Subjective Questions
Q1:
If f= {(1,3), (0,0), (-1,-3) and g= {(2,3), (-3,-1), (3,5)} show the function gof by an arrow diagram and find it in ordered pair form.
Type: Short
Difficulty: Easy
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Answer: <p>f= {(1,3), (0,0), (-1,-3)}, g= {(0,2), (-3,-1), (3,5)}, the above relation shown in the mapping diagram as follow:</p> <figure class="" style="width: 500px;"><img src="/uploads/a113.jpg" alt="sx" width="500" height="300"><figcaption><br></figcaption></figure><p>g<sub>o</sub>f= g(3) = 5</p> <p>g<sub>o</sub>f = g(0) =2</p> <p>g<sub>o</sub>f= g(-3) = -1</p> <p>g<sub>o</sub>f = g {(0,2), (-1,-1), (1,5)} <sub>Ans</sub></p> <p></p>
Q2:
If f = {(-2,4), (-1,8), (0,0), (1,4), (2,6)} and g = {(0,-2), (4,0), (6,1), (8,-2)}, show the function gof by an arrow diagram and find it ordered pair form.
Type: Short
Difficulty: Easy
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Answer: <p>f = {(-2,4), (-1,8), (0,0), (1,4), (2,6)},</p> <p>g = {(o,-2), (4,0), (6,1), (8,-2)}</p> <p>The mapping diagram is shown below:</p> <figure class="" style="width: 500px;"><img src="/uploads/a25.jpg" alt="sd" width="500" height="300"><figcaption><br></figcaption></figure><p>g<sub>o</sub>f(-2) = g(4) = 0</p> <p>g<sub>o</sub>f(-1) = g(8) = -2</p> <p>g<sub>o</sub>f(0) = g(0) = 0</p> <p>g<sub>o</sub>f(1) = g(4)=0</p> <p>g<sub>o</sub>f(2) = g(6) = 1</p> <p>g<sub>o</sub>f= {(-2,0), (-1,-2), (0,-2) , (1,0), (2,1)} <sub>Ans</sub></p>
Q3:
If f = {(p,1), (q, 2), (r,3), (s,4)} and g = {(1,5), (2,8), (3,-3), (4,6)}, show the function fog by an arrow diagram and find it in ordered pair form.
Type: Short
Difficulty: Easy
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Answer: <p>f = {(p,1), (q,2), (r,3), (s,4)}, g = {(1,5), (2,8), (3,-3), (4,6)}</p> <p>The mapping diagram is shown below:</p> <figure class="" style="width: 500px;"><img src="/uploads/a35.jpg" alt="dsv" width="500" height="300"><figcaption><br></figcaption></figure><p>f<sub>o</sub>g = {(5,p), (8,q), (-3,r), (6,s)} <sub>Ans</sub></p>
Q4:
If f = {(x,3), (y,4), (z,7)}, g = {(3,a), (4,b), (7,c)}, show the function fog by an arrow diagram and find it in ordered pair form.
Type: Short
Difficulty: Easy
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Answer: <p>f = {(x,3), (y,4), (z,7)}, g = {(3,a), (4,b), (7,c)}</p> <p>The mapping diagram is shown below:</p> <figure class="" style="width: 500px;"><img src="/uploads/a42.jpg" alt="d" width="500" height="300"><figcaption><br></figcaption></figure><p>f<sub>o</sub>g = {(a,x), (b,y), (c,z)} <sub>Ans</sub></p>
Q5:
If f(x) = x3 and g(x) = x2 find the fg(x).
Type: Short
Difficulty: Easy
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Answer: <p>f(x) = x<sup>3</sup> and g(x) = x<sup>2</sup></p> <p>fg(x)</p> <p>= f(x<sup>2</sup>) [\(\because\) g(x) = x<sup>2</sup>]</p> <p>= (x<sup>2</sup>)<sup>3 </sup>[\(\because\) f(x) = x<sup>3</sup>]</p> <p>= x<sup>6</sup><sub>Ans</sub></p> <p></p>
Q6:
If f(x) = \(\frac 1x\), x ∈ R, x ≠ 0, find the value of ff(2).
Type: Short
Difficulty: Easy
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Answer: <p>f(x) =\(\frac 1x\)</p> <p>ff(x) = f \(\frac 1x\) [\(\because\) f(x) = \(\frac 1x\)]</p> <p>= \(\cfrac{1}{<br>\cfrac{1}{x}}\) =1× \(\frac x1\) = x</p> <p>∴ ff(2) = 2 <sub>Ans</sub></p>
Q7:
If f(x) = 3x2 + x - 28 and g(x) = 6-x are the two polinomials. Find the value of f(x) ⋅ g(x).
Type: Short
Difficulty: Easy
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Answer: <p>Here, f(x) = 3x<sup>2</sup> + x - 28 and g(x) = 6-x</p> <p>f(x)⋅ g(x)</p> <p>= (3x<sup>2</sup> + x - 28) (6-x)</p> <p>= 18x<sup>2</sup> + 6x - 168 -3x<sup>3</sup>-x<sup>2</sup> + 28x</p> <p>= -3x<sup>3</sup> + 17x<sup>2</sup>+ 34x - 168 <sub>Ans</sub></p>
Q8:
Write down the range and the inverse function of the following functions:
f = {(2, \(\frac 12\)), (3, \(\frac13\)), (4,\(\frac 14\))}
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>f = {(2,\( \frac 12\)), (3, \(\frac13\)), (4,\(\frac 14\))}</p> <p>Range = { \(\frac 12\), \(\frac 13\), \(\frac 14\)}</p> <p>Inverse function (f<sup>-1</sup>) = {(\(\frac 12\), 2), (\(\frac 13\), 3), (\(\frac 14\), 4)}<sub>Ans</sub></p>
Q9:
Find the inverse function of the following function:
- {(3,2), (-1,-7), (4,5), (6,8)}
- {x, 2x+1 : x ∈ R}
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <ol><li type="a">{(3,2), (-1,-7), (4,5), (6,8)}</li> </ol><p>Let f = {(3,2), (-1,-7), (4,5), (6,8)}</p> <p>Changing domain into range and range into domain.</p> <p>f<sup>-1</sup> = {(2,3), (-7,-1), (5,4), (8,6)}</p> <ol><li type="a">{x, 2x+1 : x∈ R}</li> </ol><p>Let y = f(x) = 2x+1</p> <p>Interchange the place of x and y</p> <p>x = 2y + 1</p> <p>x-1 = 2y</p> <p>2y = x-1</p> <p>y = \(\frac {x-1}{2}\)</p> <p>∴ f<sup>-1</sup> = \(\frac{x-1}{2}\) <sub>Ans</sub></p>
Q10:
If f(4x-15) = 8x-27 then find ff(2).
Type: Short
Difficulty: Easy
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Answer: <p>f(4x-15) = 8x-27</p> <p>Or,</p> <p>f(4x-15) = 2(4x-15) + 3</p> <p>∴ f(x) = 2x+3</p> <p>ff(x) = f(2x+3) = 2 (2x+3) +3 = 4x +6+3 = 4x+9</p> <p>ff(2) = 4 × 2 + 9 = 17 <sub>Ans</sub></p>
Q11:
If g(2x-5) = 8x-19 find gg(2).
Type: Short
Difficulty: Easy
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Answer: <p>g(2x-5) = 8x-19</p> <p>g(2x-5) = 4(2x-5) +1</p> <p>g(x) = 4x+1</p> <p>gg(x) =g(4x+1) = 4(4x+1) +1 = 16x+4+1 = 16x+5</p> <p></p> <p>gg(2) = 16×2+5 = 32+5 =37 <sub>Ans</sub></p>
Q12:
If f(x+3) = 3x+5 then find f-1.
Type: Short
Difficulty: Easy
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Answer: <p>f(x+3) = 3x+5</p> <p>f(x+3) = 3(x+3) -4</p> <p>f(x) = 3x-4</p> <p>Let y = f(x) = 3x-4</p> <p>Y = 3x-4</p> <p>Interchanging the place of x and y in the above relation</p> <p>x = 3y-4</p> <p>Or, 3y = x+4</p> <p>y = \(\frac {x+4}{3}\)</p> <p>f<sup>-1</sup> (x) = \(\frac {x+4}{3}\) <sub>Ans</sub></p>
Q13:
If g(x) = 2x and fog(x) = 6x-2 then find f(x)
Type: Short
Difficulty: Easy
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Answer: <p>fog(x) = 6x-2</p> <p>Or, f[g(x)] = 6x-2</p> <p>Or, f(2x) = 6x-2</p> <p>Or, f(2⋅\( \frac x2\) = 6 \(\frac x2\)-2</p> <p>f(x) = 3x-2 <sub>Ans</sub></p>
Q14:
If f(x) = 8x-3 and fg(x) = 20x+1 find g(x).
