JEE Mathematics Determinants and Cramers Rule MCQs Set 04

Answer: (c) \( \frac{d}{dx}\Delta_1=3\Delta_2 \)

Explanation:
1. \( \Delta_1=x^3-3abx+a^2b+ab^2 \) and \( \Delta_2=x^2-ab \).
2. \( \frac{d\Delta_1}{dx}=3x^2-3ab=3(x^2-ab)=3\Delta_2 \).

Teacher's Note:
1. Expand the 3 x 3 determinant fully before differentiating.
2. The constant terms vanish on differentiation.

Answer: (a) -4

Explanation:
1. Let \( p=e^{iA},\ q=e^{iB},\ r=e^{iC} \), so \( pqr=e^{i\pi}=-1 \).
2. Expanding along the first row: \( \frac1{p^2}\left(\frac1{q^2r^2}-p^2\right)-r\left(\frac1r-pq\right)+q\left(pr-\frac1q\right) \).
3. The three terms equal \( \frac1{(pqr)^2}-1,\ \ -1+pqr,\ \ pqr-1 \).
4. Their sum is \( -2+2pqr=-2-2=-4 \).

Teacher's Note:
1. Write each entry using p, q, r and use pqr = -1.
2. A numerical check with A = 0.7, B = 1.1 also gives -4.

Answer: (c) \( \pm1 \)

Explanation:
1. Let the matrix be M. Because the vectors are orthonormal, \( M^TM=I \).
2. So \( (\det M)^2=1 \) and \( \det M=\pm1 \).

Teacher's Note:
1. The determinant of an orthogonal matrix is +1 or -1.
2. The sign depends on whether the vectors form a right-handed set.

Answer: (d) 1

Explanation:
1. At \( x=0 \) the first two rows are both \( (1,0,1) \), so \( a=\Delta(0)=0 \).
2. Using \( e^x\approx1+x,\ \sin x\approx x,\ \cos x\approx1,\ \log(1+x^2)\approx x^2 \) and expanding, \( \Delta\approx-x+2x^2+\ldots \), so \( b=-1 \).
3. \( (a+b)^2=(0-1)^2=1 \).

Teacher's Note:
1. The constant and linear terms need only the first-order expansions of each entry.
2. Row 1 and row 2 are identical at x = 0.

Answer: (b) \( \phi \)

Explanation:
1. Expand along the third row. The cofactor of the first entry is \( \sin\theta\sin\phi(-\sin\theta)-\cos\theta\cos\theta\sin\phi=-\sin\phi \).
2. The cofactor of the second entry is \( -[\sin\theta\cos\phi(-\sin\theta)-\cos\theta\cos\theta\cos\phi]=\cos\phi \).
3. \( \Delta=(-\sin\theta\sin\phi)(-\sin\phi)+(\sin\theta\cos\phi)(\cos\phi)=\sin\theta \), which does not contain \( \phi \).

Teacher's Note:
1. The third row has a zero, so only two cofactors are needed.
2. sin^2 + cos^2 = 1 removes phi.

Answer: (d) None of these

Explanation:
1. Using \( \cos(\beta-\alpha)=\cos\alpha\cos\beta+\sin\alpha\sin\beta \) the matrix is the product \( \begin{bmatrix}\cos\alpha&\sin\alpha&0\\\cos\beta&\sin\beta&0\\\cos\gamma&\sin\gamma&0\end{bmatrix}\begin{bmatrix}\cos\alpha&\cos\beta&\cos\gamma\\\sin\alpha&\sin\beta&\sin\gamma\\0&0&0\end{bmatrix} \).
2. Both factors have a zero column or row, so the product matrix has determinant 0.
3. The value 0 is none of the first three options.

Teacher's Note:
1. Write each entry as cos a cos b + sin a sin b.
2. A matrix with a zero column has zero determinant.

Answer: (a) \( \alpha \) only

Explanation:
1. Expand along the first row. The first term is \( \cos(\alpha+\beta)(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=\cos^2(\alpha+\beta) \).
2. The second and third terms together give \( \sin(\alpha+\beta)(\sin\alpha\cos\beta+\cos\alpha\sin\beta)+\cos2\beta(\sin^2\alpha+\cos^2\alpha)=\sin^2(\alpha+\beta)+\cos2\beta \).
3. Adding: \( \Delta=\cos^2(\alpha+\beta)+\sin^2(\alpha+\beta)+\cos2\beta=1+\cos2\beta \), which has no \( \alpha \).

