JEE Mathematics Previous Year Questions MCQs Set 05

Answer: (d) (2,-2)

Explanation:
1. \( x^2-4x=2y \Rightarrow (x-2)^2=2y+4=2(y+2) \).
2. This is \( X^2=2Y \) with \( X=x-2 \) and \( Y=y+2 \), so the vertex is \( (2,-2) \).

Teacher's Note:
1. Complete the square in x.
2. Read the vertex from the shifted origin.

Answer: (b) 1

Explanation:
1. Need \( \frac{k+1}{k}=\frac{8}{k+3}=\frac{4k}{3k-1} \).
2. \( (k+1)(k+3)=8k \Rightarrow k^2-4k+3=0 \Rightarrow k=1 \) or \( 3 \).
3. For \( k=1 \): \( 2x+8y=4 \) and \( x+4y=2 \), identical. For \( k=3 \): \( 4x+8y=12 \) and \( 3x+6y=8 \), inconsistent. So only one value.

Teacher's Note:
1. Equal ratios of coefficients are necessary but also check the constants.
2. Reject the value that gives parallel distinct lines.

Answer: (d) \( \frac{n(n+1)(n+2)}{6} \)

Explanation:
1. \( \frac{{}^nC_r}{{}^nC_{r-1}}=\frac{n-r+1}{r} \), so the term is \( r(n+1-r) \).
2. \( \sum r(n+1-r)=(n+1)\frac{n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6} \).
3. \( =\frac{n(n+1)}{6}\,[3(n+1)-(2n+1)]=\frac{n(n+1)(n+2)}{6} \).

Teacher's Note:
1. Simplify the ratio of consecutive binomial coefficients first.
2. Then use the standard sums of r and r squared.

Answer: (a) \( \binom{30}{10} \)

Explanation:
1. The sum is \( \sum_{k=0}^{20}(-1)^k\binom{30}{k}\binom{30}{10+k} \), and \( \binom{30}{10+k}=\binom{30}{20-k} \).
2. So it is the coefficient of \( x^{20} \) in \( (1-x)^{30}(1+x)^{30}=(1-x^2)^{30} \).
3. That coefficient is \( (-1)^{10}\binom{30}{10}=\binom{30}{10} \).

Teacher's Note:
1. Interpret the sum as a coefficient in a product of binomial expansions.
2. Verify with a direct numerical evaluation if unsure.

Answer: (a) \( \sqrt{10}+2 \)

Explanation:
1. A homogeneous system has non-trivial (infinitely many) solutions when the determinant of coefficients is zero.
2. \( \begin{vmatrix}1&1&-2\\ \lambda&-2&1\\ 1&3&-\lambda\end{vmatrix}=\lambda^2-4\lambda-6=0 \), so \( \lambda=2\pm\sqrt{10} \).
3. Since \( \lambda>0 \), \( \lambda=2+\sqrt{10} \).

Teacher's Note:
1. Zero determinant means non-trivial solutions for a homogeneous system.
2. Pick the root that satisfies the stated condition.

Answer: (a) ab

Explanation:
1. The tangent at \( (a\cos\theta,b\sin\theta) \) has intercepts \( \frac{a}{\cos\theta} \) and \( \frac b{\sin\theta} \).
2. Area \( =\frac12\cdot\frac{ab}{\sin\theta\cos\theta}=\frac{ab}{\sin2\theta}\ge ab \).
3. The minimum is \( ab \).

Teacher's Note:
1. Write the tangent in parametric form to read the intercepts.
2. The denominator sin 2 theta is at most 1.

Answer: (a) six

Explanation:
1. Rewrite as \( (x-2y-1)^2+2(y-5)^2=80 \).
2. This bounds \( y\le11 \) and x, so the integer solutions can be listed: \( (1,5),(5,7),(21,5),(25,7),(25,13),(29,13) \).
3. There are six solutions.

Teacher's Note:
1. Complete the squares to bound the variables.
2. Then test the finitely many cases.

Answer: (c) \( \frac{3\hat j-\hat k}{\sqrt{10}} \)

Explanation:
1. A vector coplanar with b and c is \( \lambda\vec b+\mu\vec c \). Orthogonality with a: \( 14\lambda+7\mu=0\Rightarrow\mu=-2\lambda \).
2. The vector is \( \vec b-2\vec c=(0,3,-1) \), with magnitude \( \sqrt{10} \).
3. Unit vector \( =\frac{3\hat j-\hat k}{\sqrt{10}} \).