Type: Short
Difficulty: Easy
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Answer: <p>Let g(x) = ax+b</p> <p>fog(x) = f[g(x)] = f(ax+b) = 8(ax+b) -3</p> <p>fog(x) = 8ax +8b - 3 ------ (1)</p> <p>fog(x) = 20x+1 ------------(2)</p> <p>From eq<sup>n</sup> (1) and (2)</p> <p>8ax +8b -3 = 20x +1</p> <p>Equating on both sides,</p> <p>8a = 20</p> <p>a = \(\frac {20}{8}\)</p> <p>∴ a = \(\frac 52\)</p> <p>8b -3 = 1</p> <p>8b = 4</p> <p>b =\(\frac 48\) = \(\frac 12\)</p> <p>∴ b = \(\frac 12\)</p> <p>Putting the value of a and b in g(x) = ax + b</p> <p>g(x) = \(\frac 52\) x + \(\frac 12\) = \(\frac {5x+1}{2}\)<sub>Ans</sub></p>
Q15:
Find the inverse function of {(x, 2x+1) : x ∈R}.
Type: Short
Difficulty: Easy
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Answer: <p>Function f = {(x, 2x+1) : x∈R}</p> <p>Let, y = 2x + 1</p> <p>Interchanging the place of x and y</p> <p>x = 2y + 1</p> <p>or, x-1 = 2y</p> <p>or, y = \(\frac {x-1}{2}\)</p> <p>∴ f<sup>-1</sup>= {(x, \(\frac {x-1}2\)) :x∈R } <sub>Ans</sub></p>
Q16:
If f(x) = 2x + 3 and g(x) = 5x2 find fg (x) and gf (x).
Type: Short
Difficulty: Easy
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Answer: <p>f(x) = 2x + 3, g (x) = 5x<sup>2</sup></p> <p>fg (x)</p> <p>= f (5x<sup>2</sup>) [\(\because\) g(x) = 5x<sup>2</sup>]</p> <p>= 2×5x<sup>2</sup>+ 3</p> <p>= 10x<sup>2</sup> +3 <sub>Ans</sub></p> <p>Also,</p> <p>gf(x)</p> <p>= g (2x + 3) [\(\because\) f(x) = 2x + 3 ]</p> <p>= 5 (2x + 3)<sup>2</sup><sub>Ans</sub></p>
Q17:
If f(x) = x2 + 5 and g(x) = x - 3 find fog (2).
Type: Short
Difficulty: Easy
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Answer: <p>fog (2)</p> <p>= f(2x - 3) [\(\because\) g(x) = x - 3 ]</p> <p>= f(-1)</p> <p>= (-1)<sup>2</sup> + 5 [\(\because\) f(x) = x<sup>2</sup> + 5 ]</p> <p>= 1 + 5</p> <p>= 6 <sub>Ans</sub></p> <p></p>
Q18:
Using synthetic division method. Find the remainder when x3 + 3x2 - 2x + 6 is divided by x - 3.
Type: Short
Difficulty: Easy
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Answer: <p>x<sup>3</sup> + 3x<sup>2</sup> - 2x + 6÷ (x - 3)</p> <p></p> <div> <table width="275"><tbody><tr><td>3</td> <td>1</td> <td>3</td> <td>-2</td> <td>6</td> </tr><tr><td></td> <td></td> <td>3</td> <td>18</td> <td>48</td> </tr><tr><td></td> <td>1</td> <td>6</td> <td>16</td> <td>54</td> </tr></tbody></table></div> <p>The remainder (R) = 54 <sub>Ans</sub></p>
Q19:
State and proved the remainder theorem.
Type: Short
Difficulty: Easy
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Answer: <p>If the polynomial f(x) is divided by x - a, the remainder is f(a)</p> <p>Proof: Let, Q(x) be the quotient and R be the remainder when f(x) is divided by (x - a) then, f(x) = (x -a)⋅Q(x) + R.</p> <p>When x = a, f(a) = (a -a)⋅Q(x) + R ∴ f(a) = R <sub>proved</sub></p>
Q20:
State and proved the factor theorem.
Type: Short
Difficulty: Easy
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Answer: <p>If the polynomial f(x) is divided by (x - a) and R = f(a) = 0, then (x - a) is a factor of f(x).</p> <p>Proof: Let Q(x)</p> <p>Let Q(x) be the quotient when f(x) is divided by (x - a), then f(x) = (x-a) Q(x) + R = (x-a) Q(x) + 0</p> <p>f(x) = (x-a) Q(x)</p> <p>∴(x-a) is a factor of f(x). <sub>proved</sub></p>
Q21:
Using synthetic division method find the quotient and remainder of the expression x4 - x3 - 3x2 -2x + 5 where x + 1 = 0.
Type: Short
Difficulty: Easy
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Answer: <p>x<sup>4</sup> - x<sup>3</sup> - 3x<sup>2</sup> -2x + 5÷ (x + 1)</p> <table width="328"><tbody><tr><td>-1</td> <td>1</td> <td>-1</td> <td>-3</td> <td>-2</td> <td>5</td> </tr><tr><td></td> <td></td> <td>-1</td> <td>2</td> <td>1</td> <td>1</td> </tr><tr><td></td> <td>1</td> <td>-2</td> <td>-1</td> <td>-1</td> <td>6</td> </tr></tbody></table><p>∴ Quotient (Q) = x<sup>3</sup> - 2x<sup>2</sup> - x -1</p> <p>Remainder (R) = 6 <sub>Ans</sub></p>
Q22:
Using synthetic division method, find the quotient and remainder when x3 - 2x + 3 is divided by x - 1.
Type: Short
Difficulty: Easy
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Answer: <p>x<sup>3</sup> - 2x + 3 is divided by x - 1.</p> <table width="340"><tbody><tr><td>1</td> <td>1</td> <td>0</td> <td>-2</td> <td>3</td> </tr><tr><td></td> <td></td> <td>1</td> <td>1</td> <td>-1</td> </tr><tr><td></td> <td>1</td> <td>1</td> <td>-1</td> <td>2</td> </tr></tbody></table><p>Quotient (Q) = x<sup>2</sup> - x -1</p> <p>Remainder (R) = 2 <sub>Ans</sub></p>
Q23:
Divide the polynomial: f(x) = x3 +3x2 -4x +2 and g(x) = x + 1.
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>f(x)÷ g(x) = \(\frac {x^3 +3x^2 -4x +2}{x +1}\)</p> <table width="303"><tbody><tr><td>x + 1</td> <td>x<sup>3</sup> + 3x<sup>2</sup> - 4x +2</td> <td>x<sup>2</sup> + 2x - 6</td> </tr><tr><td></td> <td> <p>x<sup>3</sup> + x<sup>2</sup></p> <p>- -</p> </td> <td></td> </tr><tr><td></td> <td>2x<sup>2</sup> - 4x + 2</td> <td></td> </tr><tr><td></td> <td> <p>2x<sup>2</sup> + 2x</p> <p>- -</p> </td> <td></td> </tr><tr><td></td> <td>-6x + 2</td> <td></td> </tr><tr><td></td> <td> <p>-6x - 6</p> <p>+ +</p> </td> <td></td> </tr><tr><td></td> <td>8</td> <td></td> </tr></tbody></table><p>∴ Quotient Q(x) = x<sup>2</sup> + 2x - 6</p> <p>Remainder R(x) = 8 <sub>Ans</sub></p>
Q24:
If x - k is a factor of the polynomial x3 - kx2 - 2x + k + 4, find the value of k.
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>x - k is factor of the polynomialx<sup>3</sup> - kx<sup>2</sup> - 2x + k + 4, then x = k must satisfies the given polynomial, so that the given polynomial should be zero.</p> <p>x - k is a factor of f(k) then f(k) = 0</p> <p>f(k) = k<sup>3</sup> - k⋅ k<sup>2</sup> - 2k + k + 4</p> <p>or, 0 = k<sup>3</sup> - k<sup>3</sup> -k +4</p> <p>or, 0 = -k +4</p> <p>∴ k = 4 <sub>Ans</sub></p>
Q25:
If 2x + 1 is a factor of 2x3 + ax2 + x + 2, find the value of x.
