NEB board exam maths solved problems with proper solutions - available for free | short answered mathematics problems solved group A

 This post provides you the problems with solutions of NEB board exam of grade 11 mathematics 2076.

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Group A 

`5 \times 3 \times 2 = 30`

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1. a. Find the truth value of the biconditional statements of p and q where p represents '2 is an even number' and q represents '4 is an even number'.

Solution:-

Biconditional statements imply if one statement is true then another is also true. if one is false, then another statement is also false. So, one statement supports the other in case of biconditional statement. 

Here, p: 2 is an even number.

         q: 4 is an even number.

4 is the square of 2 and 2 is the square root of 4. 2 is even, then 4 is also even. So, both p and q represent true value (T).

p	q	p ⇔ q
T	T	T

Thus truth value of p and q is T, i.e. 2 is an even number implies 4 is also an even number.//

b. If f:`R \rightarrow R` be defined by f(x) = 2x - 3, then find `f^-1`.

Solution:-

Here, f(x) = 2x - 3  .........(1)

For a function to have it's inverse, this function should be one-one and onto function.  

Let `x_1`, `x_2` `\in` R, which are domain of f, Then

`f(x_1) = 2x_1 - 3` and 

`f(x_2) = 2x_2 - 3`

For function f to be one-one, 

`f(x_1) = f(x_2)`

or, `2x_1 - 3 = 2x_2 - 3`

`\therefore` `x_1 = x_2`.

Thus, f is one-one function.

Again, let r `\in`  R , then 

r = 2x -3

or, 2x = r + 3

`\therefore` `x = \frac{r + 3}{2}` `\in` R. 

Thus, f is onto function as well. 

Therefore, f is one-one and onto function. So, `f^-1` exists for f, where `f^-1:R \rightarrow R` so that x is the image of y under `f^-1`. i.e. `x = f^-1(y)`. 

Now solving (1) for y = f(x), 

y = 2x - 3

or, `x = \frac{y + 3}{2}`.

`\therefore` `f^-1 (y) = \frac{y + 3}{2} .....(2)`.

`\therefore` Equation (2) is the formula that defines the inverse function f given by equation (1).

c. Test the even or odd nature and  symmetricity of the function `y = 8x^2`.

Solution:-

Here, `y = f(x) = 8x^2 .........(1)` 

For a function `f:A \rightarrow B` to have even or odd nature and symmetricity, it should follow the following property when x`\in`A is replaced by by -x`\in`A. 

if f(-x) = f(x) : even nature and symmetricity

if f(-x)  = - f(x): odd nature 

So replacing x (`\in`A ) in  (1) by -x(`\in` A), we have

`f(-x) = 8 (-x)^2 = 8x^2`.

That is, f(-x) = f(x).

This shows that the function given by (1) is even in nature and has symmetricity.

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2. a. Express `tan^-1 x` in terms of inverse of sin function.

Solution:-

Let   `\theta = tan^-1x`. 

or, `x = tan \theta`

Take, `2 \theta = 2 \theta`

or, `sin2\theta = sin2\theta`

But  `sin2 \theta = \frac{2 tan \theta}{1 + tan^2 \theta}`, so

`\frac{2 tan\theta}{1 + tan^2\theta} = sin 2 \theta` `\therefore` value of `sin 2\theta` is used only in one side.

or, `sin^-1\frac{2tan\theta}{1 + tan^2\theta} = 2\theta` 

or, `sin^-1 (\frac{2x}{1 + x^2}) = 2 tan^-1x`

or, `2tan^-1x = sin^-1(\frac{2x}{(1 + x^2)})`

or, `tan^-1x = \frac{1}{2} sin^-1(\frac{2x}{1 + x^2})`

`\therefore` `tan^-1x = \frac{1}{2} sin^-1(\frac{2x}{(1 + x^2)})`.//

b. Prove by mathematical induction: 1 + 3 + 5 + ........ + (2n - 1) = `n^2`.

Solution:-

Here, 1 + 3 + 5 + .........(2n-1) = `n^2`.

The principle of Mathematical Induction says that if p(n) be the statement and if 

(i) p(1) is true

(ii) p(k + 1)  is true whenever p(k) is true.

then p(n) is true for all n `\in` N.

Let, p(n) = 1 + 3 + 5 + ......... + (2n-1)  = `n^2`

let n =1, 

LHS = p(1) = 1

Also RHS = `n^2 = 1^2 =1`

So, p(n) is true foe n =1.

