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Easy

Question

A cubic polynomial f(x) = ax³ + bx² + cx + d has a graph which touches the x-axis at 2, has another x-intercept at –1 and has y-intercept at –2 as shown. The value of, a + b + c + d equals

–2
–1
0
1

hint Hint:

$I n space t h i s space q u e s t i o n comma space w e space a r e space g i v e n space a space f u n c t i o n space w i t h space u n k n o w n space c o n s tan t s space a n d space t h e i r space g r a p h. space W e space h a v e space t o space f i n d space t h e space v a l u e s space o f space t h e s e space c o n s tan t s. space S o comma space w e space w i l l space f i n d space t h e space v a l u e s space o f space t h e space f u n c t i o n space a t space d i f f e r e n t space v a l u e s space o f space x space comma space t h e n space s o l v e space t h e m space t o space g e t space t h e space v a l u e s space o f space a comma b comma c comma d space. space T h e space p o i n t space t o space r e m e m b e r space h e r e space i s comma space sin c e comma space i t space h a s space a space tan g e n t space a t space t h e space x minus a x i s space a t space t h e space p o i n t space 2 space comma space s o comma space t h e space d e r i v a t i v e space o f space t h i s space f u n c t i o n space a t space t h e space p o i n t space 2 space w i l l space b e space z e r o.$

The correct answer is: –1

$W e space a r e space g i v e n space a space f u n c t i o n space w i t h space u n k n o w n space c o n s tan t s space f left parenthesis x right parenthesis equals a x cubed plus b x squared plus c x plus d space a n d space i t s space g r a p h.$

$T o space f i n d space t h e space v a l u e s space o f space a space comma space b space comma space c space a n d space d space. F i r s t comma space w e space w i l l space d i f f e r e n t i a t e space t h e space g i v e n space f u n c t i o n space a n d space f i n d space i t s space v a l u e space a t space x equals 2 space. O n space d i f f e r e n t i a t i n g comma space w e space g e t comma space f prime left parenthesis x right parenthesis equals 3 a x squared plus 2 b x plus c space comma space n o w comma space p u t t i n g space x equals 2 space i n space i t comma space w e space g e t comma space f prime left parenthesis x right parenthesis equals 3 a left parenthesis 2 right parenthesis squared plus 2 b left parenthesis 2 right parenthesis plus c space comma space i. e. comma space f prime left parenthesis 2 right parenthesis equals 12 a plus 4 b plus c space.$

$F r o m space t h e space g r a p h comma space i t space i s space c l e a r space t h a t comma space space f apostrophe left parenthesis 2 right parenthesis equals 0 space comma space i. e. comma space 8 a plus 4 b plus c equals 0 space horizontal ellipsis space left parenthesis 1 right parenthesis A l s o comma space f r o m space t h e space g r a p h comma space f left parenthesis 2 right parenthesis equals 0 space comma space f left parenthesis negative 1 right parenthesis equals 0 space a n d space f left parenthesis 0 right parenthesis equals negative 2 space comma space s o comma space f i r s t space p u t t i n g comma space x equals 2 space i n space t h e space g i v e n space f u n c t i o n space f left parenthesis 2 right parenthesis equals 8 a plus 4 b plus 2 c plus d equals 0 space comma space t h e n comma space p u t t i n g comma space x equals negative 1 space comma space w e space g e t comma space f left parenthesis negative 1 right parenthesis equals negative a plus b minus c plus d equals 0 space a n d space f i n a l l y comma space p u t t i n g space x equals 0 space comma space w e space g e t comma space f left parenthesis 0 right parenthesis equals d equals negative 2 space. N o w comma space sin c e space w e space h a v e space t h e space v a l u e space o f space d space comma space s o space w e space w i l l space p u t space i t space i n space a l l space e q u a t i o n s comma space t h e n space t h e space e q u a t i o n space b e c o m e s comma f left parenthesis negative 1 right parenthesis equals negative a plus b minus c equals 2 space f left parenthesis 2 right parenthesis equals 8 a plus 4 b plus 2 c equals 2 space f apostrophe left parenthesis 2 right parenthesis equals 8 a plus 4 b plus c equals 0 O n space s o l v i n g space t h e s e space e q u a t i o n s space s i m u l tan e o u s l y comma space w e space g e t comma space a equals fraction numerator negative 1 over denominator 2 end fraction space comma space b equals 3 over 2 comma space c equals 0 space a n d space d equals negative 2 space.$
$H e n c e comma space t h e space v a l u e space o f space a plus b plus c plus d equals fraction numerator negative 1 over denominator 2 end fraction plus 3 over 2 plus 0 minus 2 space equals space minus 1$

$I f space a space f u n c t i o n space h a s space a space tan g e n t space a t space a space p o i n t space x equals a space comma space t h e n comma space i t space m e a n s comma space i t space i s space t h e space c r i t i c a l space p o i n t space o f space t h a t space f u n c t i o n space a n d space t h e space d e r i v a t i v e space o f space t h e space f u n c t i o n space i s space z e r o space a t space t h a t space p o i n t. space P e r f o r m space t h e space c a l c u l a t i o n s space c a r e f u l l y comma space t o space a v o i d space s i l l y space m i s t a k e s. space O n e space s h o u l d space k n o w space h o w space t o space r e a d space a space g r a p h space t o space s o l v e space s u c h space q u e s t i o n s.$

Consider f(x) = |1–x| 1 £ x £ 2 and g(x )= f(x) + b $sin invisible function application fraction numerator pi over denominator 2 end fraction x$ , 1 £ x £ 2 then which of the following is correct?