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>f(x) =2x<sup>3</sup> + ax<sup>2</sup> + x + 2</p> <p>2x + 1 is a factor of f(x) so f(x) = 0</p> <p>2x + 1 = o</p> <p>or, 2x = -1</p> <p>∴ x = - \frac 12</p> <p>2x<sup>3</sup> + ax<sup>2</sup> + x + 2 = 0</p> <p>or, 2× (- \(\frac 12)^3\) + a⋅ (- \(\frac 12)^2\) - \(\frac 12\) + 2 = 0</p> <p>or, 2× - \(\frac 18\)+ a⋅ \(\frac 14\) - \(\frac 12\) + 2 = 0</p> <p>or, - \(\frac 14\) + \(\frac a4\) - \(\frac 12\) + \(\frac 21\) = 0</p> <p>or, \(\frac {-1 +a -2 +8}{4}\) = 0</p> <p>or, a + 5 = 0</p> <p>∴ a = -5 <sub>Ans</sub></p>
Q26:
Find the quotient and remainder when f(x) = 4x3 -6x2 -5 +3x is divided by x - 2 using systentic division method.
Type: Short
Difficulty: Easy
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Answer: <p>Arrange the given polynomial in decreasing order</p> <p>f(x) = 4x<sup>3</sup> -6x<sup>2</sup>+3x -5</p> <p>The constant term with the sign changed is +2</p> <p>∴ x - 2 = x - (+2)</p> <p>Writes the coefficient of f(x) in decreasing order:</p> <table width="329"><tbody><tr><td>2</td> <td>4</td> <td>-6</td> <td>+3</td> <td>-5</td> </tr><tr><td></td> <td>↓</td> <td>8</td> <td>4</td> <td>14</td> </tr><tr><td></td> <td>4</td> <td>2</td> <td>7</td> <td>9</td> </tr><tr><td></td> <td></td> <td>Q</td> <td></td> <td>R</td> </tr></tbody></table><p>∴ Quotient Q(x) = 4x<sup>2</sup> +2x +7</p> <p>Remainder R(x) = 9 <sub>Ans</sub></p>
Q27:
Find the quotient and remainder when f(t) = 7t4 -4t3 +6t +3 is divided by t -3 using synthetic method of division.
Type: Short
Difficulty: Easy
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Answer: <p>Arranging the given polynomial in descending order</p> <p>f(t) = 7t<sup>4</sup> -4t<sup>3</sup> +6t +3</p> <p>The constant term with sign changed is + 3,</p> <p>t - 3 = t - (+3)</p> <p>Write the coefficient in descending order;</p> <table width="315"><tbody><tr><td>+3</td> <td>7</td> <td>-4</td> <td>0</td> <td>6</td> <td>3</td> </tr><tr><td></td> <td>↓</td> <td>21</td> <td>51</td> <td>153</td> <td>477</td> </tr><tr><td></td> <td>7</td> <td>17</td> <td>51</td> <td>159</td> <td>480</td> </tr><tr><td></td> <td>Q</td> <td></td> <td></td> <td></td> <td>R</td> </tr></tbody></table><p>∴ Quotient Q(t) = 7t<sup>3</sup> + 17x<sup>2</sup> +51x +159</p> <p>Remainder R(t) = 480 <sub>Ans</sub></p>
Q28:
(x + 2) is a factor of x3 + kx2 - 4x + 12, find the value of x.
Type: Short
Difficulty: Easy
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Answer: <p>(x + 2) is a factor of x<sup>3</sup> + kx<sup>2</sup> - 4x + 12 or x + 2 = 0 or x = -2</p> <p>Putting the value x = -2 in the given expression and f(-2) = 0</p> <p>x<sup>3</sup> + kx<sup>2</sup> - 4x + 12 =0</p> <p>or, (-2)<sup>3</sup> + k(-2)<sup>2</sup> - 4× (-2) + 12 = 0</p> <p>or, -8 + 4k + 8 + 12 = 0</p> <p>or, 4k = -12</p> <p>or, k = - \(\frac {12}{4}\) = -3</p> <p>∴ k = -3 <sub>Ans</sub></p> <p></p>
Q29:
If x - 3 is a factor of x3 + 4x2 + kx - 30, find the value of k.
Type: Short
Difficulty: Easy
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Answer: <p>x - 3 is a factor of the given expression.</p> <p>So, x=3 must satisfy the given expression, then f(3) = 0</p> <p>f(x) = x<sup>3</sup> + 4x<sup>2</sup> + kx - 30</p> <p>or, f(3) = 3<sup>3</sup> + 4×3<sup>2</sup> + k⋅3 - 30</p> <p>or, 0 = 27 + 36 +3k -30</p> <p>or, 3k + 33 = 0</p> <p>or, 3k = -33</p> <p>or k = -\(\frac {33}{3}\) = -11</p> <p>∴ k = -11 <sub>Ans</sub></p> <p></p>
Q30:
If (x - 2) is a factor of x3 - kx2 - x - 2, find the value of k.
Type: Short
Difficulty: Easy
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Answer: <p>f(x) = x<sup>3</sup> - kx<sup>2</sup> - x - 2</p> <p>If (x - 2) is a factor of f (x) then</p> <p>f(2) = 0</p> <p>f(2) =2<sup>3</sup> - k⋅2<sup>2</sup> - 2- 2</p> <p>or, 0 = 8 - 4k - 4</p> <p>or, 0 = 4 - 4k</p> <p>or, 4k = 4</p> <p>or, k = \(\frac 44\)</p> <p>∴ k = 1 <sub>Ans</sub></p>
Q31:
What do you mean by a factor theorem? If x + 1 is a factor of 2x3 - kx2 - 8x + 5, find the value of k.
Type: Short
Difficulty: Easy
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Answer: <p>If f(a0 = 0, the remainder when f(x) is divided by (x - a) is zero then (x - a) is factor of f(x).</p> <p>x + 1 = 0</p> <p>or, x = -1</p> <p>∴ f(-1) = 0</p> <p>2x<sup>3</sup> - kx<sup>2</sup> - 8x + 5</p> <p>Putting the value of x = -1 in above relation,</p> <p>f(-1) = 2 (-1)<sup>3</sup> - k (-1)<sup>2</sup> - 8 (-1) + 5</p> <p>or, - 2 - k + 8 + 5 = 0</p> <p>or, 11 - k = 0</p> <p>∴ k = 11 <sub>Ans</sub></p>
Q32:
A polynomial was divided by x - 1 and the quotient was x2 + 2x + 1 with the remainder 2. What was the original polynomial?
Type: Short
Difficulty: Easy
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Answer: <p>Quotient Q(x) =x<sup>2</sup> + 2x + 1, Remainder R(x) = 2, Original polynomial = f(x)</p> <p>We know,</p> <p>f(x) = (x + 1) Q(x) + R(x)</p> <p>or, f(x) = (x + 1) (x<sup>2</sup> + 2x + 1) + 2</p> <p>or, f(x) = x<sup>3</sup> + 2x<sup>2</sup> + x -x<sup>2</sup> - 2x - 1 + 2</p> <p>or, f(x) =x<sup>3</sup> + x<sup>2</sup>- x + 1 <sub>Ans</sub></p>
Q33:
If x - k is a factor of the polynomial x3 - kx2 - 2x + k + 4, find the value of k.
Type: Short
Difficulty: Easy
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Answer: <p>Let,</p> <p>f(k) =x<sup>3</sup> - kx<sup>2</sup> - 2x + k + 4</p> <p>If x - k is a factor of f(k) then f(k) = 0</p> <p>f(k) = k<sup>3</sup> - k⋅k<sup>2</sup> - 2k + k + 4</p> <p>or, 0 = k<sup>3</sup> - k<sup>3</sup> - k + 4</p> <p>or, 0 = -k + 4</p> <p>∴ k = 4 <sub>Ans</sub></p>
Q34:
If x + 3 is a factor of x3 - (k - 1) x2 + kx +54. find the value of k.
Type: Short
Difficulty: Easy
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Answer: <p>x + 3 is a factor of the given expression.</p> <p>When x = -3, putting the value in the given expression equal to zero.</p> <p>(-3)<sup>3</sup> - (k - 1) (-3)<sup>2</sup> + k(-3) +54 = 0</p> <p>or, -27 -9k + 9 - 3k + 54 = 0</p> <p>or, -12k + 36 = 0</p> <p>or, -12k = -36</p> <p>or k = \(\frac {-36}{-12}\)</p> <p>∴ k = 3 <sub>Ans</sub></p>
Q35:
Define the remainder theorem and find the remainder when 3x3 - 5x2 + 2x - 3 is divided by x - 2 with the help of remainder theorem.
