Mathematics Class X

Question 1 of 40

1. Question
If one root of the polynomial $f(x) = 5{x^2} + 13x + k$ is reciprocal of the other, then the value of k is?
- $0$
- $5$
- $\dfrac{1}{6}$
- $6$
Correct

Incorrect

Hint

$\begin{array}{l}\alpha \dfrac{1}{\alpha } = \dfrac{k}{5}\\1 = \dfrac{k}{5} \Rightarrow k = 5\end{array}$
Question 2 of 40

2. Question
If the product of two zeroes of the polynomial $f(x) = 2{x^3} + 6{x^2} – 4x + 9$ then its third zero is?
- $\dfrac{3}{2}$
- $\dfrac{{ – 3}}{2}$
- $\dfrac{9}{2}$
- $1$
Correct

Incorrect

Hint

If $f(x) = 2{x^3} + 6{x^2} - 4x + 9$
If a, b, g are the roots then
$abg = \dfrac{{ - 9}}{2}$
$\begin{array}{l}3g = \dfrac{{ - 9}}{2}\\ \Rightarrow g = \dfrac{{ - 3}}{2}\end{array}$
Question 3 of 40

3. Question
If $f(x) = a{x^2} + bx + c$ has no real zeros and a+b+c < 0 then?
- C = 0
- C>0
- C<0
- None of these
Correct

Incorrect

Hint

F(x)>0 and at x=1, f(x)<0
So, curve cuts negative g-axis
Question 4 of 40

4. Question
The number of real zeros of the polynomial ${x^4} + 4{x^2} + 5$is?
- One real zero
- Two real zero
- Third real zero
- No real zero
Correct

Incorrect

Hint

Polynomial ${x^4} + 4{x^2} + 5$ has its decrement negative.
Question 5 of 40

5. Question
If the polynomial $2{x^3} + a{x^2} + 3x – 5$and ${x^3} + {x^2} + 4x + a$ leave the same remainder when divided by x-2 then the value of a is?
- $\dfrac{3}{{13}}$
- $\dfrac{3}{{17}}$
- $\dfrac{{ – 13}}{3}$
- $\dfrac{{ – 3}}{{13}}$
Correct

Incorrect

Hint

$\begin{array}{l}f(2) = 2 \times {2^3} + a \times {2^2} + 3 \times 2 - 5\\ = 16 + 4a + 1\\ = {2^3} + {2^2} - 4 \times 2 + a\\ = 4 + a\\16 + 4a + 1 = 4 + a\\3a = - 13\\ \Rightarrow a = \dfrac{{ - 13}}{3}\end{array}$
Question 6 of 40

6. Question
Factors of ${x^3} + 7{x^2} + 14x + 8$are?
- (x-1)(x-2)(x-4)
- (x+1)(x+2)(x-4)
- (x+1)(x+2)(x-4)
- (x+1)(x+2)(x+4)
Correct

Incorrect

Hint

Product is equal to ${x^2} + 7{x^2} + 14x + 8$
Question 7 of 40

7. Question
When a polynomial $2{x^3} + 5{x^2} – 2x – 6$is divided by a polynomial x+1 then the remainder will be?
- -1
- 1
- 2
- 3
Correct

Incorrect

Hint

$\begin{array}{l}f( - 1) = 2 \times {( - 1)^3} + 5 \times {( - 1)^2} + 2 - 6\\ = - 1\end{array}$
Question 8 of 40

8. Question
If (x+1) and (x-1) are the factors of $p{x^3} + {x^2} – 2x + q$then the value of p and q are?
- P=-1,q=2
- P=2,q=-1
- P=2,q=1
- P=-2,q=-2
Correct

Incorrect

Hint

$\begin{array}{l}f( - 1) = p{( - 1)^3} + {( - 1)^2} - 2 \times ( - 1) + q\\ = - p + 1 + 2 + q\\q - p + 3 = 0.....(1)\\f(1) = p + 1 - 2 + q\\ \Rightarrow p + q - 1 = 0\end{array}$
Add (1)+(2)
2q+2=0; q=-1
P=2
Question 9 of 40