Teacher's Note:
1. Collect terms into sin(alpha + beta) and cos(alpha + beta).
2. The answer depends on beta only.

Answer: (b) I and II only

Explanation:
1. Expanding (or reducing with row operations) gives \( f(x)=4a^6 \), a constant.
2. So \( f'(x)=0 \) and the graph is a horizontal line (I, II true).
3. \( \int_0^2 4a^6dx=8a^6\ne32a^4 \) and \( 4a^6\ne-16a^6 \), so III and IV are false.

Teacher's Note:
1. Use C2 - xC1 and C3 - xC2 style operations to remove x.
2. A constant function has zero derivative.

Answer: (c) At least one of the roots of f(x) = 0 is repeated

Explanation:
1. This is a circulant determinant: with \( u=(1+x)^a,\ v=(1+2x)^b \), \( f=u^3+v^3+1-3uv=\tfrac12(u+v+1)[(u-v)^2+(v-1)^2+(u-1)^2] \).
2. At \( x=0 \), u = v = 1, so the second factor vanishes to second order.
3. Hence x = 0 is a repeated root. It is neither positive nor negative, so options (a) and (b) are false.

Teacher's Note:
1. Circulant determinants factor as (u + v + w)(u^2 + v^2 + w^2 - uv - vw - wu).
2. The squared factor forces a double root.

Answer: (d) \( a+b+c \)

Explanation:
1. \( D_2 \) has degree 0 + 1 + 3 = 4, so the remaining factor is linear and symmetric, i.e. \( k(a+b+c) \).
2. Put c = 0: \( D_2=ab^3-a^3b=ab(b-a)(b+a) \) while the known factor is \( ab(b-a) \), so the remaining factor reduces to \( a+b \).
3. Therefore it is \( a+b+c \).

Teacher's Note:
1. Degree counting shows the extra factor is linear.
2. Symmetry then fixes it.

Answer: (c) -2

Explanation:
1. Expand near 0 using \( (1+\sin x)^3\approx1+3x+\ldots,\ \log(1+x)\approx x-\tfrac{x^2}{2},\ \cos x\approx1-\tfrac{x^2}2 \).
2. Expanding along the third row: \( f(x)=x^2[2(1+\sin x)^3-\cos x\log(1+x)]-(1+x^2)[2x-\cos x]+0 \).
3. Near 0: \( -(1+x^2)(2x-1+\ldots)=1-2x+\ldots \) and the first part starts at \( x^2 \).
4. So \( f(x)=1-2x+\tfrac52x^2+\ldots \) and the coefficient of x is -2.

Teacher's Note:
1. Terms multiplied by x^2 cannot affect the coefficient of x.
2. Keep only the lowest-order terms of the remaining part.

Answer: (b) 10

Explanation:
1. Expanding along the first row: \( 1(1-3\sin\theta\cos\theta)-3\cos\theta(\sin\theta-3\cos\theta)+1(\sin^2\theta-1) \).
2. This simplifies to \( 9\cos^2\theta+\sin^2\theta-6\sin\theta\cos\theta=5+4\cos2\theta-3\sin2\theta \).
3. The maximum of \( 4\cos2\theta-3\sin2\theta \) is \( \sqrt{16+9}=5 \), so the maximum value is 10.

Teacher's Note:
1. Reduce to the form a cos 2 theta + b sin 2 theta.
2. Its range is plus or minus the square root of a^2 + b^2.

Answer: (b) 0

Explanation:
1. Apply \( R_2\to R_2-R_1,\ R_3\to R_3-R_1 \): the new rows are \( (b-a,\ x^2-2,\ 0) \) and \( (c-a,\ 2x^2-3,\ 0) \).
2. Expanding along the third column: \( f(x)=(b-a)(2x^2-3)-(c-a)(x^2-2) \).
3. This contains only even powers of x, so \( f'(x) \) is an odd function and \( f'(0)=0 \).

Teacher's Note:
1. Row operations create zeros in the last column.
2. An even function has zero derivative at 0.

Answer: (b) -2

Explanation:
1. Apply \( C_2\to C_2-C_1,\ C_3\to C_3-C_1 \): \( \begin{vmatrix}x+2&1&3\\x+4&2&5\\x+8&3&7\end{vmatrix} \).
2. Apply \( R_3\to R_3-R_2,\ R_2\to R_2-R_1 \): \( \begin{vmatrix}x+2&1&3\\2&1&2\\4&1&2\end{vmatrix} \).
3. Expanding: \( (x+2)(2-2)-1(4-8)+3(2-4)=0+4-6=-2 \).