Type: Short
Difficulty: Easy
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Answer: <p>If a number C is substituted for x in the polynomial p(x) of degree n, then P(C) is the remainder that would be obtained by dividing p(x) by x - c.</p> <p>i.e. P(x) = Q(x)⋅ (x - c) + P(c)</p> <p>where, Q(x) is a polynomial of degree n -1</p> <p>f(x) = 3x<sup>3</sup> - 5x<sup>2</sup> + 2x - 3 and g(x) = x - 2 = x - (-2)</p> <p>If f(x)÷ g(x), quotient = Q and remainder (R) = ?</p> <p>f(2) = 3(2)<sup>3</sup> - 5(2)<sup>2</sup> + 2×2 - 3 = 24 - 20 + 4 -3 = 28 - 23 = 5</p> <p>Remainder (R) = 5 <sub>Ans</sub></p>
Q36:
State remainder theorem and use it to find the remainder when: x4 - 3x3 -2x2 +x +5 is divided by x + 1.
Type: Short
Difficulty: Easy
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Answer: <p>Remainder theorem: The remainder theorem states that if f(x) is divided by (x - a), then f(a) will be the remainder.</p> <p>x + 1 is a factor ofx<sup>4</sup> - 3x<sup>3</sup> -2x<sup>2</sup> +x +5</p> <p>f(-1) =(-1)<sup>4</sup> - 3(-1)<sup>3</sup> -2(-1)<sup>2</sup> +(-1) +5 = 1 +3 -2 -1 +5 = 6 <sub>Ans</sub></p> <p></p>
Q37:
Calculate the value of p if x + 2 is a factor of 3x2 + px2 - 2x - 8.
Type: Short
Difficulty: Easy
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Answer: <p>x + 2 is a factor of 3x<sup>2</sup> + px<sup>2</sup> - 2x - 8, so the given expression is satisfies by x = -2</p> <p>3x<sup>2</sup> + px<sup>2</sup> - 2x - 8 = 0</p> <p>or, 3(-2)<sup>2</sup> + p(-2)<sup>2</sup> - 2(-2) - 8 = 0</p> <p>or, -24 + 4p +4 -8 = 0</p> <p>or, 4p = 28</p> <p>∴ p = \(\frac {28}{4}\) = 7 <sub>Ans</sub></p>
Q38:
Define "Factor Theorem". Given that the function f(x) = 2x4 - 3x2 + 6x + k. if f(1) = 0, find the value of k.
Type: Short
Difficulty: Easy
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Answer: <p>Factor Theorem: If a polynomial f(x) is divided by (x - a) and remainder R = f(a) = 0 then x - a is a factor of f(x). This theorem is known as the factor theorem.</p> <p>Given,</p> <p>f(x) = 2x<sup>4</sup> - 3x<sup>2</sup> + 6x + k and f(1) = 0</p> <p>If x = 1 then</p> <p>2 × 1<sup>4</sup> - 3 × 1<sup>2</sup> + 6 × 1 + k = f(1)</p> <p>or, 2 - 3 + 6 + k = 0</p> <p>or, 5 + k = 0</p> <p>∴ k = -5 <sub>Ans</sub></p>
Q39:
What do you mean by constant function? The range of the function f(x) = 7x - 8 is 12, find its domain.
Type: Short
Difficulty: Easy
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Answer: <p>Constant Function: A function f:A → B is called a constant function if there exists an element C∈B such that f(x) = C for all X∈A. A set containing only one element.</p> <p>Let: Y = f(x) = 7x - 8</p> <p>Y = 7x -8 [∴ Range = 13]</p> <p>or, 7x = 13 + 8</p> <p>or, x = \(\frac {21}{7}\) = 3</p> <p>∴ Domain = 3 <sub>Ans</sub></p>
Q40:
Define the remainder theorem and find the remainder when 2x3 - 7x2 + 5x + 4 is divided by (x - 3).
Type: Short
Difficulty: Easy
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Answer: <p>Remainder Theorem: When a polynomial f(x) is divided by a linear polynomial x - a then the remainder R is given by the value f(a) of the polynomial, R = f(a).</p> <table width="303"><tbody><tr><td>x - 3</td> <td>2x<sup>3</sup> - 7x<sup>2</sup> + 5x + 4</td> <td>2x<sup>2</sup> - x + 2</td> </tr><tr><td></td> <td> <p>2x<sup>3</sup> - 6x<sup>2</sup> - 445</p> <p>- + -</p> </td> <td></td> </tr><tr><td></td> <td>- x<sup>2</sup> + 5x + 4</td> <td></td> </tr><tr><td></td> <td> <p>- x<sup>2</sup> + 3x</p> <p>+ -</p> </td> <td></td> </tr><tr><td></td> <td>2x + 4</td> <td></td> </tr><tr><td></td> <td> <p>2x - 6</p> <p>- +</p> </td> <td></td> </tr><tr><td></td> <td>10</td> <td></td> </tr></tbody></table><p>∴ Remainder = 10 <sub>Ans</sub></p>
Q41:
Give the definition of remainder theorem. If a + 1 is a factor of a4 - 3a3 - 2a2 + a + p, find the value of p.
Type: Short
Difficulty: Easy
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Answer: <p>Remainder Theorem: If f(x) is a polynomial of degree n in x and if f(x)is divided by x - a, then the remainder is f(a). This theory is known as Remainder Theorey.</p> <p>Let: f(x) =a<sup>4</sup> - 3a<sup>3</sup> - 2a<sup>2</sup> + a + p</p> <p>If a + 1 is factor of f(x) then f(-1) = 0</p> <p>f(-1) =(-1)<sup>4</sup> - 3(-1)<sup>3</sup> - 2(-1)<sup>2</sup> + (-1) + p</p> <p>or, 0 = 1 + 3 - 2 - 1 + p</p> <p>or, p + 1 = 0</p> <p>∴ p = -1 <sub>Ans</sub></p>
Q42:
If f(2) = 8 and f(x) = 2x3 + 3x2 + k (1 - \(\frac {3x}{k}\) ), find the value of k.
Type: Short
Difficulty: Easy
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Answer: <p>f(2) = 8</p> <p>f(2) =2×2<sup>3</sup> + 3×2<sup>2</sup> + k (1 - \(\frac {3×2}{k}\) )</p> <p>or, 8 = 16 + 12 +k (1- \(\frac 6k\))</p> <p>or, 28 + k \(\frac {k-6}{k}\) = 8</p> <p>or, k - 6 = 8 - 28</p> <p>or, k = -20 + 6</p> <p>∴ k = -14 <sub>Ans</sub></p>
Q43:
Using the remainder theorem, find the remainder when 3x3 - 2x2 + 4x - 1 is divided by 3x + 2?
Type: Short
Difficulty: Easy
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Answer: <p>f(x) =3x<sup>3</sup> - 2x<sup>2</sup> + 4x - 1 and g(x) = 3x + 2</p> <p>g(x) = x + \(\frac 23\) = x - (- \(\frac 23\))</p> <p>In f(x)÷ g(x), remainder R = f(a) where a = - \(\frac 23\)</p> <p>f(-\( \frac 23\)) = 3 (-\( \frac 23)^3\) + 2 (-\( \frac 23)^2\) + 4×(-\(\frac 23\)) - 1</p> <p>= -\(\frac 89\) + \(\frac 89\) - \(\frac 83 \)- 1</p> <p>= \(\frac {-8-3}{3}\)</p> <p>= -\(\frac {11}{3}\)</p> <p>∴ Remainder R =f(- \(\frac 23\)) =-\(\frac {11}{3}\) <sub>Ans</sub></p> <p></p>
Q44:
If 2x - 5 is a factor of 6x3 - (k + 6)x2 + 2kx - 25, find the value of k.
Type: Short
Difficulty: Easy
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Answer: <p>f(x) =6x<sup>3</sup> - (k + 6)x<sup>2</sup> + 2kx - 25</p> <p>If (2x - 5) is a factor of f(x) then f(\(\frac 52\)) = 0</p> <p>f(\frac 52) = 6(\(\frac 52)^3\) - (k + 6)(\(\frac 52)^2\) + 2k(\(\frac 52\)) - 25</p> <p>or, 0 = \(\frac {375}{4}\) - \(\frac{25}{4}\) (k + 6) + 5k - 25</p> <p>or, 0 = \(\frac {375 - 25k - 150 + 20k - 100}{4}\)</p> <p>or, -5k + 125 = 0</p> <p>or, -5k = -125</p> <p>or, k = \(\frac {-125}{-5}\)</p> <p>∴k = 25 <sub>Ans</sub></p> <p></p>
Q45:
Define inverse function. Find the inverse function of f = {(2, 5), (3, 6), (-4, 1). (7, 4)}.
