JEE Main 2026 B.Planning Memory based paper (Available) - Download PDF with Solutions

Step 1: Analyze the first equation
\[ x^2-2ax+a^2-1=0 \]

Rewrite: \[ (x-a)^2=1 \]

So the roots are: \[ x=a\pm 1 \]

Given that both roots lie in \((-2,2)\):
\[ -2
From these: \[ -1
Combining: \[ -1
Step 2: Analyze the second equation
\[ x^2-(a^2+1)x-(a^2+2)=0 \]

Sum of roots: \[ \alpha+\beta=a^2+1>0 \]

Product of roots: \[ \alpha\beta=-(a^2+2)<0 \]

Hence, one root is positive and the other is negative.

Step 3: Locate the roots

Evaluate the polynomial at key points (using \(|a|<1\)):
\[ f(0)=-(a^2+2)<0 \]
\[ f(2)=4-2(a^2+1)-(a^2+2)=-3a^2<0 \]
\[ f(3)=9-3(a^2+1)-(a^2+2)=4-4a^2>0 \]

Thus, the positive root lies in \((2,3)\).

Now check negative side:
\[ f(-2)=4+2(a^2+1)-(a^2+2)=4+a^2>0 \]

Since \(f(-2)>0\) and \(f(0)<0\), the negative root lies in \((-2,0)\).

Final Conclusion:

One root lies in \((2,3)\) and the other lies in \((-2,0)\).
\[ \boxed{Option (3)} \] Quick Tip: When roots are restricted to an interval, first determine the parameter range, then use sign analysis of the polynomial to locate roots in specific subintervals.

For a system of three linear equations to have more than one solution, the determinant of the coefficient matrix must be zero and the system must be consistent.

Step 1: Determinant of coefficient matrix
\[ \begin{vmatrix} 2 & 1 & p
3 & -2 & 1
5 & -8 & 9 \end{vmatrix} =2((-2)\cdot 9-1\cdot(-8)) -1(3\cdot 9-1\cdot 5) +p(3\cdot(-8)-(-2)\cdot 5) \]
\[ =2(-18+8)-1(27-5)+p(-24+10) =-20-22-14p \]
\[ \Rightarrow -42-14p=0 \Rightarrow p=-3 \]

Step 2: Consistency condition

With \(p=-3\), observe that the rows satisfy: \[ -2R_1+3R_2=R_3 \]

For consistency, the constants must satisfy the same relation: \[ -2(-1)+3q=5 \Rightarrow 2+3q=5 \Rightarrow q=1 \]

Step 3: Compute \(q-p\)
\[ q-p=1-(-3)=4 \]
\[ \boxed{4} \] Quick Tip: For infinitely many solutions, ensure both: determinant of coefficient matrix is zero, constants follow the same linear dependence as the equations.

Step 1: Arrange letters in alphabetical order

Letters of the word GOING are: \[ G,\ O,\ I,\ N,\ G \]

Alphabetical order: \[ G,\ G,\ I,\ N,\ O \]

Step 2: Find rank of the word OGGIN

We calculate the number of words that come before OGGIN.

First letter: O

Letters smaller than \(O\) are \(G, G, I, N\).

Number of permutations using remaining 4 letters (with two G’s): \[ \frac{4!}{2!}=12 \]

Number of such letters before \(O\): \(4\)
\[ \Rightarrow 4 \times 12 = 48 \]

Step 3: Remaining letters

After fixing \(O\), the remaining word is GGIN.
This is already the smallest possible arrangement of these letters.

So, no additional words are added.

Step 4: Final rank
\[ Rank = 48 + 1 = 49 \]
\[ \boxed{49^{th}} \] Quick Tip: For rank problems with repeated letters: always divide by factorial of repetitions and count only letters \textbf{strictly smaller} at each position.

Step 1: Find the general form of \(f(x)\)

Given: \[ f(x+y)=f(x)+f(y)-xy \]

Assume \(f(x)\) is a polynomial of degree \(\le 2\): \[ f(x)=ax^2+bx+c \]

Substitute into the functional equation: \[ a(x+y)^2+b(x+y)+c = ax^2+bx+c + ay^2+by+c - xy \]

Comparing coefficients: \[ 2a=-1 \Rightarrow a=-\frac12 \]

Constant term: \[ c=0 \]

So, \[ f(x)=-\frac{x^2}{2}+bx \]

Step 2: Use the given limit
\[ \lim_{h\to 0}\frac{f(h)}{h} =\lim_{h\to 0}\left(-\frac{h}{2}+b\right)=b \]

Given the limit is \(3\): \[ b=3 \]

Hence, \[ f(x)=3x-\frac{x^2}{2} \]

Step 3: Compute the required sum
\[ \sum_{n=1}^{10} f(n) =\sum_{n=1}^{10}\left(3n-\frac{n^2}{2}\right) \]
\[ =3\sum_{n=1}^{10}n-\frac12\sum_{n=1}^{10}n^2 \]
\[ =3\cdot\frac{10\cdot11}{2}-\frac12\cdot\frac{10\cdot11\cdot21}{6} \]
\[ =165-\frac{385}{2} =\frac{330-385}{2} =-\frac{55}{2} \]
\[ \boxed{-\dfrac{55}{2}} \] Quick Tip: Functional equations involving \(f(x+y)\) often lead to polynomial solutions. Always use the given limit to fix remaining constants.

Step 1: Find the first derivative
\[ f(x)=\sin 2x+2\cos x \]
\[ f'(x)=2\cos 2x-2\sin x \]

Set \(f'(x)=0\): \[ 2(\cos 2x-\sin x)=0 \Rightarrow \cos 2x=\sin x \]

Using \(\cos 2x=1-2\sin^2 x\): \[ 1-2\sin^2 x=\sin x \]
\[ 2\sin^2 x+\sin x-1=0 \]
\[ (2\sin x-1)(\sin x+1)=0 \]
\[ \sin x=\frac12 \quad or \quad \sin x=-1 \]

In the interval \(\left(-\frac{3\pi}{4},\frac{3\pi}{4}\right)\), \[ \sin x=\frac12 \Rightarrow x=\frac{\pi}{6} \]

(\(\sin x=-1\) gives \(x=-\frac{\pi}{2}\), which is a boundary point for critical behaviour.)

Step 2: Second derivative test
\[ f''(x)=-4\sin 2x-2\cos x \]

At \(x=\frac{\pi}{6}\): \[ f''\!\left(\frac{\pi}{6}\right) =-4\sin\frac{\pi}{3}-2\cos\frac{\pi}{6} =-4\cdot\frac{\sqrt3}{2}-2\cdot\frac{\sqrt3}{2}<0 \]

Hence, \(x=\frac{\pi}{6}\) is a point of local maxima.

Step 3: Check for inflection point

At \(x=-\frac{\pi}{2}\), \[ f'(x)=0 \]
but \(f''(x)\) changes sign across this point, hence it is a point of inflection.

Final Conclusion:

The function has a point of local maxima and a point of inflection.
\[ \boxed{Option (3)} \] Quick Tip: A point where \(f'(x)=0\) but no sign change in monotonicity occurs is not an extremum — always check the second derivative or sign of \(f'(x)\).