9. Question
Minimum value of the expression ${x^2} + x – 2$is?
- $\dfrac{9}{4}$
- $\dfrac{{ – 9}}{4}$
- $\dfrac{4}{9}$
- $\dfrac{3}{5}$
Correct

Incorrect

Hint

Minimum value $= \dfrac{{ - D}}{{4a}} = \dfrac{{ - 9}}{4}$
Question 10 of 40

10. Question
If we draw the graph of a cubic polynomial, then it will intersect the axis of x at least in?
- Zero point
- One point
- Two points
- Three points
Correct

Incorrect

Hint

Minimum one time as if its roots are complex they occur in pair i.e. on ereal root atleast.
Question 11 of 40

11. Question
Factors of ${(x + 1)^3} – {(x – 1)^3}$are?
- $2({x^2} + 1)$
- $3{x^2} + 1$
- $2(2{x^2} + 1)$
- $2(3{x^2} + 1)$
Correct

Incorrect

Hint

$\begin{array}{l}{(x + 1)^3} - {(x - 1)^3}\\ = (x + 1 - x + 1)\left[ {{{(x + 1)}^2} + ({x^2} - 1) + {{(x - 1)}^2}} \right]\\ = 2\left[ {{x^2} + 1 + {x^2} - 1 + {x^2} + 1} \right]\\ = 2\left[ {3{x^2} + 1} \right]\end{array}$
Question 12 of 40

12. Question
The value of $({55^3} + {45^3})({55^2} – 55 \times 45 + {45^2})$is?
- 100
- 105
- 125
- 75
Correct

Incorrect

Hint

$\begin{array}{l}\dfrac{{{{55}^3} + {{45}^3}}}{{{{55}^2} - 55 \times 45 + {{45}^2}}}\\ = \dfrac{{(55 + 45)({{55}^2} - 55 \times 45 + {{45}^2})}}{{{{55}^2} - 55 \times 45 + {{45}^2}}}\\ = 100\end{array}$
Question 13 of 40

13. Question
If $x + \dfrac{1}{x} = 3$then ${x^2} + \dfrac{1}{{{x^2}}}$is equal to?
- 9
- $\dfrac{{82}}{9}$
- 7
- 11
Correct

Incorrect

Hint

$\begin{array}{l}x + \dfrac{1}{x} = 3\\ \Rightarrow {\left( {x + \dfrac{1}{x}} \right)^2} = {x^2} + \dfrac{1}{{{x^2}}} + 2\\ \Rightarrow 7 = {x^2} + \dfrac{1}{{{x^2}}}\end{array}$
Question 14 of 40

14. Question
If (x-2) is a factor of $({x^2} + 3qx – 2q)$then the value of q is?
- 2
- -2
- 1
- -1
Correct

Incorrect

Hint

$\begin{array}{l}{x^2} + 3qx - 2q = 0\\ \Rightarrow {(2)^2} + 3q \times 2 - 2q = 0\\ \Rightarrow 4 + 6q - 2q = 0\\ \Rightarrow 4q = - 4\\ \Rightarrow q = - 1\end{array}$
Question 15 of 40

15. Question
If $\left( {x + \dfrac{1}{x}} \right) = 6$then ${x^4} + \dfrac{1}{{{x^4}}}$is?
- 1148
- 1150
- 1154
- 1156
Correct

Incorrect

Hint

$\begin{array}{l}x + \dfrac{1}{x} = 6\\ \Rightarrow {\left( {x + \dfrac{1}{x}} \right)^2} = 36\\ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} + 2 = 36\\ \Rightarrow {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} = {(34)^2}\\ \Rightarrow {x^4} + \dfrac{1}{{{x^4}}} = 1154\end{array}$
Question 16 of 40

16. Question
If ${x^2} + \dfrac{1}{{{x^2}}} = 66$then$x – \dfrac{1}{x}$?
- 8
- -8
- $ \pm 8$
- $ \pm 4$
Correct

Incorrect

Hint

$\begin{array}{l}{x^2} + \dfrac{1}{{{x^2}}} = 66\\{\left( {x - \dfrac{1}{x}} \right)^2} = {\left( {{x^2} + \dfrac{1}{x}} \right)^2} - 4\\ = {x^2} + \dfrac{1}{{{x^2} + 2 - 4}}\\ = 66 + 2 - 4\\ = 68 - 4\\ = 64\\x - \dfrac{1}{x} = \pm 8\end{array}$
Question 17 of 40