Type: Short
Difficulty: Easy
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Answer: <p>Let, f: A→B be a one to one onto function then a function f<sup>-1</sup> : B→A is called an inverse function of f. i.e.</p> <figure class="" style="width: 500px;"><img src="/uploads/a52.jpg" alt="dsv" width="500" height="173"><figcaption><br></figcaption></figure><p>f = {(2, 5), (3, 6), (-4, 1). (7, 4)}</p> <p>f<sup>-1 </sup>={(5, 2), (6, 3), (1, -4). (4, 7)} <sub>Ans</sub></p>
Q46:
If f(x) = 2x3 + 3x2 - 3x + p and f(2) = 8, find the value of P.
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>f(x) = 2x<sup>3</sup> + 3x<sup>2</sup> - 3x + p and f(2) = 8</p> <p>Putting the value of x = 2,</p> <p>f(2) = 2 × 2<sup>3</sup> + 3 × 2<sup>2</sup> - 3 × 2 + p</p> <p>or, 8 = 16 + 12 - 6 + p</p> <p>or, p = 8 - 22</p> <p>∴ p = -14 <sub>Ans</sub></p>
Q47:
If f(x) = 2x4 - 3x2 + 6x + k and f(1) = 0, then find the value of k.
Type: Short
Difficulty: Easy
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Answer: <p>Here,</p> <p>f(x) = 2x<sup>4</sup> - 3x<sup>2</sup> + 6x + k and f(1) = 0</p> <p>Putting the value of x = 1</p> <p>f(1) = 2 × 1<sup>4</sup> - 3 × 1<sup>2</sup> + 6 × 1 + k</p> <p>or, 0 = 2 -3 + 6 + k</p> <p>or, k + 5 = 0</p> <p>∴ k = - 5 <sub>Ans</sub></p>
Q48:
Let f(x) = x2 and g(x) = 3x, find:
- fog(x)
- gof(x)
- fog(2)
- gof(2)
Type: Long
Difficulty: Easy
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Answer: <p></p> <p>Here,</p> <p>f(x) = x<sup>2</sup> and g(x) = 3x</p> <ol><li type="a">fog(x) = f(3x) = (3x)<sup>2</sup> =9x<sup>2</sup><sub>Ans</sub></li> <li type="a">gof(x) = g(x<sup>2</sup>) = 3x<sup>2</sup><sub>Ans</sub></li> <li type="a">fog(2) = 9x<sup>2</sup> = 9 (2)<sup>2</sup> = 36 <sub>Ans</sub></li> <li type="a">gof(2) = 3x<sup>2</sup> = 3 (2)<sup>2</sup> = 12 <sub>Ans</sub></li> </ol>
Q49:
If f(x) = 2x - 3 and g(x) = 3x + 4 find:
- gof (x)
- fog-1 (x)
- f-1og(2)
- fog (-3)
Type: Long
Difficulty: Easy
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Answer: <p>i. gof(x)</p> <p>= g(2x - 3) [\(\because\) f(x) = 2x - 3]</p> <p>= 3(2x - 3) + 4 [\(\because\) g(x) = 3x + 4]</p> <p>= 6x - 9 + 4</p> <p>= 6x -5 <sub>Ans</sub></p> <p>Let, y = f(x) = 2x - 3</p> <p>∴ y = 2x - 3</p> <p>Interchanging the place of x and y</p> <p>x = 2y - 3</p> <p>or, 2y = x + 3</p> <p>or, y = \(\frac {x+3}{2} \)</p> <p>i.e. f<sup>-1</sup>(x) =\(\frac {x+3}{2}\)</p> <p>Let, y= g(x) = 3x + 4</p> <p>∴ y = 3x + 4</p> <p>Interchanging the place of x and y</p> <p>x = 3y + 4</p> <p>or, 3y = x - 4</p> <p>or, y = \(\frac {x-4}{3}\)</p> <p>i.e. g<sup>-1</sup>(x) =\(\frac {x-4}{3}\)</p> <p>ii. fog<sup>-1</sup>(x)</p> <p>= f\(\frac {x-4}{3}\) [\(\because\) g<sup>-1</sup>(x) =\(\frac {x-4}{3}\) ]</p> <p>= 2 \(\frac {x-4}{3}\) -3</p> <p>= \(\frac {2x- 8 - 9}{3}\)</p> <p>= \(\frac {2x - 17}{3} \)<sub>Ans</sub></p> <p>iii. f<sup>-1</sup>og (2)</p> <p>= f<sup>-1</sup>(3× 2 + 4)</p> <p>=f<sup>-1</sup>(10)</p> <p>= \(\frac {10 + 3}{2}\) [\(\because\) \(\frac {x+3}{2}\)]</p> <p>= \(\frac {13}{2}\) <sub>Ans</sub></p> <p>iv. fog(-3)</p> <p>= f(3× -3 + 4) [\(\because\) g(x) = 3x + 4]</p> <p>= f( -9 + 4)</p> <p>= f(-5)</p> <p>= 2× (-5) - 3</p> <p>= -10 - 3</p> <p>= -13 <sub>Ans</sub></p> <p></p>
Q50:
If f(x) = 8 - 3x, evaluate f-1 (-4) and ff(2).
Type: Long
Difficulty: Easy
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Answer: <p>let, y= f(x) = 8 - 3x</p> <p>Interchanging the place of x and y</p> <p>x = 8 - 3y</p> <p>or, 3y = 8 - x</p> <p>∴ y = \(\frac {8 - x}{3}\)</p> <p>i.e. f<sup>-1</sup>(x) = \(\frac {8 - x}{3}\)</p> <p>i. f<sup>-1</sup>(-4) = \(\frac {8 - (-4)}{3}\) = \(\frac {12}{3}\) = 4 <sub>Ans</sub></p> <p>ff(x) = f(8 - 3x) = 8 - 3(8- 3x) = 8 - 24 + 9x = 9x - 16</p> <p>ff(2) = 9× 2 - 16 = 18 - 16 = 2 <sub>Ans</sub></p> <p></p> <p></p>
Q51:
It is given that the function f(x) = 4x + 7 and g(x) = 3x - 5. If fg-1(x) = 15, find the value of x.
Type: Long
Difficulty: Easy
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Answer: <p>Let, y = g(x) = 3x - 5</p> <p>y = 3x - 5 ---------- (1)</p> <p>Interchanging the place of xand y in equation (1)</p> <p>x = 3y - 5</p> <p>or, 3y = x + 5</p> <p>∴ y = \(\frac {x + 5}{3}\)</p> <p>i.e. g<sup>-1</sup>(x) = \(\frac {x + 5}{3}\)</p> <p>fg<sup>-1</sup>(x) = 15</p> <p>or, f\(\frac {x + 5}{3}\)=15</p> <p>or, 4\(\frac {x + 5}{3}\) = 15</p> <p>or, \(\frac {4x + 20 + 21}{3}\) = 15</p> <p>or, 4x + 41 = 45</p> <p>or, 4x = 45 - 41 = 4</p> <p>or, x = \(\frac 44\)</p> <p>∴ x = 1 <sub>Ans</sub></p>
Q52:
If f(x) = 3x + 5, find ff-1(2) and f-1(3).
Type: Long
Difficulty: Easy
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Answer: <p>Let, y = f(x) = 3x + 5</p> <p>or, y= 3x + 5</p> <p>Interchanging the position of x and y,</p> <p>x = 3y + 5</p> <p>or, 3y = x - 5</p> <p>or, y = \(\frac {x - 5}{3}\)</p> <p>∴ f<sup>-1</sup>(x) =\(\frac {x - 5}{3}\)</p> <p>f<sup>-1</sup>(2) =\(\frac {2 - 5}{3}\) = - \(\frac 33\) = -1</p> <p>ff<sup>-1</sup>(2) = f(-1) = 3× (-1) + 5 = - 3 + 5 = 2</p> <p>f<sup>-1</sup>f(3) = f<sup>-1</sup> (3× 3 +5) = f<sup>-1</sup> (14) = \(\frac {14 - 5}{3}\) = \(\frac 93\) = 3</p> <p>Hence, ff<sup>-1</sup>(2) = 2</p> <p>f<sup>-1</sup> f(3) = 3 <sub>Ans</sub></p> <p></p>
Q53:
If f(x) = 3x - 4, g(x) = x + 3, h(x) = - 2x + 1, find ghf(x) and g-1hf(x).