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If CP and CD are semi – conjugate diameters of an ellipse $fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 comma$ then CP² + CD² =

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The normals to the curve x = a ( $theta$ + sin $theta$ ), y = a (1 – cos $theta$ ) at the points $theta$ = (2n + 1) $pi$ , n $element of$ I are all -

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If the tangent at ‘t’ on the curve y = 8t3 –1, x = 4t2 + 3 meets the curve again at t¢ and is normal to the curve at that point, then value of t must be -

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The tangent at (t, t2 – t3) on the curve y = x2 – x3 meets the curve again at Q, then abscissa of Q must be -

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If $fraction numerator x over denominator a end fraction plus fraction numerator y over denominator b end fraction$ = 1 is a tangent to the curve x = 4t,y = $fraction numerator 4 over denominator t end fraction$ , t $element of$ R then -

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The line $fraction numerator x over denominator a end fraction$ + $fraction numerator y over denominator b end fraction$ = 1 touches the curve y = be-x/a at the point :

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An object moves in a straight line. It starts from the rest and its acceleration is . 2 ms².After reaching a certain point it comes back to the original point. In this movement its acceleration is -3 ms², till it comes to rest. The total time taken for the movement is 5 second. Calculate the maximum velocity.

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For the ellipse $fraction numerator x to the power of 2 end exponent over denominator 2 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 1 end fraction equals 1 comma$ the foci are

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For the ellipse $fraction numerator x to the power of 2 end exponent over denominator 1 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 4 end fraction equals 1$ , the latus rectum is

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A cubic polynomial f(x) = ax3 + bx2 + cx + d has a graph which touches the x-axis at 2, has another x-intercept at –1 and has y-intercept at –2 as shown. The value of, a + b + c + d equals

–2 –1 0 1

The correct answer is: –1

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Consider f(x) = |1–x| 1 £ x £ 2 and g(x )= f(x) + b , 1 £ x £ 2 then which of the following is correct?

Consider f(x) = |1–x| 1 £ x £ 2 and g(x )= f(x) + b , 1 £ x £ 2 then which of the following is correct?

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A cubic polynomial f(x) = ax³ + bx² + cx + d has a graph which touches the x-axis at 2, has another x-intercept at –1 and has y-intercept at –2 as shown. The value of, a + b + c + d equals

–2
–1
0
1

Consider f(x) = |1–x| 1 £ x £ 2 and g(x )= f(x) + b $sin invisible function application fraction numerator pi over denominator 2 end fraction x$ , 1 £ x £ 2 then which of the following is correct?

Consider f(x) = |1–x| 1 £ x £ 2 and g(x )= f(x) + b $sin invisible function application fraction numerator pi over denominator 2 end fraction x$ , 1 £ x £ 2 then which of the following is correct?

The value of c in Lagrange’s theorem for the function f(x) = log sin x in the interval $open square brackets fraction numerator pi over denominator 6 end fraction comma fraction numerator 5 pi over denominator 6 end fraction close square brackets$ is -

The value of c in Lagrange’s theorem for the function f(x) = log sin x in the interval $open square brackets fraction numerator pi over denominator 6 end fraction comma fraction numerator 5 pi over denominator 6 end fraction close square brackets$ is -

The relation between time and displacement of a moving particle is given by $t equals 2 a x squared$ where $alpha$ is a constant. The shape of the graph is $x not stretchy rightwards arrow y$ is....

The relation between time and displacement of a moving particle is given by $t equals 2 a x squared$ where $alpha$ is a constant. The shape of the graph is $x not stretchy rightwards arrow y$ is....

The normals to the curve x = a ( $theta$ + sin $theta$ ), y = a (1 – cos $theta$ ) at the points $theta$ = (2n + 1) $pi$ , n $element of$ I are all -

The normals to the curve x = a ( $theta$ + sin $theta$ ), y = a (1 – cos $theta$ ) at the points $theta$ = (2n + 1) $pi$ , n $element of$ I are all -

The normal to the curve x = 3 cos $theta$ – cos3 $theta$ , y = 3 sin $theta$ – sin3 $theta$ at the point $theta$ = $pi$ /4 passes through the point -

The normal to the curve x = 3 cos $theta$ – cos3 $theta$ , y = 3 sin $theta$ – sin3 $theta$ at the point $theta$ = $pi$ /4 passes through the point -

The normal of the curve given by the equation x = a (sin $theta$ + cos $theta$ ), y = a (sin $theta$ – cos $theta$ ) at the point Q is -

The normal of the curve given by the equation x = a (sin $theta$ + cos $theta$ ), y = a (sin $theta$ – cos $theta$ ) at the point Q is -

If $fraction numerator x over denominator a end fraction plus fraction numerator y over denominator b end fraction$ = 1 is a tangent to the curve x = 4t,y = $fraction numerator 4 over denominator t end fraction$ , t $element of$ R then -

If $fraction numerator x over denominator a end fraction plus fraction numerator y over denominator b end fraction$ = 1 is a tangent to the curve x = 4t,y = $fraction numerator 4 over denominator t end fraction$ , t $element of$ R then -

The line $fraction numerator x over denominator a end fraction$ + $fraction numerator y over denominator b end fraction$ = 1 touches the curve y = be-x/a at the point :

The line $fraction numerator x over denominator a end fraction$ + $fraction numerator y over denominator b end fraction$ = 1 touches the curve y = be-x/a at the point :

For the ellipse $fraction numerator x to the power of 2 end exponent over denominator 2 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 1 end fraction equals 1 comma$ the foci are

For the ellipse $fraction numerator x to the power of 2 end exponent over denominator 2 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 1 end fraction equals 1 comma$ the foci are

For the ellipse $fraction numerator x to the power of 2 end exponent over denominator 1 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 4 end fraction equals 1$ , the latus rectum is

For the ellipse $fraction numerator x to the power of 2 end exponent over denominator 1 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 4 end fraction equals 1$ , the latus rectum is