JEE maths question bank, MCQ CBSE Notes, Lectures

1 Area of Bounded Regions

If the length of sub-normal at any point P(x, y) on the curve passing through M(0, 1) is unity, then the area bounded by the curves equals

1) 1/3
2) 2/3
3) 4/3
4) 8/3

Option 1

Option 2

Option 3 Solution:

Solution

Option 4

2 Matrices

If A=$$\displaystyle \begin{bmatrix} -3 & 5 \\ 5 & 0 \\ -7 & 4 \end{bmatrix}$$ and B=$$\displaystyle \begin{bmatrix} 3 & -5 &7 \\ -5 & 0 & -4 \end{bmatrix}$$ then find $$\displaystyle A+B^{T}$$

1) 2A^T
2) 2B
3) 0
4) 2B^T

Option 1

Option 2

Option 3 Solution:

Solution

Given, $$ B = \begin{bmatrix} 3 & -5 & 7 \\ -5 & 0 & -4 \end{bmatrix} $$
$$ => {B}^{T} = \begin{bmatrix} 3 & -5 \\ -5 & 0 \\ 7 & -4 \end{bmatrix} $$

And, $$ A + {B}^{T} = \begin{bmatrix} -3 & 5 \\ 5 & 0 \\ -7 & 4 \end{bmatrix} + \begin{bmatrix} 3 & -5 \\ -5 & 0 \\ 7 & -4 \end{bmatrix} = \begin{bmatrix} -3+3 & 5-5 \\ 5-5 & 0+0 \\ -7+7 & 4-4 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{bmatrix} $$
As A has only one column, $$ => a = 1 $$

Thus $$ {B}^{T} $$ has the order $$ a \times n $$ or $$ 1 \times 3 $$,

As $$ {B}^{T} $$ is of order $$ 1 \times 3 $$, then $$ {B} $$ will have the order $$ 3 \times 1 $$

So, B will also have one column

⇒ 1 - cω - aω + acω ≠ 0

⇒ (1 - cω) (1 - aω) ≠ 0

$$a\ \ne \ \cfrac{1}{\omega },\ c\ \ne \ \cfrac{1}{\omega }$$

⇒ a = ω, c = ω and b⊂ {ω, ω2}

Therefore, number of distinct solutions = 2

Option 4

3 Matrices

If $$A=\begin{bmatrix} 2 & 1 \\ 4 & 1 \end{bmatrix}$$, $$B=\begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix}$$and $$C=\begin{bmatrix} 3 & -4 \\ -2 & 3 \end{bmatrix}$$, then $$\displaystyle tr(A)+tr\left(\frac{ABC}{2} \right) +tr\left(\frac{A{(BC)}^{2}}{4} \right)+tr\left(\frac{A{(BC)}^{3}}{8} \right) +........+\infty =$$

1) 6
2) 9
3) 12
4) 15

Option 1 Solution:

Solution

$$B=\left[ \begin{matrix} 3 & 4 \\ 2 & 3 \end{matrix} \right] $$ $$;$$ $$C=\left[ \begin{matrix} 3 & -4 \\ -2 & 3 \end{matrix} \right] $$

$$\Rightarrow BC=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] $$ $$\Rightarrow BC=I$$

$$\therefore ABC=A.I=A$$

$$A\left( BC\right)^2=AI^2=A$$

$$\therefore tr\left( A\right) +tr\left( \dfrac{ABC}{2}\right) +tr. \left( \dfrac{A\left( BC\right) ^2}{4}\right) +tr. \left( \dfrac{A\left( BC\right)^3}{8}\right)+........$$

$$=3+\dfrac{3}{2}+\dfrac{3}{4}+\dfrac{3}{8}+......+A$$

$$=3\left( \dfrac{1}{1-\dfrac{1}{2}}\right) $$

$$=\dfrac{3}{\dfrac{1}{2}}$$

$$=6$$

Hence, the answer is $$6.$$

Option 2

Option 3

Option 4

4 Differentiation

Given, f(x)=−x^3?^/3 + x²sin1.5 a−xsina⋅sin2a−5arcsin(a²−8a+17), then

1) f(x) is not defined at x=sin8
2) f′(sin8)>0
3) f′(x) is not defined at x=sin8
4) f′(sin8) < 0

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

f(x)=−x^3?^/3 + x²sin1.5 a−xsina⋅sin2a−5arcsin(a²−8a+17)

(x)=−x^3?^/3 + x²sin6 −xsin4sin8−5sin⁻¹((a−4)²+1)

f′(x)=−x²+2xsin6−sin4.sin8 .......( a=4 considering the domain of inverse sine function)
f′(sin8)=−sin²8+2sin6sin8−sin4sin8
=sin8(−sin8+2sin6−sin4)
=−sin8(sin8+sin4−2sin6)
=−sin8(2sin6cos2−2sin6)
=2sin8sin6(1−cos2)

