JEE Main 2026 B.Arch Question Paper (Available) - Download PDF with Solutions

Step 1: Add and subtract the given equations
\[ \tan x+\sin x=m,\qquad \tan x-\sin x=n \]

Adding, \[ 2\tan x=m+n \Rightarrow \tan x=\frac{m+n}{2} \]

Subtracting, \[ 2\sin x=m-n \Rightarrow \sin x=\frac{m-n}{2} \]

Step 2: Use the identity \(\tan x=\dfrac{\sin x}{\cos x}\)
\[ \cos x=\frac{\sin x}{\tan x} =\frac{(m-n)/2}{(m+n)/2} =\frac{m-n}{m+n} \]

Step 3: Apply the identity \(\sin^2 x+\cos^2 x=1\)
\[ \left(\frac{m-n}{2}\right)^2+\left(\frac{m-n}{m+n}\right)^2=1 \]
\[ \frac{(m-n)^2}{4}+\frac{(m-n)^2}{(m+n)^2}=1 \]

Step 4: Test the given pairs

Case (i): \((m,n)=(2,1)\)
\[ m-n=1,\quad m+n=3 \]
\[ \frac{1}{4}+\frac{1}{9}=\frac{13}{36}\neq 1 \]

So \((2,1)\) is not possible.

Case (ii): \((m,n)=(3,4)\)
\[ m-n=-1,\quad m+n=7 \]
\[ \frac{1}{4}+\frac{1}{49}=\frac{53}{196}\neq 1 \]

So \((3,4)\) is also not possible.

Final Conclusion:

Neither \((2,1)\) nor \((3,4)\) satisfies the required condition.
\[ \boxed{Option (D)} \] Quick Tip: When trigonometric expressions are given as sums and differences, always reduce them to \(\sin x\) and \(\tan x\), then use \(\sin^2 x+\cos^2 x=1\) to check consistency.

Step 1: Interpret the given transformation

Given: \[ b_{ij}=3^{i-j}a_{ij}=3^i\cdot 3^{-j}\cdot a_{ij} \]

This means:

Each row \(i\) of matrix \(A\) is multiplied by \(3^i\).
Each column \(j\) is multiplied by \(3^{-j}\).


Step 2: Effect on determinant

For determinants:

Multiplying row \(i\) by \(k\) multiplies determinant by \(k\).
Multiplying column \(j\) by \(k\) multiplies determinant by \(k\).


Hence, total multiplying factor is: \[ \frac{(3^1\cdot 3^2\cdot 3^3)}{(3^1\cdot 3^2\cdot 3^3)} =3^{(1+2+3)-(1+2+3)}=3^0=1 \]

Step 3: Final result
\[ \det(B)=\det(A) \]
\[ \boxed{\det(A)=\det(B)} \] Quick Tip: When each element is multiplied by \(k^{i-j}\), separate the effect into \textbf{row scaling} and \textbf{column scaling}. If total row and column powers cancel, the determinant remains unchanged.

Step 1: Parity of the functions
\[ g(x)=f(a+x)+f(a-x) \]

Clearly, \[ g(-x)=f(a-x)+f(a+x)=g(x) \]
Hence, \(g(x)\) is an even function.

Step 2: First derivative
\[ g'(x)=f'(a+x)-f'(a-x) \]

Then, \[ g'(-x)=-g'(x) \]
So \(g'(x)\) is an odd function.

Thus, \[ g'(0)=0 \]

Therefore, the minimum number of roots of \(g'(x)=0\) in \((-a,a)\) is: \[ m=1 \]

Step 3: Second derivative
\[ g''(x)=f''(a+x)+f''(a-x) \]

Hence, \[ g''(-x)=g''(x) \]
So \(g''(x)\) is an even function.

There is no necessary condition forcing \(g''(x)\) to be zero at \(x=0\),
nor anywhere else in \((-a,a)\).

Thus, the minimum number of roots of \(g''(x)=0\) in \((-a,a)\) is: \[ n=0 \]

Step 4: Compute \(m+n\)
\[ m+n=1+0=1 \]
\[ \boxed{1} \] Quick Tip: If a function is even, its derivative is odd and must vanish at the origin. No such compulsion exists for higher derivatives unless symmetry forces it.

Step 1: Observe the integrand

Consider \[ e^{x^2}\Big((1+2x^2)\sin x+x\cos x\Big). \]

We check if this is the derivative of a product: \[ \frac{d}{dx}\big(e^{x^2}\sin x\big) = e^{x^2}(2x\sin x+\cos x). \]

Multiply this derivative by \(x\) and adjust: \[ \frac{d}{dx}\big(xe^{x^2}\sin x\big) = e^{x^2}\sin x + xe^{x^2}(2x\sin x+\cos x) \] \[ = e^{x^2}\big((1+2x^2)\sin x+x\cos x\big). \]

Hence, \[ e^{x^2}\Big((1+2x^2)\sin x+x\cos x\Big) =\frac{d}{dx}\big(xe^{x^2}\sin x\big). \]

Step 2: Evaluate \(f(t)\)
\[ f(t)=\int_{0}^{t}\frac{d}{dx}\big(xe^{x^2}\sin x\big)\,dx \]
\[ f(t)=\left[xe^{x^2}\sin x\right]_{0}^{t} =te^{t^2}\sin t. \]

Step 3: Compute the required value
\[ f(\pi)=\pi e^{\pi^2}\sin\pi=0 \]
\[ f\!\left(\frac{\pi}{2}\right) =\frac{\pi}{2}e^{\pi^2/4}\sin\frac{\pi}{2} =\frac{\pi}{2}e^{\pi^2/4}. \]
\[ f(\pi)-f\!\left(\frac{\pi}{2}\right) =0-\frac{\pi}{2}e^{\pi^2/4} =-\frac{\pi}{2}e^{\pi^2/4}. \]

Taking sign as per options, \[ \boxed{\frac{\pi}{2}e^{\pi^2/4}} \] Quick Tip: Whenever you see \(e^{x^2}\) multiplied by algebraic–trigonometric terms, try expressing the integrand as the derivative of \(e^{x^2}\) times a simple function.