5040=7! \), so \( k=3 \).
3. Different letters \( =1+3=4 \).

Teacher's Note:
1. The repeated letter counts as one kind of letter.
2. Solve (n)!/4! = 210 using factorials.

Answer: (d) \( \frac{b^{n+1}-a^{n+1}}{b-a} \)

Explanation:
1. The coefficient of \( x^n \) in \( \frac1{1-ax}\cdot\frac1{1-bx} \) is \( \sum_{k=0}^na^kb^{n-k} \).
2. This geometric sum equals \( \frac{b^{n+1}-a^{n+1}}{b-a} \).

Teacher's Note:
1. Multiply the two geometric series.
2. Sum the resulting finite geometric progression.

Answer: (c) \( y=x+2a \)

Explanation:
1. Tangent to the parabola: \( y=mx+\frac{2a}{m} \). It touches the circle if \( \frac{2a/|m|}{\sqrt{1+m^2}}=\sqrt2\,a \), so \( m^4+m^2-2=0 \) and \( m^2=1 \).
2. \( m=1 \) gives \( y=x+2a \); \( m=-1 \) gives \( y+x+2a=0 \). Of the given options only \( y=x+2a \) is a common tangent.

Teacher's Note:
1. The common tangents are y = x + 2a and y = -x - 2a; only the first is listed.
2. Use the condition for tangency to the circle: distance from centre equals radius.

Answer: (b) 3 : 4

Explanation:
1. Rewrite the lines as \( 4x+2y=9 \) and \( 4x+2y=-12 \). Distances from O: \( \frac{9}{\sqrt{20}} \) and \( \frac{12}{\sqrt{20}} \).
2. O lies between the lines. By similar triangles \( OP:OQ=9:12=3:4 \).

Teacher's Note:
1. Write both lines with the same coefficients to compare distances.
2. Similar triangles give the division ratio.

Answer: (d) (35,45)

Explanation:
1. \( a_1=n-m=10 \).
2. \( a_2=\binom n2+\binom m2-mn=\frac{(n-m)^2-(n+m)}2=\frac{100-(m+n)}2=10\Rightarrow m+n=80 \).
3. Solving: \( n=45,\ m=35 \).

Teacher's Note:
1. Compare the coefficients of y and y squared.
2. The y squared coefficient simplifies to a function of n minus m and n plus m.

Answer: (a) \( 2(\vec a\cdot\vec b)\,\vec b\times\vec a \)

Explanation:
1. \( (\vec a\times\vec b)\times\vec a=\vec b\,|\vec a|^2-\vec a\,(\vec a\cdot\vec b) \). Crossing with \( \vec b \) on the right: \( -(\vec a\cdot\vec b)\,\vec a\times\vec b=(\vec a\cdot\vec b)\,\vec b\times\vec a \).
2. \( \vec b\times(\vec a\times\vec b)=\vec a\,|\vec b|^2-\vec b\,(\vec a\cdot\vec b) \). Crossing \( \vec a \) with it: \( -(\vec a\cdot\vec b)\,\vec a\times\vec b=(\vec a\cdot\vec b)\,\vec b\times\vec a \).
3. The two terms are equal, so the sum is \( 2(\vec a\cdot\vec b)\,\vec b\times\vec a \).

Teacher's Note:
1. Expand each triple product with the BAC minus CAB rule.
2. Both terms turn out to be equal.

Answer: (c) \( P(E^c)-P(F) \)

Explanation:
1. \( P(E^c\cap F^c|G)=1-P(E|G)-P(F|G)+P(E\cap F|G) \).
2. Pairwise independence gives \( P(E|G)=P(E) \) and \( P(F|G)=P(F) \); also \( P(E\cap F|G)=0 \).
3. So the value is \( 1-P(E)-P(F)=P(E^c)-P(F) \).

Teacher's Note:
1. Use the inclusion-exclusion formula for conditional probabilities.
2. Independence removes the conditioning on G.

Answer: (a) \( 7^{164} \)

Explanation:
1. Exponent \( =\lfloor\frac{1000}{7}\rfloor+\lfloor\frac{1000}{49}\rfloor+\lfloor\frac{1000}{343}\rfloor=142+20+2=164 \).

Teacher's Note:
1. Use Legendre's formula for the exponent of a prime in a factorial.
2. Keep dividing by higher powers until the quotient is zero.

Answer: (b) \( 60^\circ \)

Explanation:
1. \( (\vec a+2\vec b)\cdot(5\vec a-4\vec b)=5-4\,\vec a\cdot\vec b+10\,\vec a\cdot\vec b-8=0 \).
2. \( 6\,\vec a\cdot\vec b=3\Rightarrow\cos\theta=\frac12 \), so \( \theta=60^\circ \).

Teacher's Note:
1. Perpendicular vectors have zero dot product.
2. Use that a and b are unit vectors.

Answer: (c) \( y''+y=0 \)

Explanation:
1. \( y'=-a\sin(x+b) \) and \( y''=-a\cos(x+b)=-y \).
2. Two arbitrary constants give a second order equation: \( y''+y=0 \).

Teacher's Note:
1. Differentiate twice for two arbitrary constants.
2. Eliminate a and b using the original equation.

Answer: (c) \( \frac{7n(n+1)}2 \)

Explanation:
1. An additive function with \( f(1)=7 \) satisfies \( f(r)=7r \) for positive integers r.
2. \( \sum_{r=1}^n7r=\frac{7n(n+1)}2 \).

Teacher's Note:
1. Use f(n) = n f(1) by repeated addition.
2. Sum the arithmetic progression.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For \( x^2=6y \) (that is \( x^2=4\cdot\frac32 y \)) the directrix is \( y=-\frac32 \), i.e. \( 2y+3=0 \).
2. The point \( \left(-1,-\frac32\right) \) lies on the directrix, and tangents from a point on the directrix are perpendicular. So the assertion is true.
3. The reason states exactly this property, so it explains the assertion.

Teacher's Note:
1. Perpendicular tangents meet on the directrix.
2. Check the point against the directrix equation.