JEE 2018 :: Mathematics (2018) :: Solved questions (2018)

• Mathematics (2018) - Solved questions (2018)

If $\sum_\left(i=1\right)^9\left(x_{i}-5\right)=9$ and $\sum_\left(i=1\right)^9\left(x_{i}-5\right)^{2}=45$

then the standard deviation of the 9 items x₁, x₂,........ x₉  is 

Answer:

Explanation:

key idea  Standard deviation is remain unchanged, if observations are added or substracted by a fixed number.

  We have,

$\sum_\left(i=1\right)^9\left(x_{i}-5\right)=9$ and $\sum_\left(i=1\right)^9\left(x_{i}-5\right)^{2}=45$

$SD=\sqrt{\frac{\sum_\left(i=1\right)^9\left(x_{1}-5\right)^{2}}{9}-\left(\frac{\sum_\left(i=1\right)^9\left(x_{1}-5\right)}{9}\right)^{2}}$

$\Rightarrow$   SD= $\sqrt{\frac{45}{9}-( \frac{9}{9})^{2}}$

$\Rightarrow$    SD= $\sqrt{5-1}=\sqrt{4}=2$

If  $\begin{bmatrix}x-4 & 2x & 2x\\2x & x-4 & 2x\\ 2x & 2x &x-4  \end{bmatrix}=\left(A+Bx\right)\left(x-A\right)^{2}$  then the ordered pair ( A,B) is equal to 

Answer:

Explanation:

Given

$\begin{bmatrix}x-4 & 2x & 2x\\2x & x-4 & 2x\\ 2x & 2x &x-4  \end{bmatrix}=\left(A+Bx\right)\left(x-A\right)^{2}$

$\Rightarrow$ Apply  C1$\rightarrow$  C₁+ C₂+C₃

$\begin{bmatrix}5x-4 & 2x & 2x\\5x-4 & x-4 & 2x\\ 5x-4 & 2x &x-4  \end{bmatrix}=\left(A+Bx\right)\left(x-A\right)^{2}$

Taking common (5x-4) from C_{1 ,}we get  

$\left(5x-4\right)\begin{bmatrix}1 & 2x & 2x\\1 & x-4 & 2x\\ 1 & 2x &x-4  \end{bmatrix}=\left(A+Bx\right)\left(x-A\right)^{2}$

Apply  R₂ $\rightarrow$ R₂ - R₁ and  R₃ $\rightarrow$ R₃ - R₁

$\left(5x-4\right)\begin{bmatrix}1 & 2x & 0\\0 & -x-4 & 0\\ 0 & 0 & -x-4  \end{bmatrix}=\left(A+Bx\right)\left(x-A\right)^{2}$

Expanding Along C₁ , we get

   ( 5x-4) ( x+4)² = ( A+Bx)  ( x -A)²

Equaring We get,

   A= -4, B=5

Let A be the sum of the first 20 terms and B be the sum of first 40 terms of the series

1²+2.2²+3²+2.4²+5²+2.6²+........  If B-2A=100λ  then λ is equal to

Answer:

Explanation:

We have

   1²+2.2²+3²+2.4²+5²+2.6²+  .......

A= Sum of first 20 terms

B= Sum of first 40terms

$\therefore$   A= 1²+2.2²+3²+2.4²+5²+2.6²+........+2.20²

A= (1²+2²+3²⁺.........+20²) + (2²+4²+6²+........+20²)

A= (1²+2²+3²+.......+20²)+4(1²+2²+3²+........+10²)

$A= \frac{20\times 21\times 41}{6}+\frac{4\times10 \times 11\times21}{6}$

$A= \frac{20\times 21}{6}(41+22)=\frac{20\times 41\times63}{6}$

Similarly , B= (1²+2²+3²+........+40²)+ 4 ( 1²+2²+3²+.......+20²)

      $B= \frac{40\times41\times81}{6}+\frac{4\times20\times21\times41}{6}$

     $B= \frac{40\times41}{6}(81+42)=\frac{40\times41\times123}{6}$

Now, B-2A=100λ

$\therefore$    $\frac{40\times41\times123}{6}-\frac{2\times20\times21\times63}{6}=100\lambda$

$\Rightarrow$        $\frac{40}{6}(5043-1323)=100\lambda$

$\Rightarrow$        $\frac{40}{6}\times 3720=100\lambda$

$\Rightarrow$        $40\times620=100\lambda$

$\Rightarrow$     $\lambda =\frac{40\times620}{100}=248$

Let  $S= ( t\in R:f( x)=\mid x-\pi\mid .e^{\mid x\mid}-1)sin\mid x\mid$ is not  differentiable at t). Then, the set S is equal to

Answer:

Explanation:

We have

  $f(x)=\mid x-\pi\mid.(e^{\mid x\mid}-1)\sin \mid x\mid$

$\begin{cases}\left(x-\pi\right)\left(e^{-x}-1\right)\sin x, & x < 0\\ -\left(x-\pi\right)\left(e^{x}-1\right)\sin x, & 0 \leq x <\pi \\ \left(x-\pi\right)\left(e^{x}-1\right)\sin x, & x\geq\pi\end{cases}$

We check the differentiability at x=0 and  $\pi$

 We have,

$\begin{cases}\left(x-\pi\right)\left(e^{-x}-1\right)\cos x +\left(e^{-x}-1\right)\sin x + \left(x-\pi\right) \sin xe^{-x}\left(-1\right), & x < 0\\ -\left[\left(x-\pi\right)\left(e^{-x}-1\right)\cos x +\left(e^{-x}-1\right)\sin x + \left(x-\pi\right) \sin xe^{-x}\right],& 0 < x <\pi \\ \left(x-\pi\right)\left(e^{-x}-1\right)\cos x +\left(e^{-x}-1\right)\sin x + \left(x-\pi\right) \sin xe^{-x}, & x>\pi\end{cases}$

Clearly,      $\lim_{x \rightarrow 0^{-}}f'(x)=0=\lim_{x \rightarrow 0^{+}}f'(x)$

and             $\lim_{x \rightarrow \pi^{-}}f'(x)=0=\lim_{x \rightarrow \pi^{+}}f'(x)$

$\therefore$   f is differentiable  at x=0 and x= $\pi$

  Hence, f is differentiable for all x

Let u be a vector coplanar with the vectors $a=2\hat{i}+3\hat{j}+\hat{k}$ and  $b=\hat{j}+\hat{k}$

.If u is perpendicular to a and u. b=24, then $\mid u\mid ^{2}$ is equal to 

Answer:

Explanation:

Key Idea  If any vector x is coplanar with the vector y and z then x=λy +μz Here, u is coplanar with a and b.

$\therefore$     u= λa+µb

  Dot product with a, we get

 u.a= λ (a.a)+µ (b.a)

$\Rightarrow$    0= 14λ+2 µ  ........(i)

    $[\therefore a= 2\hat{i}+3\hat{j}-\hat{k},b=\hat{j}+\hat{k},u.a=0]$

  Dot product with b, we get 

        $u.b=\lambda (a.b)+\mu (b.b)$

     $24=2\lambda+2\mu$   .....(ii)   $[\because u.b=24]$

Solving Eqs. (i) and(ii), we get

    λ =-2, μ = 14

Dot product with u, we get

$\mid u\mid^{2}=\lambda (u.a)+\mu (u.b)$

$\mid u\mid^{2}=-2 (0)+14 (24)$

$\Rightarrow$       $\mid u\mid^{2}=-336$

• Mathematics (2018) - Solved questions (2018)