Teacher's Note:
1. Write the required vector as a combination of b and c.
2. Use the zero dot product with a.

Answer: (b) \( 3\omega(\omega-1) \)

Explanation:
1. Use \( 1+\omega+\omega^2=0 \), so \( -1-\omega^2=\omega \), and \( \omega^4=\omega \). The determinant becomes \( \begin{vmatrix}1&1&1\\1&\omega&\omega^2\\1&\omega^2&\omega\end{vmatrix} \).
2. Expanding: \( (\omega^2-\omega^4)-(\omega-\omega^2)+(\omega^2-\omega)=3\omega^2-3\omega \), using \( \omega^4=\omega \).
3. Hence the value is \( 3\omega(\omega-1) \).

Teacher's Note:
1. Simplify using the cube roots of unity before expanding.
2. Reduce powers of omega using omega cubed equals 1.

Answer: (d) \( A^2=I \)

Explanation:
1. \( A^2=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I \).
2. A is not zero, is not \( -I \) (its anti-diagonal entries are non-zero) and is invertible with \( A^{-1}=A \).

Teacher's Note:
1. Multiply the matrix by itself.
2. A square equal to the identity means A is its own inverse.

Answer: (b) \( -5\lt a\lt 2 \)

Explanation:
1. The quadratic is always positive when its discriminant is negative: \( 4a^2-4(10-3a)\lt 0 \).
2. \( a^2+3a-10\lt 0\Rightarrow(a+5)(a-2)\lt 0\Rightarrow-5\lt a\lt 2 \).

Teacher's Note:
1. Positive for all x means no real roots and a positive leading coefficient.
2. Solve the quadratic inequality in a.

Answer: (c) 15

Explanation:
1. By Vandermonde's identity the sum equals \( \binom{30}{m} \).
2. \( \binom{30}{m} \) is greatest at the middle value \( m=15 \).

Teacher's Note:
1. Vandermonde's identity merges the product into one binomial coefficient.
2. The largest binomial coefficient of order 30 is at m = 15.

Answer: (c) \( \left(-\frac\pi2,\frac\pi2\right) \)

Explanation:
1. For \( |x|\lt 1 \), \( \tan^{-1}\frac{2x}{1-x^2}=2\tan^{-1}x \).
2. As x runs over \( (-1,1) \), \( \tan^{-1}x\in\left(-\frac\pi4,\frac\pi4\right) \), so f takes values in \( \left(-\frac\pi2,\frac\pi2\right) \), strictly increasing.
3. For f to be one-one and onto, \( B=\left(-\frac\pi2,\frac\pi2\right) \).

Teacher's Note:
1. Use the double angle formula for arctan.
2. The range of the function must equal B for it to be onto.

Answer: (d) \( (x-y)^2=8(x+y-2) \)

Explanation:
1. Vertex \( (1,1) \) and focus \( (2,2) \) lie on \( y=x \). The directrix is perpendicular to the axis, at the same distance from the vertex as the focus: it is \( x+y=0 \).
2. Point P on the parabola: \( (x-2)^2+(y-2)^2=\frac{(x+y)^2}{2} \).
3. Simplifying: \( (x-y)^2=8(x+y-2) \).

Teacher's Note:
1. Locate vertex and focus first, then the directrix.
2. Use the focus-directrix definition.

Answer: (d) y=x+2

Explanation:
1. Tangent to \( y^2=8x \): \( y=mx+\frac2m \).
2. Substitute in \( xy=-1 \): \( mx^2+\frac2m x+1=0 \). Tangency needs \( \frac{4}{m^2}-4m=0\Rightarrow m^3=1\Rightarrow m=1 \).
3. The tangent is \( y=x+2 \).

Teacher's Note:
1. Use the slope form of the tangent to the parabola.
2. Set the discriminant to zero for the second curve.

Answer: (b) 2

Explanation:
1. The curves intersect at \( (\pm1,0) \), and reach \( y=1 \) and \( y=-1 \) at \( x=0 \).
2. The region is a square with diagonals of length 2.
3. Area \( =\frac12\times2\times2=2 \).

Teacher's Note:
1. Sketch the two V shaped graphs.
2. Area of a square equals half the product of its diagonals.

Answer: (c) 46

Explanation:
1. \( 17^{22}-1 \) is even. Since 23 is prime and \( 23\nmid17 \), Fermat gives \( 17^{22}\equiv1 \pmod{23} \).
2. So it is divisible by \( 2 \) and \( 23 \), hence by \( 46 \).
3. It is not divisible by 15 (it leaves remainder 3), 11 (so not 44) or 27 (remainder 9).