17. Question
If ${a^2} + {b^2} + {c^2} = 16$and ab+bc+ca=10 then a+b+c is?
- 6
- -6
- $ \pm 8$
- $ \pm 6$
Correct

Incorrect

Hint

$\begin{array}{l}{(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2(ab + bc + ca)\\ = 16 + 2 \times 10\\ = 36\\(a + b + c) = \pm 6\end{array}$
Question 18 of 40

18. Question
If a+b=10, ab=21 then ${a^3} + {b^3}$is?
- 370
- 365
- 360
- 355
Correct

Incorrect

Hint

$\begin{array}{l}{a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})\\ = (a + b)\left[ {{{(a + b)}^2} - 3ab} \right]\\ = 10\left[ {100 - 3 \times 21} \right]\\ = 10 \times 37 = 370\end{array}$
Question 19 of 40

19. Question
Factors of ${x^3} + 3{x^2} + 3x – 7$are?
- (x-1) $({x^2} + 4x + 7)$
- (x+1) $({x^2} + 4x + 7)$
- (x+1) $({x^2} + 4x – 7)$
- (x-1) $({x^2} + 4x – 7)$
Correct

Incorrect

Hint

Since f(1)=0
x-1 is a factor of f(x)
$\begin{array}{l}{x^3} + 3{x^2} + 3x - 7\\ = {x^2}(x - 1) + 4 \times (x - 1) + 7(x - 1)\\ = (x - 1)({x^2} + 4x + 7)\end{array}$
Question 20 of 40

20. Question
The value of k for which x+k is a factor of ${x^3} + k{x^2} – 2x + k + 4$is?
- -5
- 2
- $\dfrac{{ – 4}}{3}$
- $\dfrac{6}{7}$
Correct

Incorrect

Hint

$\begin{array}{l} - {K^3} + {K^3} + 2K + K + 4 = 0\\3K + 4 = 0\\K = \dfrac{{ - 4}}{3}\end{array}$
Question 21 of 40

21. Question
F(x), g(x) are two polynomials with integer coefficient such that their HCF is 1 and LCM is $({x^2} – 4)({x^4} – 1)$if $f(x) = {x^3} – 2{x^2} – x + 2$then g(x) is?
- ${x^3} – 2{x^2} + x + 2$
- ${x^3} – 2{x^2} + x – 2$
- ${x^3} + 2{x^2} + x + 2$
- ${x^3} – 2{x^2} – x + 2$
Correct

Incorrect

Hint

$\begin{array}{l}g(x) = \dfrac{{HCF \times LCM}}{{f(x)}}\\ = \dfrac{{1 \times ({x^2} - 4)({x^4} - 1)}}{{{x^3} - 2{x^2} - x + 2}}\\ = \dfrac{{(x + 2)(x - 2)(x + 1)(x - 1)({x^2} + 1)}}{{(x - 1)(x + 1)(x - 2)}}\\ = (x + 2)({x^2} + 1)\\ = {x^3} + 2{x^2} + x + 2\end{array}$
Question 22 of 40

22. Question
If ${x^2} + \dfrac{1}{{{x^2}}} = 62$then the value of ${x^4} + \dfrac{1}{{{x^4}}}$is?
- ${8^4} – {2^8} – 2$
- ${8^4} + 2$
- ${8^4} – {2^8} + 2$
- ${8^4} + {2^8} – 2$
Correct

Incorrect

Hint

$\begin{array}{l}{x^2} + \dfrac{1}{{{x^2}}} = 62\\ = {62^2} - 2\\ = {(64 - 2)^2} - 2\\ = {({8^2} - 2)^2} - 2\\ = {8^4} - 4 \times {8^2} + 4 - 2\\ = {8^4} - {2^8} + 2\end{array}$
Question 23 of 40

23. Question
If x=-2, y=1 then the value of $(4{y^2} – 9{x^2})$$(16{y^4} + 36{x^2}{y^2} + 81{x^4})$is?
- 46592
- -46592
- $ \pm $46592
- None of these
Correct