Teacher's Note:
1. Subtract columns first to reduce the x terms.
2. The result is independent of x.

Answer: (a) 2

Explanation:
1. With \( t=x^2-13 \), the determinant is \( -(t-7)(t-2)(t+9) \), so \( t=7,2,-9 \), giving \( x^2=20,15,4 \).
2. The six roots are \( \pm\sqrt{20},\pm\sqrt{15},\pm2 \) and their sum is 0.
3. Removing the root x = -2, the sum of the other five is 0 - (-2) = 2.

Teacher's Note:
1. The equation is even in x, so roots occur in pairs.
2. Subtract the known root from the total sum zero.

Answer: (a) 7

Explanation:
1. Apply \( C_3\to C_3-C_2 \) and then \( C_2\to C_2-C_1 \): \( \begin{vmatrix}x&y&z\\2x&x+2y&x+y+2z\\3x&3x+3y&4x+3y+3z\end{vmatrix} \).
2. Apply \( R_2\to R_2-2R_1,\ R_3\to R_3-3R_1 \): \( \begin{vmatrix}x&y&z\\0&x&x+y\\0&3x&4x+3y\end{vmatrix} \).
3. Value \( =x[x(4x+3y)-3x(x+y)]=x\cdot x^2=x^3 \). So \( x^3=343 \) and \( x=7 \).

Teacher's Note:
1. Column operations first reduce the three-term entries.
2. The determinant collapses to x cubed.

Answer: (b) 4

Explanation:
1. Multiply row 1 by c, row 2 by a, row 3 by b (this multiplies the determinant by abc): \( \begin{vmatrix}a^2+b^2&c^2&c^2\\a^2&b^2+c^2&a^2\\b^2&b^2&c^2+a^2\end{vmatrix} \).
2. Put \( p=a^2,q=b^2,r=c^2 \): the determinant \( \begin{vmatrix}p+q&r&r\\p&q+r&p\\q&q&r+p\end{vmatrix}=4pqr \) (known result, checked by expansion).
3. So the original value is \( \frac{4a^2b^2c^2}{abc}=4abc \) and \( k=4 \).

Teacher's Note:
1. Clear the denominators by scaling the rows.
2. The simplified determinant is a standard 4pqr form.

Answer: (b) \( (a-b)^2(b-c)^2(c-a)^2 \)

Explanation:
1. The matrix is \( V V^T \) where \( V=\begin{bmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{bmatrix} \) since the (i, j) entry of \( VV^T \) is \( a^{i+j}+b^{i+j}+c^{i+j} \).
2. \( \det(VV^T)=(\det V)^2 \) and \( \det V=(a-b)(b-c)(c-a) \) (up to sign, which squares away).
3. Hence the value is \( (a-b)^2(b-c)^2(c-a)^2 \).

Teacher's Note:
1. Factor the determinant as a product of a Vandermonde matrix and its transpose.
2. Squaring removes the sign ambiguity.

Answer: (c) 0

Explanation:
1. Let \( u=(1,\ a+c,\ ac)^T \) and \( w=(1,\ b+d,\ bd)^T \).
2. Checking entry by entry shows the matrix equals \( uw^T+wu^T \) (the middle entry is \( 2(a+c)(b+d) \) and the corner entry is \( 2abcd \)).
3. A sum of two rank-one matrices has rank at most 2, so this 3 x 3 determinant is 0.

Teacher's Note:
1. Split the matrix into two rank-one pieces built from the sums and products of the pairs (a, c) and (b, d).
2. A 3 x 3 matrix of rank at most 2 has zero determinant.

Answer: (a) \( \begin{vmatrix}\int f_1&\int f_2&\int f_3\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}+c \)

Explanation:
1. Expanding along the first row: \( D(x)=f_1(x)C_1+f_2(x)C_2+f_3(x)C_3 \) where the cofactors \( C_i \) are constants.
2. Integration is linear, so \( \int D\,dx=C_1\int f_1+C_2\int f_2+C_3\int f_3+c \).
3. This is the same determinant with the first row replaced by the integrals.

Teacher's Note:
1. Constant cofactors can be pulled out of the integral.
2. Only the row containing functions is integrated.