Now,
sin8>0   (Since,2π<8<3π )
sin6<0   (Since, π<6<2π)
(1−cos2)>0
Hence, f′(sin8)<0

5 Sets relations and functions

If is equal to

1) (f(x))²
2) 2f(x)
3) - 2f(x)
4) 2f(x²)

Option 1

Option 2 Solution:

SOLUTION

= 2f(x)

Option 3

Option 4

6 Sets relations and functions

If S₁ and S₂ are respectively the sets of local minimum and local maximum points of the function,
f(x) = 9x⁴+12x³-36x² +25, x Î R , then :

1) S₁ = {-2,1}; S₂ ={0}
2) S₁ = {-2}; S₂ ={0, 1}
3) S₁ = {-2,1}; S₂ ={1}
4) S₁ = {-1}; S₂ ={0, 2}

Option 1 Solution:

SOLUTION

Option 2

Option 3

Option 4

7 Probability

An urn contain 20 tickets numbered 1 to 20. A ticket is drawn and then another ticket is drawn without replacement. The probability that both the tickets show even number is

1) 9/38
2) 1/8
3) 1/4
4) 5/8

Option 1 Solution:

Solution

Given: an urn containing 20 tickets, and 2 tickets are drawn without replacement

To find: the probability that both the tickets show even number

Sol: According to the question

$$n(S)=20$$

There are 10 even numbers between 1 to 20.

Therefore, the probability that both the tickets show even number is

$$=\dfrac {10}{20}\times \dfrac {9}{19}=\dfrac 9{38}$$

Option 2

Option 3

Option 4

8 Limits Continuity and Differentiability

Which one of the following is not correct ?

1) If f(x) is not continuous at x = a, then it is not differentiable at x = a
2) If f(x) is continuous at x = a, then it is differentiable at x = a
3) If f(x) and g(x) are differentiable at x = a, then f(x) + g(x) is also differentiable at x = a
4) If f(x) is continuous at x = a, then exists

Option 1

Option 2 Solution:

SOLUTION

Option 3

Option 4

9 Matrices

If A=[a_ij] is a square matrix of even order such that [a_ij]=i²−j², then

1) A is a skew-symmetric matrix and |A| = 0
2) A is skew-symmetric matrix and |A| is a square
3) A is symmetric matrix and |A| = 0
4) none of these

Option 1

Option 2 Solution:

Solution

Given $$\displaystyle \left [ a_{ij} \right ]=i^{2}-j^{2}$$
$$\Rightarrow a_{ii}=0$$ for all i.

Also, for $$i \ne j, a_{ij}=i^{2}-j^{2}=-(j^{2}-i^{2})=-a_{ji}$$
$$\Rightarrow a_{ij}=-a_{ji}$$

Hence, A is skew-symmetric matrix.
For order 2, $$A=\begin{bmatrix} 0 & -3 \\ 3 & 0 \end{bmatrix}$$
$$|A|=9\ne 0$$

Hence, A is a skew-symmetric matrix and |A| is a square

Option 3

Option 4

10 Differentiation

Let f be a non-negative function defined on the interval [0,1] .If $$\int_0^x \sqrt{1-(f'(t))^2} dt = \int _0^x f(t) dt, 0 \leq x \leq 2,$$ and $$f(0) =0$$ then,

1) $$f\left(\frac{1}{2} \right) < \frac{1}{2} $$ and $$f\left( \frac{1}{3} \right) > \frac{1}{3}$$
2) $$f\left(\frac{1}{2} \right) > \frac{1}{2} $$ and $$f\left( \frac{1}{3} \right) > \frac{1}{3}$$
3) $$f\left(\frac{1}{2} \right) < \frac{1}{2} $$ and $$f\left( \frac{1}{3} \right) < \frac{1}{3}$$
4) $$f\left(\frac{1}{2} \right) > \frac{1}{2} $$ and $$f\left( \frac{1}{3} \right) < \frac{1}{3}$$

Option 1

Option 2

Option 3 Solution:

Solution

Differentiating using Libnitz formula, we get

$$\sqrt{1-(f^{'}(x))^{2}}=f(x)$$

$$\sqrt{1-(f(x))^{2}}=f^{'}(x)=\cfrac{d(f(x))}{dx}$$

$$dx=\cfrac{d(f(x))}{\sqrt{1-(f(x))^{2}}}$$

Now, on integarting we get

$$x+C=\sin^{-1}(f(x))$$

Now for x to be real, we can say that

$$0\le f(x)<1 \, or\, f(x)<\cfrac{1}{2} and f(x)<\cfrac{1}{3}$$

Option 4

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