Incorrect

Hint

$\begin{array}{l}(4{y^2} - 9{x^2})(16{y^4} + 36{x^2}{y^2} + 81{x^4})\\ = {(4{y^2})^3} - {(9{x^2})^3}\\ = {(4 \times 1)^3} - {\left[ {9{{( - 2)}^2}} \right]^3}\\ = {4^3} - {(9 \times 4)^3}\\ = {4^3}({1^3} - {9^3})\\ = - 46592\end{array}$
Question 24 of 40

24. Question
If we divide $3{y^4} – {y^3} + 12{y^2} + 2$by $3{y^2} – 1$then remainder is?
- $\dfrac{y}{3} + \dfrac{{19}}{3}$
- $\dfrac{{ – 1}}{3}y + \dfrac{{19}}{3}$
- $\dfrac{{ – 1}}{3}y – \dfrac{{19}}{3}$
- $\dfrac{1}{3}y – \dfrac{{19}}{3}$
Correct

Incorrect

Hint

$\begin{array}{l}(3{y^2} - 1)3{y^4} - {y^3} + 12{y^2} + 2\left( {{y^3} - \dfrac{1}{3}y + \dfrac{{13}}{3}} \right)\\\dfrac{{ - 3{y^4}_ + ^ - {y^2}}}{{ - {y^3} + 13{y^2} + 2}}\\\dfrac{{_ + ^ - {y^3}_ - ^ + \dfrac{1}{3}y}}{{13{y^2} - \dfrac{1}{3}y + 2}}\\\dfrac{{13{y^2}_ + ^ - \dfrac{{13}}{3}}}{{\dfrac{{ - 1}}{3}y + \dfrac{{19}}{3}}}\end{array}$
Question 25 of 40

25. Question
The value of $\left( {{a^{\dfrac{1}{8}}} + {a^{\dfrac{{ – 1}}{8}}}} \right)\left( {{a^{\dfrac{1}{8}}} – {a^{\dfrac{{ – 1}}{8}}}} \right)\left( {{a^{\dfrac{1}{4}}} + {a^{\dfrac{{ – 1}}{4}}}} \right)\left( {{a^{\dfrac{1}{2}}} + {a^{\dfrac{{ – 1}}{2}}}} \right)$
- $(a + {a^{ – 1}})$
- $(a – {a^{ – 1}})$
- $({a^2} – a)$
- $\left( {{a^{\dfrac{1}{2}}} – a\dfrac{{ – 1}}{2}} \right)$
Correct

Incorrect

Hint

$\begin{array}{l}\left( {{a^{\dfrac{1}{8}}} + {a^{\dfrac{{ - 1}}{8}}}} \right)\left( {{a^{\dfrac{1}{8}}} - {a^{\dfrac{{ - 1}}{8}}}} \right)\left( {{a^{\dfrac{1}{4}}} + {a^{\dfrac{{ - 1}}{4}}}} \right)\left( {{a^{\dfrac{1}{2}}} + {a^{\dfrac{{ - 1}}{2}}}} \right)\\ = \left( {{a^{\dfrac{1}{4}}} - {a^{ - 1/4}}} \right)\left( {{a^{\dfrac{1}{4}}} + {a^{\dfrac{{ - 1}}{4}}}} \right)\left( {{a^{\dfrac{1}{2}}} + {a^{\dfrac{{ - 1}}{2}}}} \right)\\ = \left( {{a^{\dfrac{1}{2}}} - {a^{\dfrac{{ - 1}}{2}}}} \right)\left( {{a^{\dfrac{1}{2}}} + {a^{\dfrac{{ - 1}}{2}}}} \right)\\ = (a - {a^{ - 1}})\end{array}$
Question 26 of 40

26. Question
The sum of the series $5 + 13 + 21 + ….. + 181$is?
- 2139
- 6500
- 7500
- 3750
Correct

Incorrect

Hint

$\begin{array}{l}{T_n} = 181 = a + (n - 1)d\\181 = 5 + (n - 1) \times 8\\n = 23\\{S_n} = \dfrac{1}{2}\left[ {a + 1} \right] = \dfrac{{23}}{2}\left[ {5 + 181} \right]\\ = 23 \times 93\\ = 2139\end{array}$
Question 27 of 40