When n =2,

LHS = P(2) = 1 + 3 = 4

RHS = `n^` = `2^2` = 4

So, p(n) is true for n =2 as well.

let p(n) be true for n = m, for m `\in` N, then

p(m) = 1 + 3 + 5 + ......+(2m-1) = `m^2`

Now we shall show that p(m+1) is true.

p(m+1) = 1 + 3 + 5 + ...... (2` \times` (m+1) -1)

p(m + 1) = 1 + 3 + 5 + ......(2m+1) = `(m+1)^2`

Performing p(m)+ p(m+1), 

1 + 3 + 5 + .......... (2m -1) + 1+ 3+ 5 + ....... +(2m + 1) = `m^2 + (m + 1)^2` 

or, `m^2` + m^2 + (2m+1) = `m^2 + m^2 +2m +1`

or, `2m^2 + (2m+1) = 2m^2 + 2m +1`

`\therefore` `2m^2 +2m +1 = 2m^2 + 2m +1` is true.

This shows that p(m+1) is true whenever p(m) is true. Hence by principle of induction, p(n) is true for all n `\in` N. Hence proved.//

b. Express `\sqrt{3} + i` in polar form.

Solution:-

Here, `z = \sqrt{3} +i`

Comparing with  z = x + iy, we get, `x = \sqrt{3}` and y = 1. 

z in polar form will,

 `z = r (cos \theta + isin \theta) ........(1)`     

where `r  = \sqrt{x^2 + y^2}`

or, `r = \sqrt{(\sqrt{3})^2 + (1)^2}`

or, `r = \sqrt{4}`

`\therefore` `r = \pm 2`. 

and `\theta = tan^-1(\frac{y}{x})`

or, `\theta = tan^-1(\frac{1}{\sqrt{3}})`

`\therefore` `\theta = 30^o `.

Using these values of  r and `\theta` in (1),

`z = \pm 2 (cos30^o + i sin30^o)`.

Which is required form of given complex number in polar form.

c. If one root of the equation `ax^2 + bx+ c = 0` be twice the other, show that `2b^2 = 9ac`.

Solution:- 

Suppose,    `\alpha` and `\beta` be two roots of given equation,

`ax^2 + bx + c = 0 ............(1)`  

Also let `\alpha` be twice of `\beta`, 

`\alpha = 2 \beta`.

Sum of roots = `\alpha + \beta = - \frac{b}{a} => 2\beta + \beta = -\frac{b}{a}  =>  3\beta = -\frac{b}{a} => (3\beta)^2 = (\frac{b}{a})^2 ....(2)`

product of roots = `\alpha \times \beta = \frac{c}{a} => 2\beta \times \beta = \frac{c}{a} => 2\beta^2 = \frac{c}{a} ....(3) `

Dividing (3) by (2),

`\frac{2\beta^2}{9\beta^2} = \frac{\frac{c}{a}}{(\frac{b}{a})^2}`

or, `\frac{2}{9} = \frac{c a^2}{a b^2}`

or, `\frac{2}{9} = \frac{ca}{b^2}`

or, `2b^2 = 9ca`

`\therefore` `2b^2 = 9ac`

Which is required result. Hence proved.//

4. a. Find the equation of the straight line that has y-intercept '2' and is parallel to 4x - 8y + 3 = 0.

Solution:-

Here given equation is,

`4x - 8y + 3 = 0 ....................(1)`

Slope of equation (1), `m_1 = -\frac{b}{a} = -\frac{-8}{4} = 2` 

Suppose, equation of straight line parallel to (1) be

`y = m_2 x + c ............(2)`

where c = y-intercept = 2 (given in question).

Since straight lines (1) and (2) are parallel to each other, then their slopes must be equal.

So, `m_2 = m_1 = 2`

Using value of c and `m_2` in (2), 

y = 2x + 2 .......(3)

Which is required equation of straight line.//

b. Find the equation of the circle  which passes through the origin and makes equal intercepts on both axes.

Solution:-

General equation of circle is given as,

` x^2 + y^2 + 2gx + 2fy + c = 0 ........(1)`

By given in question, this circle passes through origin(0,0) and makes equal intercepts on both axes.

Let's this intercept be h. Then, the circle (1) passes through three points (0,0), (h,0) and (0,h).

When circle (1) passes through origin (0, 0),  

0 + 0 + 2g.0 + 2f.0 + c = 0

So, c = 0 ........(2)

When circle (1) passes through x-axis (h, 0),

`h^2 + 0 +2g.h + 2f.0 + c = 0`

`h^2 + 2gh = 0`   [`\therefore` use of c = 0]

or, h + 2g = 0

`\therefore` `g = - \frac{h}{2}` .....(3)

Also, when circle (1) passes through y-axis (0, h),

`0 + h^2 + 2g.0 + 2fh + c = 0`

or, `h^2 + 2fh = 0`

  `\therefore`    `f = -\frac{h}{2}` .......(4)

Now using (2), (3), (4) in (1), we get

`x^2 + y^2 + 2 \times (- \frac{h}{2})  x + 2 \times (- \frac{h}{2}) y + 0 = 0`

or, `x^2 + y^2 - hx - hy = 0`

`\therefore` `x^2 + y^2 - h(x + y) = 0 .....(5)`

which is the required equation of circle that passes through origin and makes equal intercepts on both axes. 

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c. Evaluate: `\lim_(x ->0) \frac{1 - cosmx}{1 - cosnx}`.