Type: Long
Difficulty: Easy
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Answer: <p>ghf(x)</p> <p>= gh(3x - 4)</p> <p>= g [ -2 (3x - 4) + 1]</p> <p>= g(-6x + 8 + 1)</p> <p>= g(- 6x + 9)</p> <p>= -6x + 9 +3</p> <p>= - 6x + 12</p> <p>Let,</p> <p>y = g(x) = x + 3</p> <p>∴ y = x + 3</p> <p>Interchanging the place of x and y,</p> <p>x = y + 3</p> <p>or, x - 3 = y</p> <p>i.e y = x - 3 [\(\because\) g<sup>-1</sup>(x) = x - 3]</p> <p>g<sup>-1</sup>hf(x)</p> <p>= g<sup>-1</sup>h(3x - 4)</p> <p>= g<sup>-1</sup> [ - 2(3x - 4) + 1]</p> <p>= g<sup>-1</sup> (-6x + 8 + 1)</p> <p>= g<sup>-1</sup> (-6x + 9)</p> <p>= - 6x + 9 -3</p> <p>= - 6x + 6</p> <p>Hence, ghf(x) = - 6x + 12</p> <p>and, g<sup>-1</sup>hf(x) = -6x + 6 <sub>Ans</sub></p>
Q54:
If f(x) = 3x + 4; g(x) = 2(x + 1), prove that fog = gof and find the value of f-1(2).
Type: Long
Difficulty: Easy
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Answer: <p>Given,</p> <p>f(x) = 3x + 4</p> <p>g(x) = 2(x + 1) = 2x + 2</p> <p>f<sub>o</sub>g(x) = f(2x + 2) = 3 (2x + 2) + 4 = 6x + 6 + 4 = 6x + 10</p> <p>g<sub>o</sub>f(x) = g(3x + 4) = 2 (3x + 4) + 2 = 6x + 8 +2 =6x + 10</p> <p>Hence, f<sub>o</sub>g(x) = g<sub>o</sub>f(x) <sub>proved</sub></p> <p>Let y = 3x + 4 = f(x)</p> <p>y = 3x + 4</p> <p>Interchanging the place of x and y</p> <p>x = 3y + 4</p> <p>or, 3y = x - 4</p> <p>or, y = \(\frac {x - 4}{3}\)</p> <p>f<sup>-1</sup>(x) = \(\frac {x - 4}{3}\)</p> <p>∴ f<sup>-1</sup>(2) = \(\frac {2 - 4}{3}\) = - \(\frac 23\)<sub>Ans</sub></p>
Q55:
If f(x) =\( \frac {1}{x}\), x ≠ 0.Prove that fof-1 = f-1of.
Type: Long
Difficulty: Easy
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Answer: <p>Let y = f(x) = \(\frac 1x\)</p> <p>y = \(\frac 1x\)</p> <p>Interchanging the place of x and y</p> <p>x = \(\frac 1y\)</p> <p>y = \(\frac 1x\)</p> <p>∴ f<sup>-1</sup>(x) = \(\frac 1x\)</p> <p>L.H.S</p> <p>=fof<sup>-1</sup> (x)</p> <p>= f\(\frac 1x\)</p> <p>=\( \cfrac{1}{<br>\cfrac{1}{x}}\)</p> <p>= x</p> <p>R.H.S</p> <p>= f<sup>-1</sup>of(x)</p> <p>= f<sup>-1</sup>\(\frac 1x\)</p> <p>= \(\cfrac{1}{<br>\cfrac{1}{x}}\)</p> <p>= x</p> <p>Hence, L.H.S = R.H.S <sub>proved</sub></p> <p></p> <p></p>
Q56:
If f(x) = \(\frac{x}{x - 3}\), find the value of x for which f(x) = f-1(x).
Type: Long
Difficulty: Easy
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Answer: <p>Let y = f(x) =\(\frac {x}{x-3}\)</p> <p>y = \(\frac {x}{x - 3}\)</p> <p>Interchanging the place of x and y,</p> <p>x = \(\frac {y}{y - 3}\)</p> <p>or, xy- 3x= y</p> <p>or, xy - y = 3x</p> <p>or, y (x - 1) = 3x</p> <p>or, y = \(\frac {3x}{x - 1}\)</p> <p>i.e., f<sup>-1</sup>(x) = \(\frac {3x}{x - 1}\)</p> <p>From the question, f(x) = f<sup>-1</sup>(x)</p> <p>\(\frac {x}{x - 3}\) =\(\frac {3x}{x - 1}\)</p> <p>or, x<sup>2</sup> - x = 3x<sup>2</sup> - 9x</p> <p>or, 3x<sup>2</sup> - 9x - x<sup>2</sup> + x = 0</p> <p>or, 2x<sup>2</sup> - 8x = 0</p> <p>or, 2x(x - 4) = 0</p> <p>Either, x = 0 and x - 4 = 0 or, x = 4</p> <p>∴x = 0, 4 <sub>Ans</sub></p> <p></p> <p></p>
Q57:
It is given that f(x) = x2 - 2x and g(x) = 2x + 3. If fg-1(x) = 3, calculate the value of x.
Type: Long
Difficulty: Easy
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Answer: <p>Here,</p> <p>f(x) = x<sup>2</sup> - 2x and g(x) = 2x + 3</p> <p>Let: y = g(x) = 2x + 3</p> <p>Interchanging the place of x and y</p> <p>x = 2y + 3</p> <p>or, 2y = x - 3</p> <p>or, y = \(\frac {x-3}{2}\)</p> <p>∴ g<sup>-1</sup>(x) =\(\frac {x-3}{2}\)</p> <p>fg<sup>-1</sup>(x) = 3</p> <p>f\(\frac {x-3}{2}\) = 3</p> <p>or, \((\frac {x-3}{2})^2\) - 2 \(\frac {x-3}{2}\) = 3</p> <p>or, \(\frac {x^2 - 6x + 9}{4}\) -\(\frac {x-3}{2}\) = 3</p> <p>or, \(\frac {x^2 - 6x + 9 - 4x + 12}{4}\) = 3</p> <p>or, x<sup>2</sup> - 10x + 21 =12</p> <p>or, x<sup>2</sup> -10x + 21 - 12 = 0</p> <p>or, x<sup>2</sup> - 10x + 9 = 0</p> <p>or, x<sup>2</sup> - 9x - x + 9 = 0</p> <p>or, x(x - 9) -1 (x - 9)= 0</p> <p>or, (x - 9) (x - 1) = 0</p> <p>Either, x - 9 = 0 i.e. x = 9</p> <p>or, x - 1 = 0 i.e. x = 1</p> <p>∴ x = 9 and 1 <sub>Ans</sub></p> <p></p>
Q58:
If f(x) = 1 + 2x and g(x) = \(\frac {1}{1-x}\), find the value of g-1 \(\frac 12\) and fg (-1).
Type: Long
Difficulty: Easy
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Answer: <p>Let: y =g(x) = \(\frac {1}{1-x}\)</p> <p>y = \(\frac {1}{1-x}\)</p> <p>Interchanging the place of x and y</p> <p>x =\(\frac {1}{1-y}\)</p> <p>or, 1 - y = \(\frac 1x\)</p> <p>or, y = 1 - \(\frac 1x\) = \(\frac {x - 1}{x}\)</p> <p>g<sup>-1</sup>(x) = \(\frac {x - 1}{x}\)</p> <p>g<sup>-1</sup>\(\frac 12\)</p> <p>= \(\cfrac{(\frac{1}{2})-1}{<br>\cfrac{1}{2}}\)</p> <p>= \(\frac {1-2}{2}\)× \(\frac 21\)</p> <p>= -1 <sub>Ans</sub></p> <p>Again, fg(x) = f\(\frac {1}{1-x}\) = 1 + \(\frac {2×1}{1-x}\)</p> <p>fg(x) = f\(\frac {1}{1-x}\) = 1 + \(\frac {2×1}{1-x}\)</p> <p>= f\(\frac {1}{1-x}\) = 1 + \(\frac {2×1}{1-x}\)</p> <p>= 1 + \(\frac {2×1}{1-x}\)</p> <p>= \(\frac {1 - x + 2}{1 - x}\)</p> <p>= \(\frac {3 - x}{1 - x}\)</p> <p>fg(-1)</p> <p>= \(\frac {3 - (-1)}{1 - (-1)}\)</p> <p>= \(\frac {3 + 1}{1 + 1}\)</p> <p>= \(\frac 42\)</p> <p>= 2 <sub>Ans</sub></p>
Q59:
If function f(x) = \(\frac {2x + 3}{x + 2}\) and g(x) = x - 2, find the values of f-1(x), f-1(1), fg (x) and fg(1).