27. Question
If the product of the zeroes of the polynomial $a{c^3} – 2c – 24$is $\dfrac{{ – 8}}{5}$. What is the value of a?
- 17
- 18
- 19
- 15
Correct

Incorrect

Hint

$\begin{array}{l}\dfrac{{cons\tan tterm}}{{coefficientof{x^2}}} = \dfrac{{ - 8}}{5}\\\dfrac{{ - 24}}{a} = \dfrac{{ - 8}}{5}\\a = \dfrac{{24 \times 15}}{8} = 15\end{array}$
Question 28 of 40

28. Question
Find a cubic polynomial with the sum of its zeroes, sum of the product of its zeroes taken two at a time and the product of its zeroes are 13,52 and 60 respectively?
- ${x^3} – 52{x^2} – 13x – 60$
- ${x^3} – 13{x^2} + 52x$
- ${x^3} – {x^2} + 52x – 60$
Correct

Incorrect

Hint

Let $\alpha ,\beta ,\gamma$ be the zeroes of the polynomial
$\begin{array}{l}\alpha + \beta + \gamma = 13.......(1)\\\alpha \beta + \beta \gamma + \gamma \alpha = 52.....(2)\\\alpha \beta \gamma = 60.....(3)\end{array}$
Solving 3 equations we get
${x^3} - 13{x^2} + 52x - 60$
Question 29 of 40

29. Question
If $\alpha \& \beta $are the zeroes of the polynomials ${x^2} – 8x + k$such that ${\alpha ^2} + {\beta ^2} = 34$find the value of k.
- 18
- 12
- 15
- 20
Correct

Incorrect

Hint

$\begin{array}{l}\alpha + \beta = 8\\\alpha .\beta = k\\{(\alpha + \beta )^2} = {8^2}\\{\alpha ^2} + {\beta ^2} = 2\alpha \beta = 64\\34 + 2k = 64\\2k = 30\\k = 15\end{array}$
Question 30 of 40

30. Question
If a and b are the zeroes of the quadratic polynomial ${x^2} – 4px – 2q$, find the value of$\dfrac{1}{a} + \dfrac{1}{b}$
- $\dfrac{{4p}}{q}$
- $\dfrac{{ – 2p}}{q}$
- $\dfrac{{ – 4p}}{q}$
- $\dfrac{{2p}}{q}$
Correct

Incorrect

Hint

Sum of zeroes $= a + b = - \dfrac{{ - 4p}}{1} = 4p$
Product of zeroes $= a.b = - 2q$
$\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{{b + a}}{{ab}} = \dfrac{{4p}}{{ - 2q}} = \dfrac{{ - 2p}}{q}$
Question 31 of 40

31. Question
Find a cubic polynomial whose zeroes are $1, – 8,\dfrac{5}{6}$
- $6{x^3} + 37{x^2} – 83x + 40$
- ${x^3} + 37{x^2} – 83x + 40$
- $6{x^3} + 37{x^2} + 83x + 40$
- None of these
Correct

Incorrect

Hint

$\begin{array}{l}(x - 1)(x + 8)\left( {x - \dfrac{5}{6}} \right)\\ = ({x^2} + 8x - x - 8)\left( {x - \dfrac{5}{6}} \right)\\ = ({x^2} + 7x - 8)\left( {x - \dfrac{5}{6}} \right)\\ = {x^3} - \dfrac{5}{6}{x^2} + 7{x^2} - \dfrac{{ - 35x}}{6} - 8x + \dfrac{{40}}{6}\\ = \dfrac{{6{x^3} - 5{x^2} + 42{x^2} - 35x - 48x + 40}}{6}\\ = 6{x^3} + 37{x^2} - 83x + 40\end{array}$
Question 32 of 40

32. Question
Find zeroes of the polynomial $30{y^2} – 13y – 56$
- $y = \dfrac{{ – 6}}{7}$or $\dfrac{8}{5}$
- Y=$\dfrac{{ – 6}}{7}$or $\dfrac{8}{5}$
- $y = \dfrac{{ – 7}}{6}$or $\dfrac{5}{8}$
- $y = – \dfrac{{ – 7}}{2}$or $\dfrac{8}{5}$
Correct