Answer: (b) 4

Explanation:
1. \( |\operatorname{adj}(\operatorname{adj}A)|=|A|^{(n-1)^2} \).
2. So \( (n-1)^2=9 \) and \( n-1=3 \) (n - 1 = -3 is impossible), giving n = 4.

Teacher's Note:
1. Apply |adj A| = |A|^(n-1) twice.
2. Reject the negative root.

Answer: (a) 1

Explanation:
1. Solve the first two: \( x=-\frac2{17},\ y=-\frac{24}{17} \).
2. \( 9x-5y=\frac{-18+120}{17}=6 \), so \( [4a]=6 \).
3. \( 6\le4a<7 \) gives \( 1.5\le a<1.75 \), hence \( [a]=1 \).

Teacher's Note:
1. Use two equations to fix x and y, then test the third.
2. [4a] = 6 means 4a lies in [6, 7).

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

Explanation:
1. Expanding along the first row, \( D(x)=C_1f_1(x)+C_2f_2(x)+C_3f_3(x) \) with constant cofactors \( C_i \).
2. By the property in R, \( \int D\,dx=C_1\int f_1+C_2\int f_2+C_3\int f_3+c \), which is the stated determinant.
3. So A and R are true and R explains A.

Teacher's Note:
1. Constant cofactors come out of the integral.
2. Linearity of integration is the reason.

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

Explanation:
1. In R each sum is verified with \( \sin(\theta+x)+\sin(\theta-x)=2\sin\theta\cos x \) and \( \cos\tfrac{2\pi}3=-\tfrac12 \) (similarly \( \cos\tfrac{4\pi}3=-\tfrac12 \)), so all three relations hold.
2. These say \( R_1+R_2+R_3=0 \) column by column, i.e. \( R_2+R_3=-R_1 \).
3. The rows are dependent, so the determinant is 0 for every \( \theta \). R explains A.

Teacher's Note:
1. Add the second and third rows and compare with the first.
2. Dependent rows give a zero determinant.

Answer: (c) \( 2(a_1-a_2)(a_2-a_3)(a_3-a_1)(b_1-b_2)(b_2-b_3)(b_3-b_1) \)

Explanation:
1. Expand each entry: \( (a_i-b_j)^2=a_i^2\cdot1+a_i(-2b_j)+1\cdot b_j^2 \).
2. So the matrix is the product \( \begin{bmatrix}a_1^2&a_1&1\\a_2^2&a_2&1\\a_3^2&a_3&1\end{bmatrix}\begin{bmatrix}1&1&1\\-2b_1&-2b_2&-2b_3\\b_1^2&b_2^2&b_3^2\end{bmatrix} \).
3. Each factor is a Vandermonde-type determinant; the second has the extra factor -2, giving \( 2(a_1-a_2)(a_2-a_3)(a_3-a_1)\cdot(b_1-b_2)(b_2-b_3)(b_3-b_1) \) (signs checked by symbolic expansion).

Teacher's Note:
1. Write the squared entries as a product of two 3 x 3 matrices.
2. The determinant of a product is the product of the determinants.

Answer: (d) None of these

Explanation:
1. \( \cos(A-P)=\cos A\cos P+\sin A\sin P \), so the matrix is \( \begin{bmatrix}\cos A&\sin A&0\\\cos B&\sin B&0\\\cos C&\sin C&0\end{bmatrix}\begin{bmatrix}\cos P&\cos Q&\cos R\\\sin P&\sin Q&\sin R\\0&0&0\end{bmatrix} \).
2. Each factor has a zero column or row, so the determinant of the product is 0.
3. Zero is none of the first three expressions.

Teacher's Note:
1. Split every entry using the cosine of a difference.
2. Zero column in a factor makes the product singular.

Answer: (c) \( 9\sin x\cos^2x \)

Explanation:
1. Apply \( C_2\to C_2-C_1,\ C_3\to C_3-C_2 \) and use sum-to-product: each new entry has a factor \( 2\sin\frac h2 \), giving a factor \( 4\sin^2\frac h2 \) (about \( h^2 \)).
2. As \( h\to0 \) the remaining determinant tends to \( \sin x\cos^2x\begin{vmatrix}1&1&1\\1&-2&1\\1&1&-2\end{vmatrix}=9\sin x\cos^2x \).
3. Expanding the series of the original determinant in h also gives \( \Delta=9\sin x\cos^2x\,h^2+O(h^3) \).

Teacher's Note:
1. Differences of sines give factors of sin(h/2), which behave like h/2.
2. The remaining 3 x 3 determinant is evaluated at h = 0.