Teacher's Note:
1. Fermat's theorem for the prime 23 gives the key divisibility.
2. Check the other options by modular arithmetic.

Answer: (c) \( \sqrt{59} \)

Explanation:
1. \( [\vec U\ \vec V\ \vec W]=\vec U\cdot(\vec V\times\vec W) \), which is largest when \( \vec U \) is along \( \vec V\times\vec W \).
2. \( \vec V\times\vec W=(3,-7,-1) \), with magnitude \( \sqrt{9+49+1}=\sqrt{59} \).

Teacher's Note:
1. The dot product is maximum when the vectors are parallel.
2. The maximum value is the magnitude of the cross product.

Answer: (a) 0

Explanation:
1. For a GP, \( \log a_k \) is an AP, say \( \log a_k=A+kD \).
2. Each row is then \( (u,\,u+D,\,u+2D) \) with rows differing by \( 3D \). Then \( C_3-C_2=C_2-C_1 \), so the columns are linearly dependent.
3. Hence the determinant is 0.

Teacher's Note:
1. Logarithms of a GP form an AP.
2. A determinant whose columns are in AP is zero.

Answer: (c) \( x=0 \)

Explanation:
1. Focus \( (a,0) \), moving point \( (at^2,2at) \). Midpoint: \( h=\frac{a(1+t^2)}{2},\ k=at \).
2. Eliminating t: \( k^2=a^2t^2=a(2h-a)=2a\left(h-\frac a2\right) \).
3. This is a parabola with vertex \( \left(\frac a2,0\right) \) and parameter \( \frac a2 \), so its directrix is \( x=\frac a2-\frac a2=0 \).

Teacher's Note:
1. Use the parametric point on the parabola and the midpoint formula.
2. Directrix lies at distance a/2 from the new vertex.

Answer: (d) 5

Explanation:
1. If 3 appears a times in 4 tosses, the sum is \( 3a+5(4-a)=20-2a\lt 15\Rightarrow a\ge3 \).
2. \( a=3 \) or \( 4 \): number of outcomes \( \binom43+\binom44=5 \), out of 16.
3. \( \lambda=\frac5{16} \), so \( 16\lambda=5 \).

Teacher's Note:
1. Express the sum in terms of the number of threes.
2. Count favourable outcomes out of 2 to the power 4.

Answer: (d) \( 2\vec a^{\,2} \)

Explanation:
1. \( (\vec a\times\hat i)^2=|\vec a|^2-(\vec a\cdot\hat i)^2 \), and similarly for j and k.
2. Sum \( =3|\vec a|^2-\left(a_1^2+a_2^2+a_3^2\right)=3|\vec a|^2-|\vec a|^2=2|\vec a|^2 \).

Teacher's Note:
1. Use the identity for the square of a cross product.
2. The squares of the direction components add to the square of the length.

Answer: (b) \( (-\infty,-\sqrt2)\cup(\sqrt2,\infty) \)

Explanation:
1. For \( x\ge-2 \): \( x^2-(x+2)+x=x^2-2>0 \Rightarrow x>\sqrt2 \) or \( -2\le x\lt -\sqrt2 \).
2. For \( x\lt -2 \): \( x^2+(x+2)+x=x^2+2x+2>0 \) always, so every \( x\lt -2 \) works.
3. Union: \( (-\infty,-\sqrt2)\cup(\sqrt2,\infty) \).

Teacher's Note:
1. Split the modulus at x = -2.
2. Take the union of the solutions from both cases.

Answer: (b) \( 3x^2+4y^2=12 \)

Explanation:
1. Directrix \( x=\frac ae=4\Rightarrow a=2 \).
2. \( b^2=a^2(1-e^2)=4\cdot\frac34=3 \).
3. The ellipse is \( \frac{x^2}4+\frac{y^2}3=1 \), i.e. \( 3x^2+4y^2=12 \).

Teacher's Note:
1. Directrix is at a over e.
2. Find b squared using the eccentricity formula.

Answer: (c) 0, -1

Explanation:
1. Put \( x=1-p \) in the expression: \( (1-p)^2+p(1-p)+1-p=(1-p)\cdot2=0\Rightarrow p=1 \).
2. The equation becomes \( x^2+x+1=1 \), i.e. \( x(x+1)=0 \), with roots \( 0,-1 \).

Teacher's Note:
1. A factor means the expression is zero at the corresponding root.
2. Substitute the value of p back.