Incorrect

Hint

$\begin{array}{l}f(x) = 30{y^2} - 13y - 56\\ = 30{y^2} - 48y + 35y - 56\\(6y + 7)(5y - 8)\\y = \dfrac{{ - 7}}{6}or\dfrac{8}{5}\end{array}$
Question 33 of 40

33. Question
Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial${x^2} – x – 6$?
- $K\left[ {{x^2} + \dfrac{1}{6}x + \dfrac{1}{6}} \right]$
- $K\left[ {{x^2} + \dfrac{1}{6}x – \dfrac{1}{6}} \right]$
- $K\left[ {{x^2} – \dfrac{1}{6}x + \dfrac{1}{6}} \right]$
- $K\left[ {{x^2} – \dfrac{1}{6}x – \dfrac{1}{6}} \right]$
Correct

Incorrect

Hint


$\begin{array}{l}\dfrac{{x = - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\\alpha = \dfrac{{ - b + \sqrt D }}{{2a}},\beta = \dfrac{{ - b - \sqrt D }}{{2a}}\\{\alpha ^1} = \dfrac{1}{\alpha } = \dfrac{{2a}}{{ - b + \sqrt D }},{\beta ^1} = \dfrac{1}{\beta } = \dfrac{{2a}}{{ - b - \sqrt D }}\\{x^2} - sx + p = 0\\{\alpha ^1} + {\beta ^1} = \dfrac{{2a( - b - \sqrt D ) + 2a( - b + \sqrt D )}}{{( - b + \sqrt D )(( - b - \sqrt D ))}}\\ = \dfrac{{ - 2ab - 2a\sqrt D - 2ab + 2a\sqrt D }}{{{{( - b)}^2} - {{(\sqrt D )}^2}}}\\ = \dfrac{{ - 4ab}}{{{b^2} - ({b^2} - 4ac)}}\\ = \dfrac{{ - 4ab}}{{4ac}} = \dfrac{{ - ab}}{{ac}} = \dfrac{{ - b}}{c}\\{\alpha ^1}{\beta ^1} = \left( {\dfrac{{2a}}{{ - b + \sqrt D }}} \right) \times \left( {\dfrac{{2a}}{{ - b - \sqrt D }}} \right)\\ = \dfrac{{4{a^2}}}{{{{( - b)}^2} - {{(\sqrt D )}^2}}} = \dfrac{{4{a^2}}}{{{b^2} - ({b^2} - 4ac)}}\\ = \dfrac{{4{a^2}}}{{4ac}}\\ = \dfrac{a}{c}\\{x^2} - sx + p = 0\\{x^2} - \left( {\dfrac{{ - b}}{c}} \right)x + \dfrac{a}{c} = 0\\\dfrac{{c{x^2} + bx + a}}{c} = 0\\c{x^2} + bx + a = 0\end{array}$
On solving the equation will be
$k\left[ {{x^2} + \dfrac{1}{6}x - \dfrac{1}{6}} \right]$
Question 34 of 40

34. Question
Find a quadratic polynomial whose zeroes are $\dfrac{3}{6}\& \dfrac{{ – 2}}{5}$?
- ${x^2} – 30x – 6$
- $3{x^2} – 30x – 1$
- $3{x^2} – 30x – 60$
- $30{x^2} – 15x – 12$
Correct

Incorrect

Hint

$f(x) = {x^2} -$ (sum of zeroes)x+product of the zero
$\begin{array}{l}{x^2} - \left( {\dfrac{3}{6}} \right)x + \left( {\dfrac{{ - 2}}{5}} \right)\\ = \dfrac{{30{x^2} - 15x + ( - 12)}}{{30}}\\ = 30{x^2} - 15x - 12\end{array}$
Question 35 of 40

35. Question
Find a quadratic polynomial such that sum of its zeroes is 6 and difference between zeroes is 4?
- $K\left[ {{x^2} + 6x + 5} \right]$
- $K\left[ {{x^2} – 10x + 5} \right]$
- $K\left[ {{x^2} – 6x – 5} \right]$
- $K\left[ {{x^2} – 6x + 5} \right]$
Correct