Type: Long
Difficulty: Easy
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Answer: <p>Let: y = f(x) = \(\frac {2x + 3}{x + 2}\)</p> <p>Interchanging the position of x and y</p> <p>x= \(\frac {2y + 3}{y + 2}\)</p> <p>or, xy+ 2x - 2y = 3</p> <p>or, y(x - 2) = 3 - 2x</p> <p>∴y= \(\frac {3 - 2x}{x - 2}\)</p> <ol><li type="i">f-1(x) = \(\frac {3 - 2x}{x - 2}\)</li> <li type="i">f-1(1) = \(\frac {3 - 2(1)}{1 - 2}\) = -\(\frac 11\) = -1</li> <li type="i">fg(x) = f(x - 2) =\(\frac {2(x-2) + 3}{(x-2) + 2}\) =\(\frac {2x - 4 + 3}{x}\) = \(\frac {2x-1}{x}\)</li> <li type="i">fg(1) = \(\frac {2×1-1}{1}\) = \(\frac {2-1}{1}\)<sub>Ans</sub></li> </ol>
Q60:
If f(x) = \(\frac {3x + 11}{x - 3}\), x ≠ 3 and g(x) - \(\frac {x - 3}{2}\) are given functions. Find f-1(x) and if f(x) = g-1(x) find the value of x.
Type: Long
Difficulty: Easy
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Answer: <p>Let, y= f(x) =\(\frac {3x + 11}{x - 3}\)</p> <p>Interchanging the value of x and y,</p> <p>x =\(\frac {3y + 11}{y - 3}\)</p> <p>or, xy - 3x = 3y + 11</p> <p>or, xy - 3y = 3x + 11</p> <p>or, y(x - 3) = 3x + 11</p> <p>or, y = \(\frac {3x + 11}{x - 3}\)</p> <p>∴ f<sup>-1</sup>(x) = \(\frac {3x + 11}{x - 3}\) <sub>Ans</sub></p> <p>Again,</p> <p>Let: y = g(x) = \(\frac {x - 3}{2}\)</p> <p>y =\(\frac {x - 3}{2}\)</p> <p>Interchanging the place of x and y,</p> <p>x =\(\frac {y - 3}{2}\)</p> <p>or, 2x = y - 3</p> <p>or, y = 2x + 3</p> <p>∴ g<sup>-1</sup>(x) = 2x + 3</p> <p>From the given question,</p> <p>f(x) = g<sup>-1</sup>(x)</p> <p>or,\(\frac {3x + 11}{x - 3}\) = 2x + 3</p> <p>or, 3x + 11 = 2x<sup>2</sup> + 3x - 6x - 9</p> <p>or, 2x<sup>2</sup> - 3x - 9 - 3x - 11 = 0</p> <p>or, 2x<sup>2</sup> - 6x - 20 = 0</p> <p>or, 2(x<sup>2</sup> - 3x - 10) = 0</p> <p>or, x<sup>2</sup> - 5x + 2x - 10 = 0</p> <p>or, x(x - 5) + 2(x - 5) = 0</p> <p>or, (x - 5) (x + 2) = 0</p> <p>Either, x - 5 = 0∴ x = 5</p> <p>Or, x + 2 = 0∴ x = -2</p> <p>Hence, x = 5 or -2 <sub>Ans</sub></p> <p></p>
Q61:
It is given that the function f(x) = 4x + 7 and g(x) = 3x - 5. If fg-1(x) = 15, find the value of x.
Type: Long
Difficulty: Easy
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Answer: <p>Here,</p> <p>Let: y = g(x) = 3x - 5</p> <p>y = 3x - 5 ------------------(1)</p> <p>Interchanging the position of x and y in equation (1),</p> <p>x = 3y - 5</p> <p>or, 3y = x + 5</p> <p>or, y = \(\frac {x + 5}{3}\)</p> <p>∴ g<sup>-1</sup>(x) = \(\frac {x + 5}{3}\)</p> <p>fg<sup>-1</sup>(x) = 15</p> <p>or, f \(\frac {x + 5}{3}\) = 15</p> <p>or, 4 \(\frac {x + 5}{3}\) + 7 = 15</p> <p>or, \(\frac {4x + 20}{3}\) + 7 = 15</p> <p>or, \(\frac {4x + 20 + 21}{3}\) = 15</p> <p>or, 4x + 41 = 45</p> <p>or, 4x = 45 - 41</p> <p>or, 4x = 4</p> <p>or, x = \(\frac 44\)</p> <p>∴ x = 1 <sub>Ans</sub></p>
Q62:
Solve the equation:
x3 - 4x2 + x + 6 = 0
Type: Long
Difficulty: Easy
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Answer: <p>x<sup>3</sup> - 4x<sup>2</sup> + x + 6 = 0</p> <p>(x - 2) is a factor ofx<sup>3</sup> - 4x<sup>2</sup> + x + 6</p> <p>or, x<sup>3</sup> - 2x<sup>2</sup> - 2x<sup>2</sup> + 4x - 3x + 6 = 0</p> <p>or, x<sup>2</sup>(x - 2) - 2x(x - 2) - 3(x - 2) = 0</p> <p>or, (x - 2) (x<sup>2</sup> - 2x - 3) = 0</p> <p>or, (x - 2) (x<sup>2</sup> - 3x + x - 3) = 0</p> <p>or, (x - 2) [x(x - 3) + 1(x - 3)] = 0</p> <p>or, (x - 2) (x - 3) (x + 1) = 0</p> <p>Either, x - 2 = 0 ∴ x = 2</p> <p>Or, x - 3 = 0 ∴ x = 3</p> <p>or, x + 1 = 0 ∴ x = - 1</p> <p>Hence, x = 2, 3, -1 Ans</p> <p>Rough:</p> <p>x = 2</p> <p>x<sup>3</sup> - 4x<sup>2</sup> + x + 6</p> <p>= 2<sup>3</sup> - 4⋅ 2<sup>2</sup> + 2 +6</p> <p>= 16 - 16</p> <p>= 0</p>
Q63:
Solve:
2x3 + 3x2 - 11x - 6 = 0
Type: Long
Difficulty: Easy
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Answer: <p>x - 2 is a factor of given expression</p> <p>or, 2x<sup>3</sup> - 4x<sup>2</sup> + 7x<sup>2</sup> - 14x + 3x - 6 = 0</p> <p>or, 2x<sup>2</sup> (x - 2) + 7x (x - 2) + 3 (x - 2) = 0</p> <p>or, (x - 2) (2x<sup>2</sup> + 7x + 3) = 0</p> <p>or, (x - 2) (2x<sup>2</sup> + 6x + x +3) = 0</p> <p>or, (x - 2) [2x (x + 3) + 1 (x + 3)] = 0</p> <p>or, (x - 2) (x + 3) (2x + 1) = 0</p> <p>Either, x + 3 = 0∴ x = -3</p> <p>Or, x - 2 = 0∴ x = 2</p> <p>Or, 2x + 1 = 0∴ x = -\(\frac 12\)</p> <p>∴ x = 2 , - 3 , -\(\frac12\) <sub>Ans</sub></p> <p>Rough:</p> <p>If x = 2</p> <p>2x<sup>3</sup> + 3x<sup>2</sup> - 11x - 6</p> <p>= 2(2)<sup>3</sup> + 3(2)<sup>2</sup> - 11(2) - 6</p> <p>= 16 + 12 - 22 - 6</p> <p>= 0</p>
Q64:
Solve:
3x3 - 13x2 + 16 = 0
Type: Long
Difficulty: Easy
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Answer: <p>x = -1 is correct.</p> <p>x + 1 is a factor of above equation.</p> <p>or, 3x<sup>3</sup> + 3x<sup>2</sup> - 16x<sup>2</sup> - 16x + 16x + 16 = 0</p> <p>or, 3x<sup>2</sup>(x + 1) - 16x(x + 1) + 16(x + 1) = 0</p> <p>or, (x + 1) (3x<sup>2</sup> - 16x + 16) = 0</p> <p>or, (x + 1) (3x<sup>2</sup> - 12x - 4x + 16) = 0</p> <p>or, (x + 1) [3x (x - 4) - 4 (x - 4)] = 0</p> <p>or, (x + 1) (x - 4) (3x - 4) = 0</p> <p>Either, x + 1 = 0 ∴ x = - 1</p> <p>Or, x - 4 = 0 ∴ x = 4</p> <p>Or, 3x - 4 = 0 ∴ x = \(\frac 43\)</p> <p>∴ x = -1, 4 and \(\frac 43\) <sub>Ans</sub></p> <p>Rough:</p> <p>Put x = -1</p> <p>or, 3(-1)<sup>3</sup> - 13(-1)<sup>2</sup> + 16 = 0</p> <p>or, -3 -13 + 16 = 0</p> <p>or, - 16 + 16 = 0</p> <p>∴ 0 = 0</p>