Solution:-

Here, L = `\lim_(x->0) \frac{1 - cosmx}{1 - cosnx}`.

If we put x = 0 directly, then we get

`L = \frac{1 - cos0}{1 - cos0} = \frac{1 - 1}{1 - 1} = \frac{0}{0}` = indeterminate form.`

So, we need to simplify the function `\frac{1 - cosmx}{1 - cosnx}`before we use the limit x = 0.

`\therefore` `L = \lim_(x->0)\frac{1 - cosmx}{1 - cosnx}`

 But `1 - cosmx = 2 sin^2mx` and `1 - cosnx = 2sin^2nx`.

So, `L =  \lim_(x->0) \frac{2sin^2mx}{2sin^2nx}`

 = `\lim_(x->0) \frac{sin^2mx}{sin^2nx}`

= `\lim_(x->0) \frac{(\frac{sinmx}{mx})^2 (mx)^2}{(\frac{sinnx}{nx})^2(nx)^2}`

= `\frac{m^2}{n^2} \frac{(\lim_(x->0) \frac{sinmx}{mx})^2 }{(\lim_(x->0)\frac{sinnx}{nx})^2}`

Using the result, `\lim_(x ->0) \frac{sinx}{x} =1`, (See Proof of this relation), we get

= `\frac{m^2}{n^2} (\frac{1^2}{1^2})`

= `\frac{m^2}{n^2}`

Hence, `\lim_(x->0) \frac{1 - cosmx}{1 - cosnx} = \frac{m^2}{n^2}`. //

5. a. Find `\frac{dy}{dx}` when x = 2at and `y = 2at^2`.

Solution:-

Here, x  = 2at

and `y = 2at^2`

To find: `\frac{dy}{dx}`.

Keeping 'a' as constant in both cases, we see that x and y depends on 't'. So we find `\frac{dx}{dt}` and `\frac{dy}{dt}` respectively by differentiating x and y with respect to 't'.

That is, `\frac{dx}{dt} = \frac{d}{dt}2at = 2a ......(1)`

and `\frac{dy}{dt} = \frac{d}{dt} 2at^2 = 2a.2t = 4at ......(2)`

Dividing (2) by (1), we get

`\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4at}{2a}`

or, `\frac{dy}{dt} \frac{dt}{dx} = 2t`

`\therefore` `\frac{dy}{dx} = 2t` which is required solution. //

b. Evaluate: `\int sin2x \cdot sin3x  dx`.

Solution:- 

Suppose, `I = \int  sin2x \cdot sin3x  dx   ...........(1)`.

Using the trigonometric relation,

`sinA \cdot sinB = \frac{1}{2}[cos(A - B) - cos(A + B)]`,  

`I = \int \frac{1}{2} [cos(2x - 3x) - cos(2x + 3x)] dx`

 = `\frac{1}{2} \int [cos(-x) - cos(5x)]dx`

= `\frac{1}{2} [\int cos(x) dx - \int cos(5x) dx]`

Using `\int cos(x) dx = sin(x) + c`,

= `\frac{1}{2} [sin(x) + c_1 - \frac{sin(5x)}{5} + c_2]`

= `\frac{1}{2} [sin(x) - \frac{sin(5x)}{5} + c]`

Where `c_1` and `c_2` are integration constants and `c = c_1 + c_2`.

=`\frac{1}{10} [5cos(x) - cos(5x)] + c`

`\therefore` `\int sin2x \cdot sin3x dx  = \frac{1}{10} [5cos(x) - cos(5x)] + c` which is the required solution. //

c. Determine the interval in which the function f(x) = `\frac{1}{2} x^2 - x` is increasing or decreasing.

Solution:-

Here given function is, `f(x) = \frac{1}{2} x^2 - x  ..........(1)`

Differentiating (1) with respect to x,

` f \prime(x) = \frac{df(x)}{dx} = \frac{d}{dx} (\frac{1}{2} x^2 - x) = x - 1`.

`\therefore` `f \prime (x) = x - 1   ............(2)`

For function f(x) to be increasing, the required condition is `f \prime(x) \gt 0`. 

{For x = 0, `f \prime(x)` = -1

for x = -1, `f \prime(x)` = -2 

for x = -2, `f \prime(x)` = -3 } `f \prime(x)` < 0 

{ for x = 2, `f \prime(x)` = 1

for x = 3, `f \prime(x)` = 2 } `f \prime(x)` > 0 and so on.

So, in equation (2), `f \prime(x) \gt 0` for `x \gt 1` 

Thus, the interval for f(x) to be increasing is `(1, \infty)`.

Also, in (2), `f \prime(x) \lt 0` for  `x \lt 1` .

Thus, the interval for f(x) to be decreasing is `(- \infty, 1)`.

Hence, the function `f(x) = \frac{1}{2}x^2 - x` is increasing in interval `(1, \infty)` and decreasing in interval `(- \infty, 1)`. //

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