Incorrect

Hint

$\begin{array}{l}\alpha + \beta = 6\\\alpha - \beta = 4\\2\alpha = 10 \Rightarrow a = 5\\\beta = 1\\f(x) = k\left[ {{x^2} - 6x + 5} \right]\end{array}$
Question 36 of 40

36. Question
What are the zeroes of the polynomial$p(x) = 7{x^3} – 27{x^2} + 32x – 12$?
- $1,2,\dfrac{7}{6}$
- 1,2,6
- $1,2,\dfrac{6}{7}$
- 1,2,3
Correct

Incorrect

Hint

$\begin{array}{l}f(x) = 7{x^3} - 27{x^2} + 32x - 12\\f(1) = 7 \times {1^3} - 27 \times {(1)^2} + 32 \times 1 - 12\\ = 7 - 27 + 32 - 12\\ = 0\\f(2) = 7 \times {2^3} - 27 \times {2^2} + 32 \times 2 - 12\\ = 7 \times 8 - 27 \times 4 + 32 \times 2 - 12\\ = 0\\f\left( {\dfrac{6}{7}} \right) = 7 \times {\left( {\dfrac{6}{7}} \right)^3} - 27 \times {\left( {\dfrac{6}{7}} \right)^2} + 32 \times \dfrac{6}{7} - 12\\ = \dfrac{{216 - 972 + 1344 - 588}}{{49}} = \dfrac{0}{{49}} = 0\end{array}$
Question 37 of 40

37. Question
What is the value of $6{A^2} + {B^2} + 2AB$ where A and B are the digits of the number 39A80562B which is multiple of 72?
- 144
- 464
- 96
- 432
Correct

Incorrect

Hint

39A80562B will be divisible by 72,79 I.e. $9 \times 8$
$9.3 + 9 + A + 8 + 0 + 5 + 6 + 2 + B = 9K$ let
$= 33 + A + B = 9K......(1)$
For divisible by 8
Last digit should be 2,4,6,8 or 0
S0, 33+A+B=9K
$A = 9K - (33 + B)$
If b is 0 then A is 3
B is 2 then A is 1
B is 4, A is 8
$\begin{array}{l}6{A^2} + {B^2} + 2AB = 6 \times {8^2} + {4^2} + 2 \times 8 \times 4\\ = 464\end{array}$
Question 38 of 40

38. Question
Find the number of integral solution for x and y if 3x+4y=17 and x>0,y>0
- 1
- 2
- 3
- 4
Correct

Incorrect

Hint

$\begin{array}{l}3x + 4y = 17\\4y = 17 - 3X\\y = \dfrac{{17 - 3x}}{4}\end{array}$
Let x is 1 then y is $\dfrac{{14}}{4}$
X is 2 then y is $\dfrac{{14}}{4}$
X is 3 then y is $\dfrac{8}{4} = 2$
X is 4 then y is $\dfrac{2}{4}$
X is 5 then y is $\dfrac{{14}}{4}$
Only x is 3 and y is 2 can be taken so, number of integral solution is 1.
Question 39 of 40

39. Question
If $\dfrac{{ – 1}}{2} \le x \le \dfrac{1}{3}\& \dfrac{{ – 1}}{3} \le y \le \dfrac{1}{2}$then what will be the smallest value of ${x^2} – {y^3}$?
- $\dfrac{{ – 1}}{8}$
- $\dfrac{{ – 1}}{{27}}$
- $\dfrac{{ – 1}}{{51}}$
- $\dfrac{{ – 1}}{{72}}$
Correct

Incorrect

Hint

For minimum of ${x^2} - {y^3}$ , y should be maximum that is y is $\dfrac{1}{2}$
And x should be minimum i.e., x is 0
Then $\begin{array}{l}{x^2} - {y^3} = {0^2} - {\left( {\dfrac{1}{2}} \right)^3} = 0 - \dfrac{1}{8}\\ = \dfrac{{ - 1}}{8}\end{array}$
Question 40 of 40

40. Question
A number when divided by 1207 gives a remainder 88. What remainder should be obtained by dividing the same number by 17.
- 3
- 8
- 9
- 6
Correct

Incorrect

Hint

N = 1207K+88
$\dfrac{{1207K + 88}}{{17}} = \dfrac{{17 \times 71K + 17 \times 5 + 3}}{{17}}$