Q65:
Solve:
x3 - 19x - 30 = 0
Type: Long
Difficulty: Easy
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Answer: <p>x + 2 is a factor of given equation.</p> <p>x<sup>3</sup> + 2x<sup>2</sup> - 2x<sup>2</sup> - 4x - 15x - 30 = 0</p> <p>or, x<sup>2</sup>(x + 2) - 2x(x + 2) - 15(x + 2) = 0</p> <p>or, (x + 2) (x<sup>2</sup> - 2x - 15) = 0</p> <p>or, (x + 2) (x<sup>2</sup> - 5x + 3x - 15) = 0</p> <p>or, (x + 2) [x(x - 5) + 3(x - 5)] = 0 </p> <p>or, (x + 2) (x - 5) (x + 3) = 0</p> <p>Either, x + 2 = 0 ∴ x = -2</p> <p>Or, x - 5 = 0 ∴ x = 5 </p> <p>Or, x + 3 = 0 ∴ x = -3</p> <p>∴ x = -2, -3, 5 <sub>Ans</sub></p>
Q66:
Solve:
x3 - 3x - 2 = 0
Type: Long
Difficulty: Easy
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Answer: <p>x<sup>3</sup> - 3x - 2 = 0</p> <p>x + 1 is a factor of given equation,</p> <p>or, x<sup>3</sup> + x<sup>2</sup> - x<sup>2</sup> - x - 2x - 2 = 0</p> <p>or, x<sup>2</sup> (x + 1) - x (x + 1) - 2 (x + 1) = 0</p> <p>or, (x + 1) (x<sup>2</sup> - x - 2) = 0</p> <p>or, (x + 1) [x(x - 2) + 1 -(x - 2)] = 0</p> <p>or, (x + 1) (x - 2) (x + 1) = 0</p> <p>Either, x + 1 = 0 ∴ x = -1</p> <p>Or, x - 2 = 0 ∴ x = 2</p> <p>∴ x = -1 and 2 <sub>Ans</sub></p>
Q67:
Solve:
(x + 1) (x + 2) (x + 3) (x + 4) - 8 = 0
Type: Long
Difficulty: Easy
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Answer: <p>(x + 1) (x + 2) (x + 3) (x + 4) - 8 = 0</p> <p>or, (x<sup>2</sup> + x + 4x + 4) ( x<sup>2</sup> + 2x + 3x + 6) - 8 = 0</p> <p>or, (x<sup>2</sup> + 5x + 4) (x<sup>2</sup> + 5x + 6) - 8 = 0</p> <p>Let: x<sup>2</sup> + 5x = k</p> <p>(k + 4) (k + 6) - 8 = 0</p> <p>or, k<sup>2</sup> + 6k + 4k + 24 - 8 = 0</p> <p>or, k<sup>2</sup> + 10k + 16 = 0</p> <p>or, k<sup>2</sup> + 8k + 2k + 16 = 0</p> <p>or, k(x + 8) + 2(x + 8) = 0</p> <p>or, (k + 2) (x + 8) = 0</p> <p>Putting the value of k</p> <p>(x<sup>2</sup> + 5x + 8) (x<sup>2</sup> + 5x + 2) = 0</p> <p>Either, x<sup>2</sup> + 5x + 8 = 0 --------------(1)</p> <p>Or, x<sup>2</sup> + 5x + 2 = 0 ------------------(2)</p> <p>Taking eq<sup>n</sup>(1)</p> <p>x =\(\frac {-5 ± (\sqrt{5^{2}-4×1×8})}{2×1}\) =\(\frac {-5 ± \sqrt{-7}}{2×1}\) (Impossible)</p> <p>Taking eq<sup>n</sup>(2)</p> <p>x =\(\frac {-5 ± (\sqrt{5^{2}-4×1×2})}{2×1}\) =\(\frac {-5 ± \sqrt{17}}{2×1}\) <sub>Ans</sub></p> <p></p> <p></p> <p></p>
Q68:
Divide the polynomials using synthetic method.
f(x) = x4 - x3 - 3x2 - 2x + 5, where x = 2, -1, 3
Type: Long
Difficulty: Easy
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Answer: <p>Here,</p> <p>f(x) = x<sup>4</sup> - x<sup>3</sup> - 3x<sup>2</sup> - 2x + 5</p> <p>When x = 2 then,</p> <table width="446"><tbody><tr><td>2</td> <td>1</td> <td>-1</td> <td>-3</td> <td>-2</td> <td>5</td> </tr><tr><td></td> <td></td> <td>2</td> <td>2</td> <td>-2</td> <td>-8</td> </tr><tr><td></td> <td>1</td> <td>1</td> <td>-1</td> <td>-4</td> <td>-3</td> </tr></tbody></table><p>∴ Quotient = x<sup>3</sup> + x<sup>2</sup> - x - 4 and f(x) = -3 <sub>Ans</sub></p> <p>When x = -1 then,</p> <table width="449"><tbody><tr><td>-1</td> <td>1</td> <td>-1</td> <td>-3</td> <td>-2</td> <td>5</td> </tr><tr><td></td> <td></td> <td>-1</td> <td>2</td> <td>1</td> <td>1</td> </tr><tr><td></td> <td>1</td> <td>-2</td> <td>-1</td> <td>-1</td> <td>6</td> </tr></tbody></table><p>∴ Quotient = x<sup>3</sup> - 2x<sup>2</sup> - x - 1 and f(2) = 6 <sub>Ans</sub></p> <p>When x = 3 then,</p> <table width="450"><tbody><tr><td>3</td> <td>1</td> <td>-1</td> <td>-3</td> <td>-2</td> <td>5</td> </tr><tr><td></td> <td></td> <td>3</td> <td>6</td> <td>9</td> <td>21</td> </tr><tr><td></td> <td>1</td> <td>2</td> <td>3</td> <td>7</td> <td>26</td> </tr></tbody></table><p>∴ Quotient = x<sup>3</sup> + 2x<sup>2</sup> + 3x + 7 and f(3) = 26 <sub>Ans</sub></p>
Q69:
Solve:
x3 - x2 - 14x + 24 = 0
Type: Long
Difficulty: Easy
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Answer: <p>x - 2 = 0</p> <p>x - 2 is a factor of given expression</p> <p>or, x<sup>3</sup> - x<sup>2</sup> - 14x + 24 = 0</p> <p>or, x<sup>3</sup> - 2x<sup>2</sup> + x<sup>2</sup> - 2x - 12x + 24 = 0</p> <p>or, x<sup>2</sup>(x - 2) + x(x- 2) - 12(x - 2) = 0</p> <p>or, (x - 2) (x<sup>2</sup> + x -12) = 0</p> <p>or, (x - 2) (x<sup>2</sup> + 4x - 3x - 12) = 0</p> <p>or, (x - 2) [x(x + 4) - 3(x + 4)] = 0</p> <p>or, (x - 2) (x + 4) (x -3) = 0</p> <p>Either, x - 2 = 0 ∴x = 2</p> <p>Or, x + 4 = 0 ∴ x = - 4</p> <p>Or, x - 3 = 0 ∴ x = 3</p> <p>∴ x = 2, 3, -4 Ans</p> <p>Rough:</p> <p>x = 2</p> <p>i.e 2<sup>3</sup> - 2<sup>2</sup> - 14× 2 + 24</p> <p>= 8 - 4 -28 + 24</p> <p>= 0</p>
Q70:
Solve:
2x3 - 3x2 - 3x + 2 = 0
Type: Long
Difficulty: Easy
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Answer: <p>2x<sup>3</sup> - 3x<sup>2</sup> - 3x + 2 = 0</p> <p>or, 2x<sup>3</sup> + 2 - 3x<sup>2</sup> - 3x = 0</p> <p>or, 2(x<sup>3</sup> + 13) - 3x (x + 1) = 0</p> <p>or, 2(x + 1) (x<sup>2</sup> - x + 1) - 3x (x + 1) = 0</p> <p>or, (x + 1)[2x<sup>2</sup> - 2x + 2 - 3x] = 0</p> <p>or, (x + 1) (2x<sup>2</sup> - 5x + 2) = 0</p> <p>or, (x + 1) (2x<sup>2</sup> - 4x - x + 2) = 0</p> <p>or, (x + 1) [2x(x - 2) - 1(x - 2)] = 0</p> <p>or, (x + 1) (x - 2) (2x - 1) = 0</p> <p>Either, x + 1 = 0 ∴x = -1</p> <p>Or, x - 2 = 0 ∴x = 2</p> <p>Or, 2x - 1 = 0 ∴ x = \(\frac 12\)</p> <p>∴ x = -1, 2 , \(\frac12\) <sub>Ans</sub></p> <p></p>
